# Asymptotic Theory for Estimation of the Hüsler-Reiss Distribution via Block
Maxima Method

Hank Flury  University of North Carolina at Chapel Hill, Department of
Statistics and Operations Research, Chapel Hill, North Carolina, United States
Jan Hannig  University of North Carolina at Chapel Hill, Department of
Statistics and Operations Research, Chapel Hill, North Carolina, United States
Richard Smith  University of North Carolina at Chapel Hill, Department of
Statistics and Operations Research, Chapel Hill, North Carolina, United States

###### Abstract

The Hüsler-Reiss distribution describes the limit of the pointwise maxima of a
bivariate normal distribution. This distribution is defined by a single
parameter, $\lambda$. We provide asymptotic theory for maximum likelihood
estimation of $\lambda$ under a block maxima approach. Our work assumes
independent and identically distributed bivariate normal random variables,
grouped into blocks where the block size and number of blocks increase
simultaneously. With these assumptions our results provide conditions for the
asymptotic normality of the Maximum Likelihood Estimator (MLE). We
characterize the bias of the MLE, provide conditions under which this bias is
asymptotically negligible, and discuss how to choose the block size to
minimize a bias-variance trade-off. The proofs are an extension of previous
results for choosing the block size in the estimation of univariate extreme
value distributions (Dombry and Ferreria 2019), providing a potential basis
for extensions to multivariate cases where both the marginal and dependence
parameters are unknown. The proofs rely on the Argmax Theorem applied to a
localized loglikelihood function, combined with a Lindeberg-Feller Central
Limit Theorem argument to establish asymptotic normality. Possible
applications of the method include composite likelihood estimation in Brown-
Resnick processes, where it is known that the bivariate distributions are of
Hüsler-Reiss form.

Hank Flury is the corresponding author. His email is<EMAIL_ADDRESS>Hank
Flury’s ORCiD is 0009-0004-1722-8607. Richard Smith’s ORCiD is
0000-0002-3454-3199. Jan Hannig’s ORCiD is 0000-0002-4164-0173.

Jan Hannig’s research was supported in part by the National Science Foundation
under Grant No. DMS-1916115, 2113404, and 2210337. Hank Flury was also support
in part by the National Science Foundation under Grant No. DMS-2210337.

##  1 Introduction

The Block Maxima (BM) method is one of the main techniques in Extreme Value
Theory (EVT), with the goal of estimation of the limiting distribution of the
maxima of a random variable. In this context, the distribution is limiting in
the sense that we take the maxima over an increasingly large number of
samples. In this method, we take $n$ independent and identically distributed
(i.i.d.) observations and split them into $k_{n}$ equally sized groups of
$m_{n}$ samples. Going forward, we drop the subscript $n$ for the sake of
readability. Then, we can take the maximum over our $m$ data points for each
block, now having $k$ values with which to do inference. Two questions
naturally arise from this practice: If we are trying to estimate a limiting
distribution, and our sample comes from a distribution taken under a finite
sample size, is it reasonable to estimate the parameters of the limiting
distribution with our sample? Assuming that it is reasonable, how should a
practitioner of the BM method split their $n$ samples into $k$ and $m$?

dombry19 answers these questions for the univariate case. They showed that,
under mild assumptions on the distribution of the sample and the relationship
between $k$ and $m$, with probability tending toward 1, there exists a maximum
likelihood estimator, $\hat{\theta}$ such that we have a Central Limit
Theorem-like result:

$\sqrt{k}\left(\hat{\theta}-\theta\right)\xrightarrow{d}N\left(I_{\theta}^{-1}\lambda,I_{\theta}^{-1}\right)$
(1)

where $I_{\theta}$ is the Fisher Information of the limiting distribution with
true parameters $\theta$, and $\lambda$ is a parameter that arises as the
limit of the relationship between $k$ and $m$. This work will develop similar
results for a specific bivariate case. One of the complications of the
bivariate case is that we do not have a general form for the limiting
distribution, while in the univariate case we know that the limiting
distribution is a generalized extreme value (GEV) distribution. Thus, for our
work, we need to assume a limiting distribution.

The Hüsler-Reiss distribution arises as the limit of properly scaled maxima of
bivariate normal data ([5]). The distribution is given by

$H_{\lambda}\left(x,y\right)=\exp\left\\{-e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)-e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\right\\}.$
(2)

where $\Phi\left(t\right)$ is the distribution function of a standard normal
random variable. Namely, if $\left(X_{i},Y_{i}\right)$ are i.i.d. bivariate
standard normal random variables with correlation $\rho_{m}$, and we define
$b_{m}$ to satisfy the equation $b_{m}=m\phi\left(b_{m}\right)$, where
$\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^{2}}$ is the standard normal
density, then the Hüsler-Reiss distribution will arise as the distribution of
$\left\\{b_{m}\max\left(X_{1},...,X_{m}\right)-b_{m}^{2},b_{m}\max\left(Y_{1},...Y_{m}\right)-b_{m}^{2}\right\\}$
as $m\rightarrow\infty$. The parameter $\lambda\in\left[0,\infty\right]$
arises as the limit of
$\left(1-\rho_{m}\right)\ln\left(m\right)\rightarrow\lambda^{2}$, with the
convention that $\lambda=0$ corresponds to complete dependence (making the
Hüsler-Reiss distribution equivalent to a univariate Gumbel distribution with
$X_{i}=Y_{i}$) and $\lambda=\infty$ corresponds to complete independence
(making the Hüsler-Reiss distribution equivalent to the product of two
independent Gumbel distributions). While the Hüsler-Reiss distribution is
useful in context other than bivariate maxima, this is the only case that is
studied in the literature. Our work continues to focus on this framework for
possible extension to other cases in the future. Maximum likelihood estimation
of the Hüsler-Reiss distribution was discussed by engelke2015. Their work,
however, conditions their estimator on the observation of an extreme event,
leading to entirely different results from those given here. In contrast,
tawn88 and tawn90 derive results for the MLE of dependence parameters of
bivariate extreme values (not including the Hüsler-Reiss distribution),
focusing on properties of the MLE when the true value is on the boundaries of
the parameter space, corresponding to independence or exact dependence. Our
work focuses on the properties of the MLE when $\lambda\in(0,\infty)$, leading
to new results. The uniform convergence rate of the maxima of scaled bivariate
normal random variables was derived by liao. Our work extends the
probabilistic results of [6] to the statistical estimation problem.

Our work is broken into the following sections. Section 2 discusses the
relevant details of maximum likelihood estimation in the case that our data
are distributed exactly according to a Hüsler-Reiss distribution. Section 3
handles the case of the misspecified model, where our data are the scaled
maxima of bivariate normal random variables but we approximate this joint
distribution by the Hüsler-Reiss distribution and try to estimate $\lambda$.
Section 4 provides a simulations study to support our theoretical findings.
Finally, the appendix contains details of the proofs.

##  2 Estimation Under the Hüsler-Reiss Model

We begin with some properties of the Hüsler-Reiss model, the main result being
a rigorous proof that the Fisher Information is finite. Although we do not
prove the existence of asymptotic normality of the MLE for $\lambda$ when our
data are distributed according to a Hüsler-Reiss distribution as a separate
case, we note that this arises as a consequence of Theorem 5. It will be
useful later to have analyzed the log-likelihood and know its form. We will
denote
$G\left(x,y,\lambda\right):=e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)+e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)$
so that $H_{\lambda}\left(x,y\right)=e^{-G\left(x,y,\lambda\right)}$. Where
necessary for space, we omit the arguments of $G(x,y,\lambda)$ when there is
no risk of ambiguity. We begin by finding the form of the derivative of the
log-likelihood,

$\displaystyle
l^{\prime}\left(x,y,\lambda\right):=\sum_{i=1}^{k}\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial y}H_{\lambda}\left(x_{i},y_{i}\right)\right)$
$\displaystyle=\sum_{i=1}^{k}\frac{\partial}{\partial\lambda}\ln\left(e^{-G\left(x_{i},y_{i},\lambda\right)}\left(\frac{\partial}{\partial
y}G\left(x_{i},y_{i},\lambda\right)\frac{\partial}{\partial
x}G\left(x_{i},y_{i},\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x_{i},y_{i},\lambda\right)\right)\right)$
$\displaystyle=\sum_{i=1}^{k}\frac{\frac{\partial}{\partial\lambda}\left[\frac{\partial}{\partial
y}G\frac{\partial}{\partial x}G\right]-\frac{\partial^{3}}{\partial y\partial
x\partial\lambda}G-\left(\frac{\partial}{\partial\lambda}G\right)\left(\frac{\partial}{\partial
y}G\frac{\partial}{\partial x}G-\frac{\partial^{2}}{\partial y\partial
x}G\right)}{\frac{\partial}{\partial y}G\frac{\partial}{\partial
x}G-\frac{\partial^{2}}{\partial y\partial x}G}.$

We calculate the forms of each term to be:

$\displaystyle\frac{\partial}{\partial\lambda}G\left(x,y,\lambda\right)$
$\displaystyle=2e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right),$
$\displaystyle\frac{\partial}{\partial x}G\left(x,y,\lambda\right)$
$\displaystyle=-e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right),$
$\displaystyle\frac{\partial}{\partial y}G\left(x,y,\lambda\right)$
$\displaystyle=-e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right),$
$\displaystyle\frac{\partial^{2}}{\partial x\partial
y}G\left(x,y,\lambda\right)$
$\displaystyle=-e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda},$
$\displaystyle\frac{\partial^{3}}{\partial x\partial
y\partial\lambda}G\left(x,y,\lambda\right)$
$\displaystyle=e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{1}{2\lambda^{2}}+\frac{1}{2}-\frac{\left(y-x\right)^{2}}{8\lambda^{4}}\right),$
$\displaystyle\frac{\partial}{\partial\lambda}\left[\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\right]$
$\displaystyle=e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left[\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)e^{-y}\left(1-\frac{y-x}{2\lambda^{2}}\right)\right.$
$\displaystyle\hskip
28.45274pt+\left.\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)e^{-x}\left(1-\frac{x-y}{2\lambda^{2}}\right)\right].$

Our first goal is to show that the Fisher Information for this distribution
exists. To do this we begin with a lemma:

######  Lemma 1.

$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-ax}e^{-by}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)x^{k_{1}}y^{k_{2}}\left(x-y\right)^{k_{3}}e^{-e^{-x}f\left(x-y\right)}dxdy$
(3)

is finite for $a,b\in\mathbb{Z}$, such that $a+b\geq
1,\lambda>0,k_{i}\in\mathbb{N}\cup\left\\{0\right\\},i=1,2,3$ and $f$ such
that there exists $c>0$ such that $c<f\left(t\right)$ for all real $t$.

To show that this integral is finite, we show that the integral of the
absolute value of the integrand is finite. It can be shown through integration
by parts and induction that
$\int_{-\infty}^{\infty}e^{-kx}e^{-ce^{-x}}dx=\frac{\left(k-1\right)!}{c^{k}}$.
Using the change of variable $\delta=x-y$,

$\displaystyle 0$
$\displaystyle\leq\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left|e^{-ax}e^{-by}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)x^{k_{1}}y^{k_{2}}\left(x-y\right)^{k_{3}}e^{-e^{-x}f\left(x-y\right)}\right|dxdy$
$\displaystyle\leq\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int_{-\infty}^{\infty}\left|\delta^{k_{3}+k_{2}-i}\right|e^{b\delta}\phi\left(\lambda-\frac{\delta}{2\lambda}\right)d\delta\int_{-\infty}^{\infty}\left|x^{k_{1}+i}\right|e^{-\left(a+b\right)x}e^{-ce^{-x}}dx.$

Note that the first integral depends only on the integer moments of a normal
distribution, which must be finite. To see that the second integral is finite,
we use the bounds $|x^{t}|<e^{-2tx}$ when $x\leq 0$. This gives us

$\displaystyle\int_{-\infty}^{\infty}|x^{k_{1}+i}|e^{-\left(a+b\right)x}e^{-ce^{-x}}dx$
$\displaystyle\leq\int_{-\infty}^{0}e^{-\left(2k_{1}+2i+a+b\right)x}e^{-ce^{-x}}dx+\int_{0}^{\infty}x^{k_{1}+i}e^{-\left(a+b\right)x}dx$
$\displaystyle\leq\frac{\left(2k_{1}+2i+a+b\right)!}{c^{2k_{1}+2i+a+b}}+\left(a+b\right)^{-1-k_{1}-i}\Gamma\left(1+k_{1}+i\right),$

which is surely finite. ∎

With this lemma in hand, we are ready to prove that the Fisher Information
must exist.

######  Theorem 1.

The Fisher Information for a Hüsler-Reiss distribution with parameter
$\lambda\in\left(0,\infty\right)$ is finite.

Proof.

The proof of Theorem 1 involves showing that each term of the integrand
corresponding to the Fisher Information is bounded by a form of the integrand
above, up to a constant. We take a derivative of
$l^{\prime}\left(x,y,\lambda\right)$ with respect to $\lambda$ to find:

$\displaystyle\frac{\partial}{\partial\lambda}l^{\prime}\left(x,y,\lambda\right)=\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial y}H_{\lambda}\left(x,y\right)\right)$ $\displaystyle=$
$\displaystyle-\frac{\partial^{2}}{\partial^{2}\lambda}G\left(x,y,\lambda\right)+\frac{\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]-\frac{\partial^{4}}{\partial y\partial
x\partial^{2}\lambda}G\left(x,y,\lambda\right)}{\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)}$
$\displaystyle-\left(\frac{\frac{\partial}{\partial\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]-\frac{\partial^{3}}{\partial y\partial
x\partial\lambda}G\left(x,y,\lambda\right)}{\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)}\right)^{2}.$ (4)

We begin with
$-\frac{\partial^{2}}{\partial^{2}\lambda}G\left(x,y,\lambda\right)=2e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\lambda-\frac{\left(y-x\right)^{2}}{4\lambda^{3}}\right)$.
We see that our integrand is:

$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}2e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\lambda-\frac{\left(y-x\right)^{2}}{4\lambda^{3}}\right)e^{{-e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)-e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)}}$
$\displaystyle\hskip
28.45274pt\left[e^{-x-y}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)+e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}\right]dxdy$
$\displaystyle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}2e^{-2x}\phi\left(\lambda-\frac{\delta}{2\lambda}\right)\left(\lambda-\frac{\delta^{2}}{4\lambda^{3}}\right)e^{{-e^{-x}\left(\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)+e^{\delta}\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)\right)}}$
$\displaystyle\hskip
28.45274pt\left[e^{\delta-x}\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)+\phi\left(\lambda-\frac{\delta}{2\lambda}\right)\frac{1}{2\lambda}\right]dxd\delta.$

When factored, both terms of this integral are bounded by a form of the
integral in Lemma 1 with appropriate $a$, $b$, and $k_{i},i=1,2,3$, since
$0<\Phi\left(t\right),\phi\left(t\right)<1$ for all real $t$ and
$\frac{1}{2}<\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)+e^{\delta}\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)$
for all values of $\delta$ since $\lambda>0$. Thus we are assured that the
expectation of this term is finite. We now move to the term
$\frac{\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]}{\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)}$. We begin by finding the form of
$\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]$:

$\displaystyle\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]$
$\displaystyle=e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left[2e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(1-\frac{\left(x-y\right)^{2}}{2\lambda^{2}}\right)\right.$
$\displaystyle\hskip
28.45274pt+e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left\\{-\lambda+\frac{x-y}{2\lambda}+\frac{\left(y-x\right)^{2}}{4\lambda^{3}}-\frac{\left(x-y\right)^{3}}{8\lambda^{5}}+\frac{x-y}{\lambda^{3}}\right\\}$
$\displaystyle\hskip
28.45274pt\left.+e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left\\{-\lambda+\frac{y-x}{2\lambda}+\frac{\left(x-y\right)^{2}}{4\lambda^{3}}-\frac{\left(y-x\right)^{3}}{8\lambda^{5}}+\frac{y-x}{\lambda^{3}}\right\\}\right].$

We note that our density takes the form
$e^{-G\left(x,y\right)}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)\right]$, and thus our integrand will reduce to
$\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]e^{-G\left(x,y,\lambda\right)}$. Thus our
integral is:

$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\partial^{2}}{\partial^{2}\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]e^{-G\left(x,y,\lambda\right)}dxdy$
$\displaystyle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x}\phi\left(\lambda-\frac{\delta}{2\lambda}\right)\left[2e^{-x}\phi\left(\lambda-\frac{\delta}{2\lambda}\right)\left(1-\frac{\delta^{2}}{2\lambda^{2}}\right)\right.$
$\displaystyle\hskip
28.45274pt+e^{-x}\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)\left\\{-\lambda+\frac{\delta}{2\lambda}+\frac{\delta^{2}}{4\lambda^{3}}-\frac{\delta^{3}}{8\lambda^{5}}+\frac{\delta}{\lambda^{3}}\right\\}$
$\displaystyle\hskip
28.45274pt+\left.e^{\delta-x}\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)\left\\{-\lambda-\frac{\delta}{2\lambda}+\frac{\delta^{2}}{4\lambda^{3}}+\frac{\delta^{3}}{8\lambda^{5}}-\frac{\delta}{\lambda^{3}}\right\\}\right]e^{-e^{-x}\left(\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)+e^{\delta}\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)\right)}dxd\delta.$

Again, by factoring we see each term is bounded by Lemma 1. We move to the
next term, $\frac{-\frac{\partial^{4}}{\partial y\partial
x\partial^{2}\lambda}G\left(x,y,\lambda\right)}{\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)}$. The form of the numerator is

$\displaystyle\frac{\partial^{4}}{\partial y\partial
x\partial^{2}\lambda}G\left(x,y,\lambda\right)=\frac{\partial}{\partial\lambda}\left[e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{1}{2\lambda^{2}}+\frac{1}{2}-\frac{\left(y-x\right)^{2}}{8\lambda^{4}}\right)\right]$
$\displaystyle=e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left[\frac{-\lambda}{2}-\frac{1}{2\lambda}-\frac{1}{\lambda^{3}}+\frac{\left(y-x\right)^{2}}{4\lambda^{3}}+\frac{5\left(y-x\right)^{2}}{8\lambda^{5}}-\frac{\left(y-x\right)^{4}}{32\lambda^{7}}\right].$

Once again, the denominator of the term cancels with the same term in our
density, and we see that each term is bounded by an integral of the form in
Lemma 1. Finally, we move to the squared term:
$\frac{\frac{\partial}{\partial\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]-\frac{\partial^{3}}{\partial y\partial
x\partial\lambda}G\left(x,y,\lambda\right)^{2}}{\left(\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)\right)^{2}}$. Again, one of the terms in the
denominator cancels with that in the density, and we see:

$\displaystyle
0\leq\frac{\left(\frac{\partial}{\partial\lambda}\left[\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)\right]-\frac{\partial^{3}}{\partial y\partial
x\partial\lambda}G\left(x,y,\lambda\right)\right)^{2}}{\frac{\partial}{\partial
y}G\left(x,y,\lambda\right)\frac{\partial}{\partial
x}G\left(x,y,\lambda\right)-\frac{\partial^{2}}{\partial y\partial
x}G\left(x,y,\lambda\right)}$
$\displaystyle=\frac{\left(e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left[e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(1-\frac{x-y}{2\lambda^{2}}\right)+e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(1-\frac{y-x}{2\lambda^{2}}\right)-\frac{1}{2\lambda^{2}}-\frac{1}{2}+\frac{\left(x-y\right)^{2}}{8\lambda^{4}}\right]\right)^{2}}{e^{-x}e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)+e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}}$
$\displaystyle\leq\frac{e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}\left(e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(1-\frac{x-y}{2\lambda^{2}}\right)+e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(1-\frac{y-x}{2\lambda^{2}}\right)-\frac{1}{2\lambda^{2}}-\frac{1}{2}+\frac{\left(x-y\right)^{2}}{8\lambda^{4}}\right)^{2}}{\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}}$
$\displaystyle\leq 2\lambda
e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(e^{-x}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(1-\frac{x-y}{2\lambda^{2}}\right)+e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(1-\frac{y-x}{2\lambda^{2}}\right)\right.$
$\displaystyle\hskip
28.45274pt-\left.\frac{1}{2\lambda^{2}}-\frac{1}{2}+\frac{\left(x-y\right)^{2}}{8\lambda^{4}}\right)^{2}.$

We see each of these terms is bounded by a form of the integrand in Lemma 1.
This was the final term of the integrand, and thus the Fisher information is
finite.

∎

We note that the integrand in our definition of the Fisher Information is
continuous in $\lambda$ on the interior of the parameter space. Thus by the
Leibniz Integral Rule, we may pass a derivative under the integral and, by the
usual argument, we may represent the Fisher Information as the negation of the
second moment of the first derivative of the log-likelihood function with
respect to $\lambda$. This is to say that the value of the Fisher Information
is indeed positive.

##  3 Estimation Under the Misspecified Model

######  Theorem 2.

Let ($X_{i}$, $Y_{i}$) for $i\in\left\\{1,..,m\right\\}$ be independent and
identically distributed according to a standard bivariate normal distribution
with correlation parameter $\rho_{m}$ such that
$\ln\left(m\right)\left(1-\rho_{m}\right)\rightarrow\lambda^{2}\in\left(0,\infty\right)$.
Define $l^{\prime\prime}\left(x,y,\lambda\right)$ to be the second derivative
with respect to $\lambda$ of the log-likelihood function of a Hüsler-Reiss
distribution. Namely,
$l^{\prime\prime}\left(x,y,\lambda\right):=\frac{\partial^{2}}{\partial^{2}\lambda}\left[\ln\left(h\left(x,y,\lambda\right)\right)\right]$,
where $h\left(x,y,\lambda\right)$ is the density function of the Hüsler-Reiss
distribution with parameter $\lambda$. Then the expectation of
$l^{\prime\prime}\left(x,y,\lambda\right)$ with respect to the density of the
maxima of the properly scaled $\left(X_{i},Y_{i}\right)$ converges to
$I_{\lambda}$, the finite Fisher Information of the Hüsler-Reiss distribution,
which depends only on the value of $\lambda$.

Proof. Following huslerreiss89 in their choice of scaling constants, we choose
$b_{m}$, defined by $b_{m}=m\phi(b_{m})$ for both the location and scale
constants. The first step of the proof is to show that the density of
$\left(b_{m}\max\left(X_{i}\right)-b_{m}^{2},b_{m}\max\left(Y_{i}\right)-b_{m}^{2}\right)$
converges to $h\left(x,y,\lambda\right)$. Define
$u_{m}\left(x\right)=\frac{x}{b_{m}}+b_{m}$. Recall that
$\lambda\in\left(0,\infty\right)$ implies that $\rho_{m}\rightarrow 1$. Recall
from Equations 2.11 and 2.12 of huslerreiss89 that, without loss of
generality,
$\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\rightarrow\lambda+\frac{x-y}{2\lambda}$.
Define $\Phi_{\rho_{m}}\left(x,y\right)$ and $\phi_{\rho_{m}}\left(x,y\right)$
as the CDF (cumulative distribution function) and PDF (probability density
function) of a standard normal bivariate distribution with correlation
parameter $\rho_{m}$. Throughout the proof we will utilize the following
useful equality:
$\phi_{\rho_{m}}\left(x,y\right)=\phi\left(\frac{x-\rho_{m}y}{\sqrt{1-\rho_{m}^{2}}}\right)\phi\left(y\right)\left(1-\rho_{m}^{2}\right)^{-1/2}$.
We begin by differentiating the CDF:

$\displaystyle\frac{\partial^{2}}{\partial x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}$
$\displaystyle=\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}\left[\frac{m-1}{m}e^{-y}e^{-\frac{y^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right.$
$\displaystyle\left.\hskip
28.45274pt+\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right].$

We note that
$\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right),e^{-\frac{x^{2}}{2b_{m}^{2}}},e^{-\frac{y^{2}}{2b_{m}^{2}}},\frac{m-1}{m}\rightarrow
1$,
$\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}\rightarrow
H_{\lambda}\left(x,y\right)$ and $b_{m}\sqrt{1-\rho_{m}^{2}}\rightarrow
2\lambda$ as $m\rightarrow\infty$. This gives us:

$\displaystyle\frac{\partial^{2}}{\partial x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}$
$\displaystyle=\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}\left[\frac{m-1}{m}e^{-y}e^{-\frac{y^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right.$
$\displaystyle\hskip
28.45274pt\left.+\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right]$
$\displaystyle\rightarrow
H_{\lambda}\left(x,y\right)\left[e^{-x}e^{-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)+e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}\right]=:h_{\lambda}(x,y)$

This limiting density is exactly the density of a Hüsler-Reiss distribution
with parameter $\lambda$ derived in Theorem 1, therefore the densities
converge pointwise. Thus, the limiting expectation is exactly the Fisher
Information which was proven to be finite in Theorem 1. To show that these
expectations converge we will use the Dominated Convergence Theorem (DCT). We
do this in two steps: first, we show that

$\iint_{\mathbb{R}^{2}}l^{\prime\prime}\left(x,y,\lambda\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}1_{x\geq-
b_{m}^{2}}dxdy\rightarrow I_{\lambda},$ (5)

then, we show that

$\iint_{\mathbb{R}^{2}}l^{\prime\prime}\left(x,y,\lambda\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}1_{x\leq-
b_{m}^{2}}\rightarrow 0,$ (6)

which in turn proves the theorem. To show (5), we want to show there exists a
function $f\left(x,y\right)$ that is independent of $m$ such that
$|l^{\prime\prime}\left(x,y,\lambda\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}1_{x\geq-
b_{m}^{2}}|\leq f\left(x,y\right)$ and
$\iint_{\mathbb{R}^{2}}f\left(x,y\right)1_{x\geq-b_{m}^{2}}dxdy<\infty$. We
assume

$\int_{\mathbb{R}}|l^{\prime\prime}\left(x,y,\lambda\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}|dy\leq
e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}$
(7)

for some $k>0$. With this assumption, we show (5) and (6), and then go back to
prove the assumption. Assuming the form
$e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}$,
we see

$\displaystyle\int_{0}^{\infty}e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}dx\leq\int_{0}^{\infty}e^{-kx}dx=\frac{1}{k}.$

Consider the integral over $\left(-kb_{m},0\right)$ for some fixed $k>1$. We
see

$\displaystyle\int_{-kb_{m}}^{0}e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}dx\leq$
$\displaystyle\int_{-kb_{m}}^{0}e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}dx$
$\displaystyle\leq$
$\displaystyle\int_{-kb_{m}}^{0}e^{-kx}\exp\left[-\frac{m-2}{m}\int_{x}^{\infty}e^{-t}e^{-\frac{1}{2}\left(\frac{t}{b_{m}}\right)^{2}}dt\right]dx$
$\displaystyle\leq$
$\displaystyle\int_{-kb_{m}}^{0}e^{-kx}\exp\left[-\frac{1}{3}e^{-\frac{k^{2}}{2}}\int_{x}^{0}e^{-t}dt\right]dx$
$\displaystyle\leq$ $\displaystyle
e^{\frac{1}{3}e^{-\frac{k^{2}}{2}}}\int_{-kb_{m}}^{0}e^{-kx}e^{-\frac{1}{3}e^{-\frac{k^{2}}{2}}e^{-x}}dx.$

We see that the remaining integral is of the integrable form in Lemma 1 with
$f\left(x-y\right)=\frac{1}{3}e^{-\frac{k^{2}}{2}}$. Thus we see that the
integral is bounded by a finite value over $\left[-kb_{m},\infty\right)$. Now
we need to show that the integral over $\left[-\infty,-kb_{m}\right]$ goes to
zero as $m\rightarrow 0$. Consider the integral over
$\left(-\infty,-b_{m}^{2}\right)$:

$\displaystyle\int_{-\infty}^{-b_{m}^{2}}e^{-kx}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}dx=$
$\displaystyle\sqrt{2\pi}e^{\frac{1}{2}k^{2}b_{m}^{2}}\int_{-\infty}^{-b_{m}^{2}}\phi\left(\frac{x}{b_{m}}+kb_{m}\right)\Phi\left(u_{m}\left(x\right)\right)^{m-2}dx$
$\displaystyle\leq$
$\displaystyle\sqrt{2\pi}e^{\frac{1}{2}k^{2}b_{m}^{2}}\left(\frac{1}{2}\right)^{m-2}\int_{-\infty}^{-b_{m}^{2}}\phi\left(\frac{x}{b_{m}}+kb_{m}\right)dx$
$\displaystyle\leq$
$\displaystyle\sqrt{2\pi}b_{m}m^{k^{2}}\left(\frac{1}{2}\right)^{m-2}\rightarrow
0.$

Finally, we look at the integral over the region
$\left(-b_{m}^{2},-kb_{m}\right)$:

$\displaystyle\int_{-b_{m}^{2}}^{-kb_{m}}e^{-kx}e^{-\frac{1}{2}\frac{x^{2}}{b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-2}dx.$

Taking the derivative of the integrand we see:

$\frac{\partial}{\partial
x}e^{-kx}e^{-\frac{1}{2}\frac{x^{2}}{b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-2}=e^{-kx}e^{-\frac{1}{2}\frac{x^{2}}{b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-3}\left[-\Phi\left(u_{m}\left(x\right)\right)\left(k+\frac{x}{b_{m}}\right)+\phi\left(\frac{x}{b_{m}}+b_{m}\right)/b_{m}\right].$

This is surely positive for $x\leq-kb_{m}$. Let $m$ be sufficiently large such
that $\frac{b_{m}-2}{b_{m}-1}>\frac{1}{2}$ Then, returning to our integral, we
see:

$\displaystyle\int_{-b_{m}^{2}}^{-kb_{m}}e^{-kx}e^{-\frac{1}{2}\frac{x^{2}}{b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-2}dx$
$\displaystyle\leq
b_{m}^{2}e^{k^{2}b_{m}}e^{-\frac{k^{2}}{2}}\Phi\left(b_{m}-k\right)^{m-2}$
$\displaystyle\leq
b_{m}^{2}e^{k^{2}b_{m}}e^{-\frac{k^{2}}{2}}\Phi\left(b_{m}-1\right)^{m-2}$
$\displaystyle\leq
e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\left(1-\Phi\left(b_{m}-1\right)\right)\right)^{m-2}$
$\displaystyle\leq
e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{\phi\left(b_{m}-1\right)}{b_{m}-1}\left(1-\frac{1}{\left(b_{m}-1\right)^{2}}\right)\right)^{m-2}$
$\displaystyle=e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{\phi\left(b_{m}-1\right)b_{m}\left(b_{m}-2\right)}{\left(b_{m}-1\right)^{3}}\right)^{m-2}$
$\displaystyle=e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{e^{b_{m}-\frac{1}{2}}b_{m}^{2}\left(b_{m}-2\right)}{m\left(b_{m}-1\right)^{3}}\right)^{m-2}$
$\displaystyle\leq
e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{1}{2}\frac{e^{b_{m}-\frac{1}{2}}}{m}\right)^{m-2}$
$\displaystyle=e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{1}{2m}e^{b_{m}-\frac{1}{2}}\right)^{-2}e^{-m\sum_{i=1}^{\infty}\frac{\left(\frac{1}{2}e^{b_{m}-\frac{1}{2}}\right)^{i}}{im^{i}}}$
$\displaystyle\leq
e^{-\frac{k^{2}}{2}}e^{\left(k^{2}+1\right)b_{m}}\left(1-\frac{1}{2m}e^{b_{m}-\frac{1}{2}}\right)^{-2}e^{-\frac{1}{2}e^{b_{m}-\frac{1}{2}}}\rightarrow
0.$

All that is left to prove is that (7) is satisfied. To show this, we break
$l^{\prime\prime}\left(x,y,\lambda\right)$ into its constituent terms and
bound their absolute values. Recall the form of
$l^{\prime\prime}\left(x,y,\lambda\right)$ as given in (2) and made explicit
in the following calculations. We see that the absolute value of each of these
terms may be bounded above by a term of the form:

$Ce^{-k_{i}x}e^{-j_{i}y}P_{i}\left(x-y\right),$ (8)

where $k_{i},j_{i}$ are non-negative integers and $P_{i}\left(t\right)$ is a
polynomial in $t$ for each $i\in\left\\{1,2,3,4\right\\}$. This variable $i$
is just an indexing variable which will not affect the arguments below. For
brevity, we drop the subscript in the following analysis. Consider the
integral of one of these terms.

$\displaystyle\iint_{\mathbb{R}^{2}}Ce^{-kx}e^{-jy}P\left(x-y\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}dxdy$
(9) $\displaystyle=$ $\displaystyle
C\iint_{\mathbb{R}^{2}}e^{-kx}e^{-jy}P\left(x-y\right)\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}$
$\displaystyle\hskip
28.45274pt\left[\frac{m-1}{m}e^{-y}e^{-\frac{y^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right.e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
$\displaystyle\hskip
28.45274pt+\left.\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)e^{-x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right]dxdy$
$\displaystyle\leq$ $\displaystyle
C\iint_{\mathbb{R}^{2}}e^{-\left(k+1\right)x}e^{-jy}P\left(x-y\right)\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}$
$\displaystyle\hskip
28.45274pt\left[e^{-y}e^{-\frac{y^{2}}{2b_{m}^{2}}}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right.\noindent$
$\displaystyle\hskip
28.45274pt\left.+\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right]dxdy.$

For further analysis, we break the integral across its sum. We also assume
that $m$ is sufficiently large such that $\rho_{m}>0$. To begin the analysis,
we first notice that $b_{m}\sqrt{1-\rho_{m}^{2}}$ is a convergent sequence and
thus its inverse must achieve a maximum value which we will denote $M_{1}$.
Likewise, denote the maximum of $b_{m}\sqrt{1-\rho_{m}^{2}}$ as $M_{2}$.
Additionally, denote
$\lambda\left(m\right)=b_{m}\sqrt{\left(1-\rho_{m}\right)/\left(1+\rho_{m}\right)}$.
There must exist $l_{1},l_{2},C$ such that $P\left(\delta\right)\leq
C\left(e^{-l_{1}\delta}+e^{l_{2}\delta}\right)$. Using this bound and the
transformation of variables $\delta=x-y$, we analyze the second term:

$\displaystyle\iint_{\mathbb{R}^{2}}e^{-\left(k+1\right)x}e^{-jy}P\left(x-y\right)\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}\left[\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right]dxdy$
$\displaystyle\leq M_{1}\int
e^{-\left(k+j+1\right)x}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}\left(\int
e^{j\delta}P\left(\delta\right)\phi\left(\frac{\delta-\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right)d\delta\right)dx$
$\displaystyle\leq CM_{1}\int
e^{-\left(k+j+1\right)x}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}\left(\int\left(e^{j-l_{1}\delta}+e^{\left(l_{2}+j\right)\delta}\right)\phi\left(\frac{\delta-\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)}{b_{m}^{2}\sqrt{1-\rho_{m}^{2}}}\right)d\delta\right)dx$
$\displaystyle=CM_{1}\int
e^{-\left(k+j+1\right)x}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}$
$\displaystyle\hskip
28.45274pt\left(e^{\left(j-l_{1}\right)\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)+\left(j-l_{1}\right)^{2}/2b_{m}\left(1-\rho_{m}^{2}\right)}+e^{\left(l_{2}+j\right)\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)+\left(l_{2}+j\right)^{2}/2b_{m}\left(1-\rho_{m}^{2}\right)}\right)dx$
$\displaystyle\leq
CM_{1}e^{\max\left(|j-l_{1}|,l_{2}+j\right)M_{1}^{2}+\max\left(|j-l_{1}|,l_{2}+j\right)^{2}M_{2}}$
$\displaystyle\hskip 28.45274pt\int
e^{-\left(k+j+1\right)x}\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}\left(e^{-\left(j-l_{1}\right)\left(1-\rho_{m}\right)x}+e^{-\left(l_{2}+j\right)\left(1-\rho_{m}\right)x}\right)dx,$

where the transition from the third to fourth line is dependent on the moment
generating function of the normal distribution. But for sufficiently large
$m$, we know that
$-\left(l_{2}+j\right)\left(1-\rho_{m}\right)-\left(k+j+1\right)<0$ and
$-\left(j-l_{1}\right)\left(1-\rho_{m}\right)-\left(k+j+1\right)<0$ as
$\rho_{m}\rightarrow 1$. Thus, for sufficiently large $m$ we are left with the
integral:

$\int\left(e^{-s_{1}x}+e^{-s_{2}x}\right)\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\frac{x^{2}}{2b_{m}^{2}}}dx.$

This works for any term in our integrand. Throughout this process, all
inequalities have come from bounding the integrand itself, not the integral.
Thus, noting that this last integral is of the form in (7), we see that these
terms satisfy all requirements for this theorem. We still must prove the same
requirements are met for the first term in the sum in (9).

It is easy to check $\Phi\left(x\right)\leq 2\phi\left(x\right)$ for $x<0$.
This leads us to
$\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(x-\delta\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\leq
2\phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(x-\delta\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
when $\delta>-\frac{1-\rho_{m}}{\rho_{m}}\left(x+b_{m}^{2}\right)$ and
$\Phi\left(\frac{u_{m}\left(x-\delta\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\leq
2\phi\left(\frac{u_{m}\left(x-\delta\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
when $\delta<\left(1-\rho_{m}\right)\left(x+b_{m}^{2}\right)$. For the purpose
of the following analysis, we assume $m$ is sufficiently large such that
$\rho_{m}>\frac{1}{2}$. Recalling that
$\left(1-\rho_{m}\right)b_{m}^{2}\rightarrow\lambda^{2}$, we note that this
quantity must have a maximum, which we denote $M_{3}$. Then, we can relax our
conditions to
$\delta>-\frac{1-\rho_{m}}{\rho_{m}}\left(x+b_{m}^{2}\right)>-\frac{1}{2}x+2M_{2}$
and
$\delta<\left(1-\rho_{m}\right)\left(x+b_{m}^{2}\right)<\frac{1}{2}x+M_{2}$.
With this, we may finally make the calculations:

$\displaystyle\iint_{\mathbb{R}^{2}}Ce^{-kx}e^{-jy}P\left(x-y\right)\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}e^{-y-x}e^{-\frac{y^{2}+x^{2}}{2b_{m}^{2}}}$
$\displaystyle\hskip
42.67912pt\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)dxdy$
$\displaystyle\leq$
$\displaystyle\int\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\left(2+k+j\right)x}e^{-\frac{x^{2}}{2b_{m}^{2}}}$
$\displaystyle\hskip 42.67912pt\left(\int
e^{(j+1)\delta}P\left(\delta\right)\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(x-\delta\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\Phi\left(\frac{u_{m}\left(x-\delta\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)d\delta\right)dx$
$\displaystyle\leq$
$\displaystyle\int\Phi\left(u_{m}\left(x\right)\right)^{m-2}e^{-\left(2+k+j\right)x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\left(\int_{\left(1-\rho_{m}\right)x+M}^{\infty}2e^{\left(j+1\right)\delta}P\left(\delta\right)\phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(x-\delta\right)}{\sqrt{1-\rho_{m}^{2}}}\right)d\delta\right.$
$\displaystyle+$
$\displaystyle\left.\int^{\left(1-\rho_{m}\right)x+M}_{-\frac{1-\rho_{m}}{\rho_{m}}x-2}e^{\left(j+1\right)\delta}P\left(\delta\right)d\delta+\int_{-\infty}^{-\frac{1-\rho_{m}}{\rho_{m}}x-2M}2e^{\left(j+1\right)\delta}P\left(\delta\right)\phi\left(\frac{u_{m}\left(x-\delta\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)d\delta\right)dx.$

Note that the terms including
$\int_{-\infty}^{-\frac{1-\rho_{m}}{\rho_{m}}x-2M}2e^{\left(j+1\right)\delta}P\left(\delta\right)\phi\left(\frac{u_{m}\left(x-\delta\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)d\delta$
and
$\int_{\left(1-\rho_{m}\right)x+M}^{\infty}2e^{\left(j+1\right)\delta}P\left(\delta\right)\phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(x-\delta\right)}{\sqrt{1-\rho_{m}^{2}}}\right)d\delta$
are of the same form shown to satisfy our conditions in our analysis of the
first term in (9). Seeing out the calculation of the remaining term and
utilizing $P\left(\delta\right)\leq
C_{1}e^{k_{1}\delta}+C_{2}e^{-k_{2}\delta}$:

$\displaystyle\int^{\left(1-\rho_{m}\right)x+M}_{-\frac{1-\rho_{m}}{\rho_{m}}x-2M}e^{\left(j+1\right)\delta}P\left(\delta\right)d\delta$
$\displaystyle\leq$
$\displaystyle\int^{\left(1-\rho_{m}\right)x+M}_{-\frac{1-\rho_{m}}{\rho_{m}}x-2M}C_{1}e^{\left(k_{1}+j+1\right)\delta}+C_{2}e^{\left(j+1-k_{2}\delta\right)}d\delta$
$\displaystyle=$
$\displaystyle\frac{C_{1}}{k_{1}+j+1}\left[e^{\left(k_{1}+j+1\right)\left(1-\rho_{m}\right)x+M\left(k_{1}+j+1\right)}-e^{-\left(k_{1}+j_{1}+1\right)\frac{\left(1-\rho_{m}\right)}{\rho_{m}}x-2M\left(k_{1}+j+1\right)}\right]$
$\displaystyle\hskip
28.45274pt+\frac{C_{2}}{j+1-k_{2}}\left[e^{\left(j+1-k_{2}\right)\left(1-\rho_{m}\right)x+M\left(j+1-k_{2}\right)}-e^{\left(k_{2}+j+1\right)\frac{\left(1-\rho_{m}\right)}{\rho_{m}}x-2M\left(j+1-k_{2}\right)}\right].$

Once again, we see that this leaves the integral in the form of (7). Thus all
three of the final terms can be bounded and by Dominated Convergence Theorem,
the expectations converge.

We note that this immediately implies that the variance of this term is
convergent and thus also bounded. Since each term in
$\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(h\left(x,y,\lambda\right)\right)$
was treated as $Ce^{-k_{i}x-j_{i}y}P_{i}\left(x-y\right)$, taking the square
of or product of any two terms yields the same form. Thus, the same argument
above applies to the variance. ∎

######  Theorem 3.

Denote
$\tilde{L}\left(h\right)=\displaystyle\sum_{i=1}^{k}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)$
with $h\in H_{k}$ where $H_{k}$ is a ball of radius
$r_{k}=o\left(\sqrt{k}\right)$. Then, we have:

$\frac{\partial^{2}}{\partial^{2}h}\tilde{L}\left(h\right)-I_{\lambda}=o_{p}\left(1\right).$
(10)

Proof. Denote
$\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)=l^{\prime\prime}\left(h,x,y\right)$
and note that since $h$ must grow slower than $\sqrt{k}$,
$\lambda+\frac{h}{\sqrt{k}}$ stays within an open interval of $\lambda$.
Throughout this proof we will need to take expectations with respect to the
limiting Hüsler-Reiss distribution and with respect to the distribution of the
maxima of scaled bivariate normal random variables. We will denote the former
$\mathbb{E}_{\lambda}$ and the latter $\mathbb{E}_{\phi_{m}}$. Applying the
triangle inequality, we break the difference into three other differences:

$\displaystyle\left|\frac{\partial^{2}}{\partial^{2}h}\tilde{L}\left(h\right)-I_{\lambda}\right|$
$\displaystyle=\left|\frac{1}{k}\sum_{i=1}^{k}\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H\left(x_{i},y_{i}\right)_{\lambda+\frac{h}{\sqrt{k}}}\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)\right.$
$\displaystyle\hskip
28.45274pt\left.+\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)+\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)-I_{\lambda}\right|$
$\displaystyle\leq\left|\frac{1}{k}\sum_{i=1}^{k}\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H\left(x_{i},y_{i}\right)_{\lambda+\frac{h}{\sqrt{k}}}\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)\right|$
$\displaystyle\hskip
28.45274pt+\left|\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)\right|$
$\displaystyle\hskip
28.45274pt+\left|\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)-I_{\lambda}\right|.$

Denote
$|\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)-I_{\lambda}|=a_{m}$.
As shown in Theorem 2,
$\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)\rightarrow
I_{\lambda}$, and thus $a_{m}\rightarrow 0$. Denote
$|\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)|=c_{m}$.
Since $h\in H_{n}$, we know that
$\lambda+\frac{h}{\sqrt{k}}\rightarrow\lambda$. Consider, then,

$\displaystyle\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(0,X,Y\right)\right)$
$\displaystyle=\iint_{\mathbb{R}^{2}}\left(l^{\prime\prime}\left(h,x,y\right)-l^{\prime\prime}\left(0,x,y\right)\right)\frac{\partial^{2}}{\partial
x\partial
y}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m}dxdy.$

Since $l$ is continuous, this integrand converges to 0 as
$n\rightarrow\infty$. We also note that
$c_{m}\leq\mathbb{E}_{\phi_{m}}\left|l^{\prime\prime}\left(h,X,Y\right)\right|+\mathbb{E}_{\phi_{m}}\left|l^{\prime\prime}\left(0,X,Y\right)\right|$.
The second of these terms can be bounded by DCT as per Theorem 2. To take care
of the first term, we need to show that we can once again use DCT to show the
convergence of this integral, even when $\lambda$ is not exactly equal to its
true value, but instead converges to it. We see that this crucial aspect is to
once again bound each term by the form of (8). We notice that any $\Phi$ or
$\phi$ is immediately negated as these are bounded functions, and do not
appear in the form of the bound. Thus, the only time
$\lambda+\frac{h}{\sqrt{k}}$ appears is in the polynomial $P_{i}(x-y)$. Since
$\lambda+\frac{h}{\sqrt{k}}$ converges to a non-zero quantity, for
sufficiently large $k$ both $\lambda+\frac{h}{\sqrt{k}}$ and
$(\lambda+\frac{h}{\sqrt{k}})^{-1}$ will have finite suprema. Thus, wherever
$\lambda+\frac{h}{\sqrt{k}}$ appears in this bound, the we may bound it by a
value which does not depend on $m$ and use the same form in (8). Thus, the
same argument present in Theorem 2 applies, and by DCT the quantity will
converge. As both of these values converge, we can use DCT to say
$c_{m}\rightarrow 0$. Fix $\epsilon>0$. Using Chebyshev’s Inequality, we can
now show:

$\displaystyle\mathbb{P}\left(\left|\frac{\partial^{2}}{\partial^{2}h}\tilde{L}\left(h\right)-I_{\lambda}\right|>\epsilon\right)$
$\displaystyle\leq\mathbb{P}\left(\left|\frac{1}{k}\sum_{i=1}^{k}\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H\left(X_{i},Y_{i}\right)_{\lambda+\frac{h}{\sqrt{k}}}\right)-\mathbb{E}_{\phi_{m}}\left(l^{\prime\prime}\left(h,X,Y\right)\right)\right|>\epsilon-
a_{m}-c_{m}\right)$
$\displaystyle\leq\frac{Var_{\phi}\left(\frac{\partial^{2}}{\partial^{2}\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H\left(X_{i},Y_{i}\right)_{\lambda+\frac{h}{\sqrt{k}}}\right)\right)}{\left(\epsilon-
a_{m}-c_{m}\right)^{2}k}.$

Since the variance in the numerator is convergent and thus bounded due to the
note following Theorem 2, this converges to zero as $k=k_{n}\rightarrow\infty$
and the theorem is proven. ∎

######  Theorem 4.

Assume the same conditions as Theorem 3 and define

$A:=\lim_{m\rightarrow\infty}\sqrt{k}\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(X,Y\right)\right)\right).$

Then, $A$ is finite if and only if

$\lim_{m\rightarrow\infty}\frac{\sqrt{k}}{b_{m}^{2}},\hskip
14.22636pt\lim_{m\rightarrow\infty}\sqrt{k}\left(\lambda-
b_{m}\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}\right)$

are both finite. When $A$ is finite, we have

$\frac{\partial}{\partial h}\tilde{L}\left(h\right)=\frac{\partial}{\partial
h}\tilde{L}\left(0\right)-hI_{\lambda}+o_{p}\left(1\right)$ (11)
$\tilde{L}\left(h\right)=\tilde{L}\left(0\right)+h\frac{\partial}{\partial
h}\tilde{L}\left(0\right)-\frac{1}{2}h^{2}I_{\lambda}+o_{p}\left(1\right)$
(12)

as $n\rightarrow\infty$. Here, $\frac{\partial}{\partial
h}\tilde{L}\left(0\right)=\frac{1}{\sqrt{k}}\sum_{i=1}^{k}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)\xrightarrow{d}N\left(A,I_{\lambda}\right)$.

Proof. Equations 11 and 12 are a direct result of integrating (10). What
remains to show is the convergence of $\frac{\partial}{\partial
h}\tilde{L}\left(0\right)$. Note from the form of
$\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)$
(as made explicit in Section 2) each term may once again be bounded by a term
of the form $Ce^{-kx}e^{-jy}P\left(x-y\right)$, where $P\left(t\right)$ is
some polynomial of $t$. Then the first and second moments of
$\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)$
are convergent and bounded, which can be shown by the same method as used in
Theorem 2. That is, the relevant variance and expectation are bounded, and we
show that this implies that the Lindeberg condition for triangular arrays is
satisfied ([4]). Define
$Var\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(X,Y\right)\right)\right)=\sigma^{2}_{k,i}=\sigma^{2}_{k}$,
$s_{k}^{2}=\sum_{i=1}^{k}\sigma^{2}_{k,i}=k\sigma^{2}_{k}$ and
$\mu_{k}=\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(X,Y\right)\right)\right)$.
In our case, the Lindeberg condition for triangular arrays reduces to

$\lim_{n\rightarrow\infty}\frac{1}{\sigma_{k}^{2}}\mathbb{E}_{\phi_{m}}\left(\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(X,Y\right)\right)-\mu_{k}\right)^{2}1_{|\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(X,Y\right)\right)-\mu_{k}|>\epsilon
s_{k}^{2}}\right)=0.$

Since $s_{k}^{2}\rightarrow\infty$ as $k\rightarrow\infty$, we know that the
pointwise limit of this function must be $0$. We also know that, as stated
above, we may use the same DCT argument in Theorem 2 to show that this
expectation will converge to that pointwise limit, and thus the condition is
satisfied. This gives us:

$\frac{1}{\sqrt{k}}\sum_{i=1}^{k}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda+\frac{h}{\sqrt{k}}}\left(x_{i},y_{i}\right)\right)-\sqrt{k}\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H_{\lambda}\left(X,Y\right)\right)\right)\xrightarrow{D}N\left(0,I_{\lambda}\right)$

Thus, our only remaining point is to show the conditions for $A$ to be finite.
We first notice that
$\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial y}H_{\lambda}\left(x,y\right)\right)$ is continuous in $\lambda$ for
all $\lambda\in\left(0,\infty\right)$. By the Leibniz Integral Rule, we may
pass the partial derivative with respect to $\lambda$ inside the expectation,
giving us
$\mathbb{E}_{\lambda}\left(\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial y}H_{\lambda}\left(X,Y\right)\right)\right)=0$. As previously
stated, we know that
$\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial
y}H_{\lambda}\left(X,Y\right)\right)\right)\rightarrow\mathbb{E}_{\lambda}\left(\frac{\partial}{\partial\lambda}\ln\left(\frac{\partial^{2}}{\partial
x\partial y}H_{\lambda}\left(X,Y\right)\right)\right)=0$ by DCT, thus we need
only to find the balance between the growth rates of $k$ and $m$.

We begin by rewriting the Hüsler-Reiss distribution as

$H_{\lambda}\left(x,y\right)=\exp\left\\{-e^{-x}-\displaystyle\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}dz\right\\},$
(13)

and the distribution of the maxima of the rescaled normals as

$\displaystyle\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)=$
$\displaystyle\exp\left\\{-\frac{1}{m}\left[\int_{y}^{\infty}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-z-\frac{z^{2}}{2b_{m}^{2}}}dz+\right.\int_{x}^{\infty}e^{-t-\frac{t^{2}}{2b_{m}^{2}}}dt\right]$
$\displaystyle\left.-\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}\right\\}.$

We write:

$\Biggr{\rvert}\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(X,Y\right)\right)\right)-0\Biggr{\rvert}=\Biggr{\rvert}\mathbb{E}_{\phi_{m}}\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(X,Y\right)\right)\right)-\mathbb{E}_{\lambda}\left(\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(X,Y\right)\right)\right)\Biggr{\rvert}.$

We further split this into the two terms:

$\displaystyle\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)$
$\displaystyle\left[\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}e^{-x-\frac{1}{2}\left(\frac{x}{b_{m}}\right)^{2}}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right.$
(14) $\displaystyle-$
$\displaystyle\left.H_{\lambda}\left(x,y\right)e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}\right]dxdy$

and

$\displaystyle\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)$
$\displaystyle\left[\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}\frac{m-1}{m}e^{-x-y-\frac{1}{2}\left(\frac{x}{b_{m}}\right)^{2}-\frac{1}{2}\left(\frac{y}{b_{m}}\right)^{2}}\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right.$
$\displaystyle-\left.H_{\lambda}\left(x,y\right)e^{-x-y}\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\right]dxdy.$
(15)

We begin with (14). We write

$e^{-\frac{1}{2}\left(\frac{t}{b_{m}}\right)^{2}}=1-\frac{t^{2}}{2b_{m}^{2}}+e^{-\nu_{t}}\frac{t^{4}}{4b_{m}^{4}},$

where $\frac{t^{2}}{2b_{m}^{2}}>\nu_{t}>0$ is the value such that the final
term is the Lagrange Remainder of order 2. Then we rewrite

$\displaystyle\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}=\exp\left\\{-\frac{m-1}{m}\left[\int_{y}^{\infty}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-z-\frac{z^{2}}{2b_{m}^{2}}}dz\right.\right.$
$\displaystyle\hskip
28.45274pt+\left.\int_{x}^{\infty}e^{-t-\frac{t^{2}}{2b_{m}^{2}}}dz\right]\left.-(m-1)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}\right\\}$
$\displaystyle=\exp\left\\{-\frac{m-1}{m}\left[\int_{y}^{\infty}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-z}\left(1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right)dz+\right.\right.$
$\displaystyle\hskip
28.45274pt\left.\left.e^{-x}\left(1-\frac{x^{2}+2x+2}{2b_{m}^{2}}\right)+\frac{1}{4b_{m}^{4}}\int_{x}^{\infty}e^{-t-\nu_{t}}t^{4}dt\right]-\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}\right\\}.$

We introduce the notation

$\Phi\left(a\right)=\Phi\left(b\right)+\phi\left(b\right)\left(b-a\right)-\frac{1}{2}\gamma_{a,b}\phi\left(\gamma_{a,b}\right)\left(b-a\right)^{2}$
(16)

where $\min\left\\{a,b\right\\}<\gamma_{a,b}<\max\left\\{a,b\right\\}$ make
the final term the Lagrange Remainder of the expansion of $\Phi\left(a\right)$
about $b$ of order 2. Going forward, we drop the subscripts from $\gamma$ for
the sake of notation. Using this formula to continue our calculations we see

$\displaystyle\int_{y}^{\infty}\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-z}\left(1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right)dz$
$\displaystyle=\int_{y}^{\infty}\left(\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)+\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt\left.-\frac{1}{2}\gamma_{z}\phi\left(\gamma_{z}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)^{2}\right)e^{-z}\left(1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right)dz$
$\displaystyle=\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}dz+\frac{1}{4b_{m}^{4}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}e^{-\nu_{z}}z^{4}dz$
$\displaystyle\hskip
28.45274pt-\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz+\int_{y}^{\infty}\left(\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt\left.-\frac{1}{2}\gamma_{z}\phi\left(\gamma_{z}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)^{2}\right)e^{-z}\left(1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right)dz.$

This leads us to the result

$\displaystyle\frac{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}}{H_{\lambda}\left(x,y\right)}=\exp\left\\{f\left(x,y,m\right)\right\\}:=$
$\displaystyle=\exp\left\\{-\frac{m-1}{m}\left[-\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz+\frac{1}{4b_{m}^{4}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}e^{-\nu_{z}}z^{4}dz\right.\right.$
$\displaystyle\hskip
28.45274pt+\int_{y}^{\infty}\left(\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt\left.-\frac{1}{2}\gamma_{z}\phi\left(\gamma_{z}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)^{2}\right)e^{-z}\left(1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right)dz$
$\displaystyle\hskip
28.45274pt\left.\left.-e^{-x}\frac{x^{2}+2x+2}{2b_{m}^{2}}+\frac{1}{4b_{m}^{4}}\int_{x}^{\infty}e^{-t-\nu_{t}}t^{4}dt\right]+\frac{1}{m}\left[e^{-x}+\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}dz\right]\right.$
$\displaystyle\hskip
28.45274pt\left.-\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}\right\\}.$

Returning to (14),

$\displaystyle\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}e^{-x}$
$\displaystyle\hskip
28.45274pt\left[\frac{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)2\lambda
e^{\frac{-x^{2}}{2b_{m}^{2}}}}{H_{\lambda}\left(x,y\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)b_{m}\sqrt{1-\rho_{m}^{2}}}-1\right]dxdy$
$\displaystyle=\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}e^{-x}$
$\displaystyle\hskip
28.45274pt\left[\exp\left\\{f\left(x,y,m\right)-\frac{x^{2}}{2b_{m}^{2}}-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}\right.\right.$
$\displaystyle\hskip
28.45274pt\left.\left.-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)\right\\}-1\right]dxdy$
$\displaystyle=\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}e^{-x}\left[f\left(x,y,m\right)-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}\right.$
$\displaystyle\hskip
28.45274pt+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)-\frac{x^{2}}{2b_{m}^{2}}+e^{\zeta_{a}}\left(f\left(x,y,m\right)-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}\right.$
$\displaystyle\left.\left.\hskip
28.45274pt+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)-\frac{x^{2}}{b_{m}^{2}}\right)^{2}\right]dxdy,$

where $\zeta_{a}$ is the appropriate Lagrange remainder for the expansion of
the exponential term about 0. To show the convergence rate, we need only to
assure that each term in the integral is integrable and converges to zero. We
will show the absolute value of each term is integrable to simplify
calculations. In the body of this proof, we will show this only for terms
which may determine the rate of convergence, relegating any terms which
necessarily converge faster to the appendix. This means that the terms
accounted for here are:

  * Ia

$=\displaystyle\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz$,

  * IIa

$=-\displaystyle\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)e^{-z}dz$,

  * IIIa

$=\displaystyle e^{-x}\frac{x^{2}+2x+2}{2b_{m}^{2}}$,

  * IVa

$=-\displaystyle\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)$,

  * Va

$=-\displaystyle\frac{x^{2}}{2b_{m}^{2}}$,

  * VIa

$\displaystyle=-\frac{1}{2}\left[\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}-\left(\lambda-\frac{x-y}{2\lambda}\right)^{2}\right]$.

To show that the above 6 terms are finite, we argue that each term is of the
form $Ce^{-ax-by}x^{k_{1}}y^{k_{2}}$. We once again notice that each term in
$\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)$
can be bounded by a term of the form $Ce^{-ax-
by}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)P\left(x-y\right)$ where
$a,b\geq 0$ and $P$ is a polynomial. These two facts combined will mean that
each term above will be integrable by Lemma 1. Terms IIIa, Va and VIa are
already in this form, and term IVa is a constant with respect to our integral,
and thus does not affect the integrability of the integrand.

To handle term Ia, we use the bounds:

Ia
$\displaystyle=\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz$
$\displaystyle\leq\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}e^{-z}z^{2}dz$
$\displaystyle=e^{-y}\frac{y^{2}+2y+2}{2b_{m}^{2}}.$

To analyze term IIa we first note

$\displaystyle\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}$
$\displaystyle=b_{m}\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}-\lambda+\left(x-y\right)\left[\frac{1}{b_{m}\sqrt{\left(1-\rho_{m}^{2}\right)}}-\frac{1}{2\lambda}\right]+x\sqrt{\frac{1-\rho_{m}}{1+\rho}}b_{m}^{-1}.$

We notice that the first and second pairs of terms need not converge at the
rate $\frac{1}{log\left(m\right)}$ or faster as all other terms have so far.
Thus if this term converges slower than $1/b_{m}^{2}$, this will dictate our
asymptotic bias. We continue, showing that this term is indeed integrable:

IIa
$\displaystyle=\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)e^{-z}dz$
$\displaystyle\leq\int_{y}^{\infty}e^{-z}\left(b_{m}\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}-\lambda+\left(x-z\right)\left[\frac{1}{b_{m}\sqrt{\left(1-\rho_{m}^{2}\right)}}-\frac{1}{2\lambda}\right]+z\sqrt{\frac{1-\rho_{m}}{1+\rho}}b_{m}^{-1}\right)dz$
$\displaystyle=e^{-y}\left(b_{m}\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}-\lambda+\left(x-y-1\right)\left[\frac{1}{b_{m}\sqrt{\left(1-\rho_{m}^{2}\right)}}-\frac{1}{2\lambda}\right]+(y+1)\sqrt{\frac{1-\rho_{m}}{1+\rho}}b_{m}^{-1}\right),$

which again allows us to utilize the form of Lemma 1.

The analysis of (3) is generally similar, but contains a few unique details.
First, we may ignore the term $\frac{m-1}{m}=1-\frac{1}{m}$, as factoring out
$\frac{1}{m}$ is already a faster rate of convergence than other terms. So,
understanding that this term doesn’t affect our analysis, we drop it in
further analysis. Adopting the notation that $f^{\prime}\left(x,y,m\right)$ is
the same as $f\left(x,y,m\right)$ but with the quantity $m-2$ in place of
$m-1$, our equation becomes:

$\displaystyle\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)e^{-x-y}\left[-\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt+\left.e^{f^{\prime}\left(x,y,m\right)-\frac{1}{2}\left(\frac{x}{b_{m}}\right)^{2}-\frac{1}{2}\left(\frac{y}{b_{m}}\right)^{2}}\Phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\right]dxdy$
$\displaystyle=\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)e^{-x-y}\left[-\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt+\left(1+f^{\prime}\left(x,y,m\right)-\frac{1}{2b_{m}^{2}}\left(x^{2}+y^{2}\right)+\frac{1}{2}\left(f^{\prime}\left(x,y,m\right)-\frac{1}{2b_{m}^{2}}\left(x^{2}+y^{2}\right)\right)^{2}e^{\zeta_{b}}\right)$
$\displaystyle\hskip
28.45274pt\left(\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)+\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\right.$
$\displaystyle\left.\hskip
28.45274pt-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)^{2}\gamma_{1}\phi\left(\gamma_{1}\right)\right)$
$\displaystyle\hskip
28.45274pt\left(\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)+\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)\phi\left(\lambda+\frac{x-y}{2\lambda}\right)\right.$
$\displaystyle\hskip
28.45274pt-\left.\left.\frac{1}{2}\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)^{2}\gamma_{2}\phi\left(\gamma_{2}\right)\right)\right]dxdy,$

where
$\min\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}},\lambda+\frac{y-x}{2\lambda}\right)\leq\gamma_{1}\leq\max\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}},\lambda+\frac{y-x}{2\lambda}\right)$
and $\gamma_{2}$ is defined similarly. Note that, once factored out,
$\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)$
will cancel. Again, in the body of this proof we only show the integrability
of terms which may affect our convergence rate. The terms we consider are:

  * Ib

$=\displaystyle\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\frac{1}{2b_{m}^{2}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz$,

  * IIb

$=-\displaystyle\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)$
$\displaystyle\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right)e^{-z}dz$,

  * IIIb

$=\displaystyle\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)e^{-x}\frac{x^{2}+2x+2}{2b_{m}^{2}}$,

  * IVb

$=\displaystyle\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{-x^{2}}{2b_{m}^{2}}$,

  * Vb

$=\displaystyle\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{-y^{2}}{2b_{m}^{2}}$,

  * VIb

$=\displaystyle\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)$,

  * VIIb

$=\displaystyle\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)\phi\left(\lambda+\frac{x-y}{2\lambda}\right)$.

Terms Ib through Vb are the same terms (or the same term with $y$ in place of
$x$) as in the previous analysis, with a leading
$\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)$.
Each of these terms may be written as
$\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)e^{-ax-
by}x^{k_{1}}y^{k_{2}}$. Lemma 5 shows that each of these terms is finite.
Similarly, VIb and VIIb already contain the term
$\phi(\lambda+\frac{x-y}{2\lambda})$, so we can use the same argument as in
Lemma 1 and thus the theorem is proven. We note that this theorem also gives
the form of $A$, which is made explicit in the appendix. ∎

######  Theorem 5.

There exists a sequence of estimators $\hat{\lambda}_{n}$ such that

$\lim_{n\rightarrow\infty}\mathbb{P}\left(\hat{\lambda}_{n}\text{is an
MLE}\right)=1$

and

$\sqrt{k}\left(\hat{\lambda}_{n}-\lambda\right)\xrightarrow{d}N\left(I_{\lambda}^{-1}A,I_{\lambda}^{-1}\right)$

where $A\in\mathbb{R}$ is the limiting bias defined in Theorem 4.

The proof of this theorem is nearly identical to the proof of Theorem 2.2 in
dombry19, the only difference being the parameter space in dombry19 is
$\mathbb{R}^{3}$, while ours is $\mathbb{R}$. What follows is a simplified
version of their Theorem 2.2, reduced to 1-dimensional case.

Proof. Consider the processes

$M_{n}\left(h\right)=\tilde{L}\left(h\right)-\tilde{L}\left(0\right)$ (17)

and

$M\left(h\right)=h\left(A+G\right)-\frac{1}{2}h^{2}I_{\lambda}$ (18)

where $G$ is a 0-mean Gaussian random variable with covariance $I_{\lambda}$.
Let $H_{n}$ be a closed ball in $\mathbb{R}$ centered at 0 and with radius
$r_{k}\rightarrow\infty$ such that $r_{k}=o\left(k^{-\frac{1}{2}}\right)$.
Define $\hat{h}_{n}=\text{argmax}M_{n}\left(h\right)$ for $h\in H_{n}$. By
Theorem 4, for any compact $K\subset\mathbb{R}$, $M_{n}$ converges in
distribution to $M$ in $l^{\infty}\left(K\right)$ as $n\rightarrow\infty$. $M$
is continuous and has the unique maximizer

$\hat{h}=\text{argmax}M\left(h\right)=I_{\lambda}^{-1}\left(A+G\right).$ (19)

By a corollary of the Argmax theorem ([9], Corollary 5.58), if $\hat{h}_{n}$
is tight, then it converges in distribution to $\hat{h}$. Thus all that
remains is to show that $\hat{h}_{n}$ is tight. Fix $\epsilon>0$. Note that
the form of $\hat{h}$ assures us that there must exist some $R>0$ such that
$\mathbb{P}\left(|\hat{h}|<R\right)>1-\epsilon$. The form of $M$ gives us

$M\left(h\right)=M\left(\hat{h}\right)-\frac{1}{2}\left(h-\hat{h}\right)^{2}I_{\lambda},$

and thus

$\max_{|\hat{h}-h|\geq
1}M\left(h\right)=M\left(\hat{h}\right)-\frac{1}{2}I_{\lambda}.$

For $|\hat{h}|<R$

$M\left(\hat{h}\right)-\max_{|h|=R+1}M\left(h\right)=\max_{|h|\leq
R}M\left(h\right)-\max_{|h|=R+1}M(h)\geq\frac{1}{2}I_{\lambda},$

which, based on the selection of $R$, occurs with probability at least
$1-\epsilon$. Define $K=\left\\{h:|h|\leq R+1\right\\}$. Since $M_{n}$
converges weakly to $M$ in $l^{\infty}\left(K\right)$, for sufficiently large
$n$,

$\max_{|h|\leq
R}M_{n}\left(h\right)-\max_{|h|=R+1}M_{n}\left(h\right)\geq.25I_{\lambda}$

with probability at least $1-2\epsilon$. For sufficiently large $n$, $K\subset
H_{n}$. Hence $M_{n}$ is strictly concave on $H_{n}$ by Theorem 2 with
probability at least $1-\epsilon$. Then, from the equation above, and the fact
that $M_{n}$ is strictly concave, we see that
$\mathbb{P}\left(|\hat{h}_{n}|\leq R\right)\geq 1-3\epsilon$. Thus, $\hat{h}$
is tight. Also note, that $\hat{h}_{n}$ is on the interior of $H_{n}$, thus

$\frac{\partial}{\partial
h}M_{n}\left(\hat{h}_{n}\right)=0\implies\frac{\partial}{\partial
h}\tilde{L}\left(\hat{h}_{n}\right)=0.$

Define $\hat{\lambda}_{n}=\lambda+\frac{\hat{h}_{n}}{\sqrt{k}}$. Then if
$\hat{h}_{n}$ maximizes the localized loglikelihood $\tilde{L}$,
$\hat{\lambda}$ must maximize the loglikelihood $L$. Thus, with probability
tending to one, $\hat{\lambda}$ is an MLE. ∎

##  4 Simulation Study

To illustrate the practical implications of the results, we run a simulation
study. A simulation of the MLE (labelled $\hat{\lambda}$) was repeated 10,000
times. As shown below, the convergence to the limiting results is very slow.
We mitigate this by first considering a sample size $n=100,000$. After
selecting a sample size, to determine the values for $m$ and $k$ we must
select a formula for $\rho_{m}$. For convenience, we select a formula for
$\rho_{m}$ such that $\lambda_{m}=\lambda$ at each step:

$\rho_{m}=\frac{b_{m}^{2}-\lambda^{2}}{b_{m}^{2}+\lambda^{2}}.$

This assures us that the expectation of the score function will converge to
zero at a rate of $1/b_{m}^{2}$, up to a constant. To balance this, we then
choose $k=b_{m}^{4}$. We can then use the definition of $b_{m}$ and $mk=n$ to
solve for $b_{m}$ and subsequently $m$ and $k$. With this method we arrive at
the values $m=1,044$ and $k=95$. Selecting $\lambda=.5$, this gives us
$\rho_{m}=0.95$. The results of this simulation are shown in Figure 1. We
report both the simulated $\hat{\lambda}$s as well as the values of the score
functions, whose distribution (as made explicit in Theorem 4) depends on the
same parameters $A$ and $I_{\lambda}$. Table 4 numerically summarizes the
results of this simulation. We see close agreement between theoretical and
experimental values for the variances of both the score function and
$\hat{\lambda}$. The empirical bias term, however seems not to initially agree
with the theoretical values. To investigate this disagreement, we repeat this
simulation with increased sample sizes of $n=1,000,000$ and $n=10,000,000$,
the results of which are also summarized in Table 4. We see that it takes
until the final experiment for a 95% confidence interval to contain our
theoretical value for both the bias of the score function and
$\hat{\lambda}$s. We additionally see that with each step the confidence
interval shifts right toward our predicted theoretical value. This implies
that convergence towards the theoretical bias value takes large sample sizes.
We also note that the variance of these distributions are quite large compared
to the size of the bias terms, and as such a noisy estimates of these means
are to be expected. This is supported by repeating these simulations, which
tend to produce fairly different values for the means in many runs.

This underpins the importance of finding the correct balance between $k$ and
$m$. We see through the previous simulations that an increase in $k$ does
reduce the variability, but only slightly, and that large samples of $m$ are
necessary to find accurate estimates of the bias term. To that end, we run the
same simulation, with $n=1,000,000$ but redefine the relationship between $k$
and $m$ to be $k=\frac{1}{4}b_{m}^{4}$. This gives us the values $k=56$ and
$m=17,732$. This simulation shows the closest agreement with the theoretical
values thus far. While the performance of this simulation can at least
partially be attributed to random variation, it exemplifies the importance of
striking the right balance in the bias variance tradeoff, as we were able to
get similar or better performance to the simulation with $n=10,000,000$ with
one tenth the sample size.

Figure 1:  Left: 10,000 replications of simulated $\hat{\lambda}$ with
$n=100,000$. Right: Simulated values of the score function from the same
sample. Dashed lines represent theoretical distributions, solid lines
represent fitted distributions

Figure 2:  Left: 10,000 replications of simulated $\hat{\lambda}$ with
$n=1,000,000$. Right: Simulated values of the score function from the same
sample. Dashed lines represent theoretical distributions, solid lines
represent fitted distributions

Figure 3:  Left: 10,000 replications of simulated $\hat{\lambda}$ with
$n=10,000,000$ Right: Simulated values of the score function from the same
sample. Dashed lines represent theoretical distributions, solid lines
represent fitted distributions

Table 1: Results from four simulations, each with 10,000 replications.

|  | Sample
---
Size
| Theoretical
---
Mean
| Simulated
---
Mean
95% C.I. |  | Theoretical
---
Variance
| Simulated
---
Variance
$\displaystyle k=b_{m}^{4}$ | $10^{5}$ | 0.0117 | -0.0081 | (-0.0173,0.0013) | 0.2196 | 0.2255
$10^{6}$ | 0.0117 | -0.0029 | (-0.0121,0.0063) | 0.2196 | 0.2209
$10^{7}$ | 0.0117 | 0.0035 | (-0.0057,0.0127) | 0.2196 | 0.2187
$\displaystyle k=\frac{1}{4}b_{m}^{4}$ | $10^{6}$ | 0.0058 | 0.0038 | (-0.0055,0.0131) | 0.2196 | 0.2258

Statements and Declarations:Jan Hannig’s research was supported in part by the
National Science Foundation under Grant No. DMS-1916115, 2113404, and 2210337.
Hank Flury was also support in part by the National Science Foundation under
Grant No. DMS-2210337. The authors have no competing interests to report.

## References

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##  Appendix A Technical Proofs

######  Lemma 2.

Integrals of the form

$\iint_{\mathbb{R}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}dxdy$

are finite for all $m$ such that $\rho_{m},a+b+1,k_{1},k_{2}>0$.

Proof. We first establish the bound

$1-\Phi(u_{m}(x))=\int_{u_{m}(x)}^{\infty}\phi(t)dt=\frac{1}{b_{m}}\phi(b_{m})\int_{x}^{\infty}e^{-s-\frac{s^{2}}{2b_{m}^{2}}}ds\leq\frac{1}{m}e^{-x}$

With this bound, and the trivial bound
$1-\Phi_{\rho_{m}}(u_{m}(x),u_{m}(y))\leq 1-\Phi(u_{m}(x))+1-\Phi(u_{m}(y))$,
we derive the following bound.

$\displaystyle\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}$
$\displaystyle\leq\left(m-1\right)\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{2}\sum_{i=0}^{\infty}\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}$
$\displaystyle=\left(m-1\right)\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{2}}{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)}$
$\displaystyle\leq\frac{(m-1)}{m^{2}}\left(e^{-x}+e^{-y}\right)^{2}\frac{1}{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)}$
$\displaystyle\leq\left(e^{-x}+e^{-y}\right)^{2}\frac{1}{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)},$

For $x,y>0$, $\Phi_{\rho_{m}}\left(x,y\right)>\frac{1}{4}$. For $x,y$ such
that $\min\left(x,y\right)<0$ and $\rho_{m}>0$

$\displaystyle\Phi_{\rho_{m}}\left(x,y\right)$
$\displaystyle\geq\Phi_{\rho_{m}}\left(\min\left(x,y\right),\min\left(x,y\right)\right)$
$\displaystyle=\mathbb{P}(X<\min(x,y)|Y<\min(x,y))\mathbb{P}(Y<\min(x,y))$
$\displaystyle\geq\mathbb{P}(X<\min(x,y))\mathbb{P}(Y<\min(x,y))$
$\displaystyle=\Phi(\min(x,y))^{2}.$

With this, we may return to the integral in the statement of Lemma 2. We drop
the term $(e^{-x}+e^{-y})^{2}$. The only affect of this term is to replace
increase the coefficients $a$ or $b$ by 2, or each by 1. This has no affect on
the technique going forward, and so the term is dropped. We write the integral
as:

$\displaystyle\iint x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\frac{1}{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)}dxdy$
$\displaystyle\leq
4\iint_{\mathbb{R}^{2}\setminus\left(-b_{m}^{2},\infty\right)^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)dxdy$
$\displaystyle+\iint_{\left(-b_{m}^{2},\infty\right)^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\frac{1}{\Phi\left(u_{m}\left(\min\left(x,y\right)\right)\right)^{2}}dxdy$
$\displaystyle\leq
4\iint_{\mathbb{R}^{2}\setminus\left(-b_{m}^{2},\infty\right)^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)dxdy$
$\displaystyle+\iint_{\left(-b_{m}^{2},\infty\right)^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\frac{1}{\Phi\left(\min\left(x,y\right)\right)^{2}}dxdy.$

The first integral is finite by Lemma 1. To see that the second integral is
finite, we split the integral into the subsets $\left\\{x\leq y\right\\}$ and
$\left\\{y<x\right\\}$. Over the region including $\\{x<0,x\leq y\\}$, we have
the bound
$\Phi\left(\min\left(x,y\right)\right)=1-\Phi(-x)\geq\phi(x)\frac{x^{2}-1}{-x^{3}}$.
With this bound, we have

$\displaystyle\iint_{\mathbb{R}^{2}\setminus\left(-b_{m}^{2},\infty\right)\cap\left\\{x\leq
y\right\\}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\frac{1}{\Phi\left(\min\left(x,y\right)\right)^{2}}dxdy$
$\displaystyle\leq$
$\displaystyle\iint_{\mathbb{R}^{2}\setminus\left(-b_{m}^{2},\infty\right)\cap\left\\{x\leq
y\right\\}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)H_{\lambda}\left(x,y\right)\frac{1}{\left[\phi\left(x\right)\frac{x^{2}-1}{-x^{3}}\right]^{2}}dxdy$
$\displaystyle\leq$ $\displaystyle
2\pi\iint_{\mathbb{R}^{2}\setminus\left(-b_{m}^{2},\infty\right)\cap\left\\{x\leq
y\right\\}}x^{k_{1}}y^{k_{2}}e^{-ax-
by+x^{2}}\phi\left(\frac{\left(x-y\right)-\mu}{\sigma}\right)e^{-\frac{1}{2}e^{-x}}x^{2}dxdy$
$\displaystyle\leq$ $\displaystyle
2\pi\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int_{-\infty}^{-b_{m}^{2}}e^{-\left(a+b\right)x+x^{2}}e^{-\frac{1}{2}e^{-x}}x^{2+k_{1}+k_{2}-i}dx\int^{\infty}_{-\infty}\left(-\delta\right)^{i}e^{b\delta}\phi\left(\frac{\delta-\mu}{\sigma}\right)d\delta,$

where $\delta=x-y$. Finally, to see the same holds for $\left\\{y<x\right\\}$,
note that $x$ and $y$ are symmetric, and redefine $\delta=y-x$.

∎

######  Lemma 3.

Integrals of the form

$\iint_{\mathbb{R}^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}\\\
\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}dxdy$
(20)

are finite for all sufficiently large $m$.

Proof. Assume $m$ large enough such that $\rho_{m}>0$. Using the bound on the
infinite sum from the previous lemma (and once again dropping the
$(e^{-x}+e^{-y})^{2}$ term), we re-write the integral as

$\displaystyle\iint_{\mathbb{R}^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-3}dxdy$

The only difference between the integral above and the integral in Theorem 2
is the difference between the power $m-2$ and $m-3$, which makes little
material difference, as well as the inclusion of the powers of $x,y$. As such
the proof that this integral is finite is similar to the proof in Theorem 2.
Define $l,j,C>0$ such that $|\delta|^{i}<Ce^{l\delta}+Ce^{-j\delta}$ for all
$i\in\left\\{0,...,k_{2}\right\\}$. Using this bound, we get

$\displaystyle\iint_{\mathbb{R}^{2}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-3}dxdy$
$\displaystyle\leq\int_{\mathbb{R}}x^{k_{1}}e^{-\left(a+b\right)x}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-3}\int_{\mathbb{R}}\left(x-\delta\right)^{k_{2}}e^{k_{2}\delta}\phi\left(\frac{\delta-\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right)d\delta
dx$ $\displaystyle\leq
C\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int_{\mathbb{R}}x^{k_{1}+i}e^{-\left(a+b\right)x-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-3}\int_{\mathbb{R}}\left(e^{l\delta}+e^{-j\delta}\right)e^{k_{2}\delta}\phi\left(\frac{\delta-\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)}{b_{m}\sqrt{1-\rho_{m}^{2}}}\right)d\delta
dx$
$\displaystyle=b_{m}\sqrt{1-\rho_{m}^{2}}C\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int_{\mathbb{R}}x^{k_{1}+i}e^{-\left(a+b\right)x-\frac{x^{2}}{2b_{m}^{2}}}\Phi\left(u_{m}\left(x\right)\right)^{m-3}$
$\displaystyle\hskip
42.67912pt\left[e^{\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)\left(l+k_{2}\right)+\frac{\left(l+k_{2}\right)^{2}}{2}\left(b_{m}^{2}\left(1-\rho_{m}^{2}\right)\right)}+e^{\left(1-\rho_{m}\right)\left(b_{m}^{2}-x\right)\left(k_{2}-j\right)+\frac{\left(k_{2}-j\right)^{2}}{2}\left(b_{m}^{2}\left(1-\rho_{m}^{2}\right)\right)}\right]dx.$

Again, we note that $\rho_{m}\rightarrow 1$ and
$b_{m}\sqrt{1-\rho_{m}^{2}}\rightarrow 2\lambda$. Thus for large enough $m$
such that $\left(1-\rho_{m}\right)\left(k_{2}+l\right)-\left(a+b\right)<0$ and
$\left(1-\rho_{m}\right)\left(k_{2}-j\right)-\left(a+b\right)<0$, we write the
remaining integral as

$\displaystyle\int_{\mathbb{R}}x^{k_{1}+i}e^{-tx-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-3}dx$

for some $t>0$. Over the interval $\left[0,\infty\right)$

$\displaystyle\int_{0}^{\infty}x^{k_{1}+i}e^{-tx-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-3}dx\leq\int_{0}^{\infty}x^{k_{1}+i}e^{-tx}dx=\frac{\Gamma\left(k_{1}+i+1\right)}{t^{k_{1}+i-1}}.$

Over the interval $\left[-\infty,0\right]$, there exist $C_{2},v>0$ such that
$x^{k_{2}+i}<Ce^{-vx}$. The integral becomes

$\displaystyle\int_{-\infty}^{0}Ce^{-\left(t+v\right)x-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-3}dx.$

The proof that this term is finite is similar to that presented in Theorem 2,
with the power $m-2$ instead of $m-3$. Thus this term is finite.

∎

######  Lemma 4.

Integrals of the form

$\iint_{\mathbb{R}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\gamma\phi\left(\gamma\right)H_{\lambda}\left(x,y\right)dxdy$

where $\gamma$ is defined to be the second order Lagrange remainder of the
Taylor series expansion of
$\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
about $\lambda+\frac{y-x}{2\lambda}$, or the same term defined symmetrically
about x and y, are finite.

Proof.

Without loss of generality, we prove the lemma for $\gamma$ defined as the
Lagrange remainder of the Taylor series expansion of
$\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
about $\lambda+\frac{y-x}{2\lambda}$, as the proof with the alternate
definition is similar. Recall that we know
$\min\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}},\lambda+\frac{y-x}{2\lambda}\right)\leq\gamma\leq\max\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}},\lambda+\frac{y-x}{2\lambda}\right)$.
Thus, we know that $|\gamma|$ is always bounded by some polynomial in $x$ and
$y$. Continuing, we absorb this term into the terms $x^{k_{1}}$ and
$y^{k_{2}}$. If
$\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}$
and $\lambda+\frac{y-x}{2\lambda}$ have the same sign, then we may write
either $\phi\left(\gamma\right)<\phi\left(\lambda-\frac{y-x}{2\lambda}\right)$
or
$\phi\left(\gamma\right)<\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$.
In either case, we see that the integral over this region is finite by the
same argument used in Lemma 2. These terms have the same sign when
$\delta:=x-y>\max\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)$
or
$\delta<\min\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)$.
Between these bounds, we have the result:

$\displaystyle\iint_{\mathbb{R}}x^{k_{1}}y^{k_{2}}e^{-ax-
by}\gamma\phi\left(\gamma\right)H_{\lambda}\left(x,y\right)dxdy$
$\displaystyle\leq\int
x^{k_{1}}e^{-\left(a+b\right)x}e^{-\frac{1}{2}e^{-x}}\int^{\max\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}_{\min\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}\left(x-\delta\right)^{k_{2}}e^{b\delta}dxd\delta$
$\displaystyle\leq C\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int
x^{k_{1}+i}e^{-\left(a+b\right)x}e^{-\frac{1}{2}e^{-x}}\int^{\max\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}_{\min\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}e^{\left(b+j\right)\delta}+e^{\left(b-l\right)\delta}dxd\delta$
$\displaystyle\leq C\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int
x^{k_{1}+i}e^{-\left(a+b\right)x}e^{-\frac{1}{2}e^{-x}}\left(e^{\left(b+j\right)\max\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}+e^{\left(b-l\right)\max\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}\right.$
$\displaystyle\left.\hskip
14.22636pt-e^{\left(b+j\right)\min\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}-e^{\left(b-l\right)\min\left(2\lambda^{2},b_{m}^{2}\left(1-\rho_{m}\right)+x\sqrt{1-\rho_{m}^{2}}\right)}\right)dx$

for some sufficiently large $j,l$. This integral is surely finite for $m$
large enough such that
$\left(b+j\right)x\sqrt{1-\rho_{m}^{2}}<\left(a+b\right)$ by the same argument
in Lemma 1.

∎

######  Lemma 5.

Integrals of the form

$\iint e^{-ax-
by}x^{k_{1}}y^{k_{2}}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)H_{\lambda}\left(x,y\right)dxdy$

or

$\iint e^{-ax-
by}x^{k_{1}}y^{k_{2}}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}dxdy$

are finite for $a+b\geq 1$.

Proof. The key to this proof is the bound

$\frac{\phi\left(x\right)}{x}\left(1-\frac{1}{x^{2}}\right)<1-\Phi\left(x\right)<\frac{\phi\left(x\right)}{x}$

for $x>0$ as stated in 26.2.12 in abramowitz+stegun. Rewriting the first
equation as

$\displaystyle\iint e^{-ax-
by}x^{k_{1}}y^{k_{2}}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)H_{\lambda}\left(x,y\right)dxdy$
$\displaystyle\leq\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int
e^{-\left(a+b\right)x}x^{k_{1}+i}e^{-\frac{1}{2}e^{-x}}dx\int
e^{b\delta}\delta^{k_{2}-i}\Phi\left(\lambda+\frac{\delta}{2\lambda}\right)\Phi\left(\lambda-\frac{\delta}{2\lambda}\right)d\delta$
$\displaystyle=\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int
e^{-\left(a+b\right)x}x^{k_{1}+i}e^{-\frac{1}{2}e^{-x}}dx\int
e^{b\delta}\delta^{k_{2}-i}\left(1-\Phi\left(-\lambda-\frac{\delta}{2\lambda}\right)\right)\left(1-\Phi\left(-\lambda+\frac{\delta}{2\lambda}\right)\right)d\delta$
$\displaystyle=\sum_{i=0}^{k_{2}}\binom{k_{2}}{i}\int
e^{-\left(a+b\right)x}x^{k_{1}+i}e^{-\frac{1}{2}e^{-x}}dx\left[\int^{-2\lambda\left(\lambda+1\right)}_{-\infty}e^{b\delta}\delta^{k_{2}-i}\phi\left(\lambda+\frac{\delta}{2\lambda}\right)d\delta\right.$
$\displaystyle\left.\hskip
28.45274pt+\int_{2\lambda\left(\lambda+1\right)}^{\infty}e^{b\delta}\delta^{k_{2}-i}\phi\left(\lambda-\frac{\delta}{2\lambda}\right)d\delta+\int_{-2\lambda\left(\lambda+1\right)}^{2\lambda\left(\lambda+1\right)}e^{b\delta}\delta^{k_{2}-i}d\delta\right].$

The rest of the proof for this form of the integral is similar to Lemma 1. The
proof for the second form is a combination of this argument and that used in
Lemma 3

We also note that inclusion or exclusion of the term
$\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}$
does not affect the integrability, which can be shown by the same technique
used in Lemma 2.

######  Theorem 6.

All terms in (14) that were excluded from the body of the proof of Theorem 4
necessarily converge to zero faster than $\frac{1}{\ln\left(m\right)}$ or
$\lambda_{m}-\lambda$, and are integrable.

We begin with the left over terms from (14). The first terms we handle are

  1. I

$\displaystyle=\frac{1}{4b_{m}^{4}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z-\nu_{z}}z^{4}dz$

  2. II

$\displaystyle=\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left[\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right]e^{-z}\left[-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right]dz$

  3. III

$\displaystyle=\int_{y}^{\infty}\gamma_{z}\phi\left(\gamma_{z}\right)\left[\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right]^{2}e^{-z}\left[1-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right]dz$

  4. IV

$\displaystyle=\frac{1}{4b_{m}^{4}}\int_{x}^{\infty}e^{-t-\nu_{t}}t^{4}dt$

  5. V

$\displaystyle=\frac{1}{m}e^{-x}$

  6. VI

$\displaystyle=\frac{1}{m}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}dz$

  7. VII

$\displaystyle=-\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}$

Our goal for terms $I$ to $VI$ is to bound each term by the form
$\frac{1}{g\left(m\right)}e^{-ax-by}x^{k_{1}}y^{k_{2}}$ for some
$a,b,k_{1},k_{2}\in\mathbb{N}\cup\left\\{0\right\\}$ and $g$ such that
$\frac{1}{g\left(m\right)}$ is dominated by at least one of
$\frac{1}{b_{m}^{2}}$ or $\lambda_{n}-\lambda$. Recalling the definition of
$\nu_{z}$ and $\nu_{t}$ to be the second order Lagrange remainder term for the
expansion of $e^{-\frac{z^{2}}{2b_{m}^{2}}}$ or
$e^{-\frac{t^{2}}{2b_{m}^{2}}}$ about 1, we know $0<\nu_{t},\nu_{z}$ and thus
$e^{-\nu_{t}},e^{-\nu_{z}}<1$. Thus we drop the term wherever it may appear.
We see that

$\displaystyle I$
$\displaystyle=\frac{1}{4b_{m}^{4}}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z-\nu_{z}}z^{4}dz$
$\displaystyle\leq\frac{1}{4b_{m}^{4}}\int_{y}^{\infty}e^{-z}z^{4}dz$
$\displaystyle=e^{-y}\frac{y^{4}+4y^{3}+12y^{2}+24y+24}{4b_{m}^{4}}$
$\displaystyle II$
$\displaystyle=\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left[\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-z}{2\lambda}\right]e^{-z}\left[-\frac{z^{2}}{2b_{m}^{2}}+e^{-\nu_{z}}\frac{z^{4}}{4b_{m}^{4}}\right]dz$
$\displaystyle\leq\int_{y}^{\infty}\left[\lambda_{n}-\lambda+\left(x-z\right)\left(\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}-\frac{1}{2\lambda}\right)+z\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}b_{m}^{-1}\right]e^{-z}\left[\frac{z^{2}}{2b_{m}^{2}}+\frac{z^{4}}{4b_{m}^{4}}\right]dz$
$\displaystyle=e^{-y}\left(\lambda_{n}-\lambda+x\left(\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}-\frac{1}{2\lambda}\right)\left(\frac{y^{2}+2y+2}{2b_{m}^{2}}+\frac{y^{4}+4y^{3}+12y^{2}+24y+24}{4b_{m}^{4}}\right)\right)$
$\displaystyle\hskip
28.45274pt-e^{-y}\left[\left(\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}-\frac{1}{2\lambda}\right)+\sqrt{\frac{1-\rho_{m}}{1+\rho_{m}}}b_{m}^{-1}\right]\left(+\frac{y^{3}+3y^{2}+6y+6}{2b_{m}^{2}}\right.$
$\displaystyle\hskip
42.67912pt+\left.\frac{y^{5}+5y^{4}+20y^{3}+60y^{2}+120y+120}{4b_{m}^{4}}\right)$
$\displaystyle IV$
$\displaystyle=\frac{1}{4b_{m}^{4}}\int_{x}^{\infty}e^{-t-\nu_{t}}t^{4}dt$
$\displaystyle\leq\frac{1}{4b_{m}^{4}}e^{-x}\left(x^{4}+4x^{3}+12x^{2}+24x+24\right)$
$\displaystyle VI$
$\displaystyle=\frac{1}{m}\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}dz$
$\displaystyle\leq\frac{1}{m}e^{-y}$

We note that term $V$ is already in the form required. To handle $III$, recall
the definition of $\gamma_{z}$ as the Lagrange remainder from the expansion of
$\Phi\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}}\right)$
about $\lambda+\frac{x-z}{2\lambda}$. Then we know
$|\gamma_{z}|<\left|\max\left\\{\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(z\right)}{\sqrt{1-\rho_{m}^{2}}},\lambda+\frac{x-z}{2\lambda}\right\\}\right|$.
Thus, including this term only includes another polynomial in $x$ and $y$.
Using this polynomial bound, and dropping $\phi\left(\gamma_{z}\right)<1$, we
see that the analysis of $III$ is similar to that of $II$. Thus, since every
term is of the form $\frac{1}{g\left(m\right)}e^{-ax-by}x^{k_{1}}y^{k_{2}}$,
the integral will take the form

$\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)H_{\lambda}\left(x,y\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}\frac{1}{g\left(m\right)}e^{-ax-
by}x^{k_{1}}y^{k_{2}}dxdy.$ (21)

Recall that each term in
$\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)$
is also of the form $e^{-ax-by}x^{k_{1}}y^{k_{2}}$. Thus each of these
integrals is finite and converges to zero by Lemma 1. Of the terms listed
above, only $VII$ reamins, which is explicitly taken care of in Lemma 2. There
yet remains one set of term we need to handle:

$e^{\zeta_{a}}\left(f\left(x,y,m\right)-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)\right)^{2}.$

We note that this is the product of any two terms (or square of any term)
analyzed either here or in the body of Theorem 4 multiplied by the term
$e^{\zeta_{a}}$. Thus, so long as the integral stays finite, each term will
necessarily converge to zero faster than either $\frac{1}{\ln\left(m\right)}$
or $\lambda_{n}-\lambda$. Recall the definition of $\zeta_{a}$ as the second
order Lagrange remainder term for the expansion of

$\exp\left\\{f\left(x,y,m\right)-\frac{x^{2}}{2b_{m}^{2}}-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)\right\\}$

about 0. If $\zeta_{a}<0$, then we may simply drop the term and the analysis
of each term is just as before. However, if
$0>\zeta_{a}>f\left(x,y,m\right)-\frac{x^{2}}{2b_{m}^{2}}-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)$
then our integrand includes the term

$\displaystyle
H_{\lambda}\left(x,y\right)e^{-x}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}\exp\left\\{f\left(x,y,m\right)-\frac{x^{2}}{2b_{m}^{2}}-\frac{1}{2}\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)^{2}+\frac{1}{2}\left(\lambda+\frac{y-x}{2\lambda}\right)^{2}\right.$
$\displaystyle\hskip
28.45274pt\left.-\ln\left(\frac{b_{m}\sqrt{1-\rho_{m}^{2}}}{2\lambda}\right)\right\\}$
$\displaystyle=\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}e^{-x}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}.$

As such, our integrals will take the form

$\iint_{\mathbb{R}^{2}}e^{-ax-
by}x^{k_{1}}y^{k_{2}}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}dxdy.$

or

$\iint_{\mathbb{R}^{2}}e^{-ax-
by}x^{k_{1}}y^{k_{2}}e^{-\frac{x^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-1}\phi\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}\right)\frac{1}{b_{m}\sqrt{1-\rho_{m}^{2}}}\\\
\left(m-1\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}dxdy$

Both of these cases are handled by Lemma 3. Thus, all terms converge to 0
faster than $\frac{1}{\ln\left(m\right)}$ or $\lambda_{n}-\lambda$.

∎

######  Theorem 7.

All terms in (3) that were excluded from the body of the proof of Theorem 4
necessarily converge to zero faster than $\frac{1}{ln\left(m\right)}$ or
$\lambda_{m}-\lambda$, and are integrable.

Proof. The first set of excluded terms we consider are:

  1. I

$=\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)\phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)$

  2. II

$=\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)^{2}\gamma_{2}\phi\left(\gamma_{2}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)$

  3. III

$=\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)^{2}\gamma_{2}\phi\left(\gamma_{2}\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)$

  4. IV

$=\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)^{2}\gamma_{1}\phi\left(\gamma_{1}\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)$

  5. V

$=\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)^{2}\gamma_{1}\phi\left(\gamma_{1}\right)\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)$

  6. VI

$=\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)^{2}\gamma_{1}\phi\left(\gamma_{1}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)^{2}\gamma_{2}\phi\left(\gamma_{2}\right)$

Recall that we have shown that
$\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}$
and
$\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}$
are polynomials in $x$ and $y$. Additionally, we know that
$|\gamma_{1}|,|\gamma_{2}|$ are bounded by a polynomial in $x$,$y$. We can
drop any $\Phi$ above as they must be between 0 and 1. Thus terms $I$, $III$,
and $IV$ all contain a term $\phi\left(\lambda+\frac{y-x}{2\lambda}\right)$ or
$\phi\left(\lambda+\frac{x-y}{2\lambda}\right)$ and are thus integrable by
Lemma 1. Terms $II$,$V$ and $VI$ all depend on either
$\phi\left(\gamma_{1}\right)$ or $\phi\left(\gamma_{2}\right)$ and are thus
integrable by Lemma 4. We note that each of these terms must converge faster
than $\frac{1}{ln\left(m\right)}$ or $\lambda_{n}-\lambda$ as each term
includes a product of
$\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}$
and/or
$\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}$.
As shown, each term in these polynomials must coverge at least as fast as one
of $\frac{1}{ln\left(m\right)}$ or $\lambda_{n}-\lambda$ and thus their
product must converge faster.

The next bundle of terms is the product of any term in
$f^{\prime}\left(x,y,m\right)$ with any one of the 6 terms above or

  1. (a)

$=\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)$

  2. (b)

$=\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left(\frac{u_{m}\left(y\right)-\rho_{m}u_{m}\left(x\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{y-x}{2\lambda}\right)$

  3. (c)

$=\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(\frac{u_{m}\left(x\right)-\rho_{m}u_{m}\left(y\right)}{\sqrt{1-\rho_{m}^{2}}}-\lambda-\frac{x-y}{2\lambda}\right)$

Recall that all terms in $f^{\prime}\left(x,y,m\right)$ are of the form
$e^{-ax-by}x^{-k_{1}}y^{-k_{2}}$. Thus, whichever Lemma showed that the terms
by themselves were finite also hold when multiplied by a term in
$f^{\prime}\left(x,y,m\right)$. Any term including the infinite sum
$\left(m-2\right)\sum_{i=2}^{\infty}\frac{\left(1-\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)\right)^{i}}{i}$
is also finite by the analogous lemma in the appendix.

Finally, we address the term including $e^{\zeta_{b}}$. Just as with
$\zeta_{a}$, if $\zeta_{b}<0$, we may simply drop the term, and all terms are
finite by the same analysis above. If, however, $\zeta_{b}$ is positive, we
have the bound

$e^{\zeta_{b}}<e^{-\frac{x^{2}}{2b_{m}^{2}}-\frac{y^{2}}{2b_{m}^{2}}}\frac{\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}}{H_{\lambda}\left(x,y\right)}.$

Thus our integrals become

$\iint_{\mathbb{R}^{2}}\frac{\partial}{\partial\lambda}\ln\left(h_{\lambda}\left(x,y\right)\right)e^{-\frac{x^{2}}{2b_{m}^{2}}-\frac{y^{2}}{2b_{m}^{2}}}\Phi_{\rho_{m}}\left(u_{m}\left(x\right),u_{m}\left(y\right)\right)^{m-2}Xdxdy,$

where $X$ is any included term from above. Again, any term in
$f\left(x,y,m\right)$ form $e^{-ax-by}x^{k_{1}}y^{k_{2}}$, so the
integrability of each term only depend on which of the 9 terms is included in
the product. Each term will be finite by one of the lemmas given. ∎

Finally, we give the form of the bias term $A$. Define

$L_{1}:=\lim_{m\rightarrow\infty}\frac{\sqrt{k}}{b_{m}^{2}}\quad\text{and}\quad
L_{2}:=\lim_{m\rightarrow\infty}\sqrt{k}(\lambda-\lambda_{m})$

and assume both are finite. $A$ is given by:

$\displaystyle
L_{1}\iint_{\mathbf{R}^{2}}\frac{\partial}{\partial\lambda}\ln(h_{\lambda}(x,y))H_{\lambda}(x,y)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}e^{-x}\left(.5\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}z^{2}dz\right.$
$\displaystyle\hskip
14.22636pt\left.+.5e^{-x}(x^{2}+2x+2)-.5x^{2}+\lambda^{2}-\frac{\lambda}{2}\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}(x+z)dz\right)dxdy$
$\displaystyle+$ $\displaystyle
L_{2}\iint_{\mathbf{R}^{2}}\frac{\partial}{\partial\lambda}\ln(h_{\lambda}(x,y))H_{\lambda}(x,y)\phi\left(\lambda+\frac{y-x}{2\lambda}\right)\frac{1}{2\lambda}e^{-x}\left(\frac{(x-y)^{2}}{4\lambda^{3}}-\lambda+\frac{1}{\lambda}\right.$
$\displaystyle\hskip
14.22636pt-\left.e^{-x}\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{x-z}{2\lambda^{2}}-1\right)dz\right)dxdy$
$\displaystyle+$ $\displaystyle
L_{1}\iint_{\mathbf{R}^{2}}\frac{\partial}{\partial\lambda}\ln(h_{\lambda}(x,y))H_{\lambda}(x,y)e^{-x-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)\left..5\int_{y}^{\infty}\Phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-x}z^{2}dz\right.$
$\displaystyle\hskip
14.22636pt-\frac{\lambda}{2}\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)e^{-z}(x+z)dz+.5e^{-x}(x^{2}+2x+2)-x^{2}$
$\displaystyle\left.\hskip
14.22636pt+\lambda\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)^{-1}\phi\left(\lambda+\frac{y-x}{2\lambda}\right)(x+y)\right)dxdy$
$\displaystyle+L_{2}\iint_{\mathbf{R}^{2}}\frac{\partial}{\partial\lambda}\ln(h_{\lambda}(x,y))H_{\lambda}(x,y)e^{-x-y}\Phi\left(\lambda+\frac{x-y}{2\lambda}\right)\left(\phi(\lambda+\frac{y-x}{2\lambda})\left(\frac{y-x}{\lambda^{2}}-2\right)\right.$
$\displaystyle\left.\hskip
14.22636pt-\Phi\left(\lambda+\frac{y-x}{2\lambda}\right)e^{-x}\int_{y}^{\infty}\phi\left(\lambda+\frac{x-z}{2\lambda}\right)\left(\frac{x-z}{2\lambda^{2}}-1\right)dz\right)dxdy.$