#  $L_{p}$ and almost sure convergence of estimation on heavy tail index under
random censoring

Yunyi Zhang, Jiazheng Liu, Zexin Pan, Dimitris N. Politis

###### Abstract

In this paper, we prove $L_{p},\ p\geq 2$ and almost sure convergence of tail
index estimator mentioned in [1] under random censoring and several
assumptions. $p$th moment of the error of the estimator is proved to be of
order $O\left(\frac{1}{\log^{m\kappa/2}n}\right)$ with given assumptions. We
also perform several finite sample simulations to quantify performance of this
estimator. Finite sample results show that the proposed estimator is effective
in finding underlying tail index even when censor rate is high.

##  1 Introduction and assumption

###  1.1 Introduction

Research on heavy tail data is relevant to numerous statistical application,
such as actuarial science [2], economics [3] and etc. Tail index is one of the
most crucial factor for long tail data since it is related with extreme
quantiles of the underlying distribution, see [4] and [5] for further
discussion. Hill [6] proposed an estimator for tail index and this estimator
has been proved convergence under several assumptions and situations, we refer
[7] and [8] as two examples. Apart from traditional Hill’s estimator, Grama
and Spokoiny [1] applied Kullback-Leibler divergence to estimate heavy tail
index and proposed an estimator based on maximization local log-likeliihood
method, Kratz and Resnick [9] proposed a qq type estimator and Politis et.al.
[10] proposed a truncated ratio statistics and proved its $L^{p}$ convergence.

On the other hand, it is common for dealing with incomplete observations,
especially right censor data, in practical researches. Klein and Moeschberger
[11] provide detail discussion and examples on this topic. For heavy tail
data, censoring is more likely to occur. For example, in clinical trial, if
the survival time obeys long tail distribution, it is more likely for the
patient to survive after trial ends. Therefore, how to estimate heavy tail
index under censoring is worth discussion. Beirlant and Guillou [12] proved
consistency of a modified Hill’s estimator under mild censoring, Einmahl
et.al. [13] applied moment estimator in this problem and proved the asymptotic
normality of the proposed estimator. Ndao et.al. [14] and Stupfler[4]
generalized the result to the conditional heavy tail index.

Instead of convergence in probability, we mainly focus on almost sure
convergence and $L^{p}$ convergence of estimator proposed in [1] for censoring
data. Similar to [13] and [10], we apply a truncated version ratio type
statistics for heavy tail index and use method proposed by Vasiliev [15] for
proving convergence of the estimator.

We will give the basic assumption and statistics in 1.2. In section 2 and 3,
we respectively discuss $L^{p}$ convergence and almost sure convergence of the
proposed estimator, and numerical examples can be seen in section 4. Finally,
we make conclusions in section 5.

###  1.2 Basic assumptions and main results

In this part, we introduce basic assumptions, frequently used notations and
the main statistics in this paper. The notations that are not listed below
will be defined when being used.

Suppose $(X_{1},Y_{1}),\ (X_{2},Y_{2}),...,(X_{n},Y_{n})$ are i.i.d data from
underlying distribution whose tail functions are respectively
$P_{X}(x)=L_{X}(x)x^{-\alpha_{X}}$, $P_{Y}(x)=L_{Y}(x)x^{-\alpha_{Y}}$, and we
further assume that $X_{i},Y_{i},\ i=1,2,...,n$ are mutually independent.
Suppose the observed data are $(Z_{i},\delta_{i}),\ Z_{i}=X_{i}\bigwedge
Y_{i}=\min(X_{i},Y_{i})$ and $\delta_{i}=\mathbf{1}_{X_{i}\leq Y_{i}}$. Under
this assumption, it is obvious that tail function of $Z_{i}$,
$P_{Z}(x)=P_{X}(x)P_{Y}(x)=L_{Z}(x)x^{-\alpha_{Z}}$, here

$L_{Z}(x)=L_{X}(x)L_{Y}(x),\ \alpha_{Z}=\alpha_{X}+\alpha_{Y}$ (1)

Moreover, we suppose $L_{X},\ L_{Y}$ are slow varying function (see [16]).

According to Karamata’s theorem[16], the slow varying function $L_{K}(x),\
K=X,Y$ satisfy

$\log(L_{K}(x))=c_{K}(x)+\int_{a}^{x}\frac{\epsilon_{K}(y)}{y}dy,\ c_{K}(x)\to
c,\ \epsilon_{K}\to 0$ (2)

as $x\to\infty$, and if we further assume that $L_{K}$ is differentiable, then
formula 2 intuitively implies that $c^{{}^{\prime}}_{K}(x)\to 0$ and

$\frac{xL_{K}^{{}^{\prime}}(x)}{L_{K}(x)}=xc_{K}^{{}^{\prime}}(x)+\epsilon_{K}(x)$
(3)

Apart from the independent censor assumption, we hope the absolute value of
derivative of $L_{K},\ K=X,Y$ to be small enough as $x$ being sufficiently
large so that the influence of $L_{K}$ on estimating tail index can be
controlled by taking logarithm. This idea leads to assumption A2. The third
assumption comes form [10].

Assumption A1: Suppose $X_{i}$ being i.i.d data and $Y_{i}$ being i.i.d censor
time, $X_{i},Y_{i}$ are mutually independent and respectively have tail
function(that is, 1-cumulative distribution)
$P_{X}(x)=L_{X}(x)x^{-\alpha_{X}}$, $P_{Y}(x)=L_{Y}(x)x^{-\alpha_{Y}}$. Thus
the tail index of the data and censor time are $\gamma_{X}=1/\alpha_{X},\
\gamma_{Y}=1/\alpha_{Y}$

Assumption A2: Suppose $L_{K},\ K=X,Y$ are differentiable and there exists a
number $\kappa>0$ such that, for $K=X,Y$, as $x\to\infty$.

$|\frac{xL_{K}^{{}^{\prime}}(x)}{L_{K}(x)}|=O\left(\frac{1}{\log^{\kappa}x}\right)$
(4)

Assumption A3: There exists a known constant $\gamma_{0}>0$ such that

$\gamma_{X}\geq 2\gamma_{0},\ \gamma_{Y}\geq 2\gamma_{0}$ (5)

From 1 this implies that $\alpha_{Z}\leq 1/\gamma_{0}$ and corresponding
$\gamma_{Z}\geq\gamma_{0}$

Table 1 displays the frequently used notations and their meanings. In order to
illustrate the main estimator, we first introduce several intermediate
statistics.

Table 1: Frequently used notations Notation | Meaning
---|---
$X_{i},Y_{i}$ |  $X_{i},Y_{i}$ respectively represents the i.i.d real data and censor time
$(Z_{i},\delta_{i})$ |  $Z_{i}=\min(X_{i},Y_{i})$ and $\delta_{i}=\mathbf{1}_{X_{i}\leq Y_{i}}$
$P_{K}(x),\ K=X,Y,Z$ | Tail function of random variable $X_{i},Y_{i},Z_{i}$
$L_{K}(x),\ K=X,Y,Z$ | Slow varying factor in the tail function
$f_{K}(x),\ K=X,Y,Z$ | Density function of random variable $K$
$\mathbf{1}_{h(K_{1},..,.K_{n})\in A},\ K=X,Y,Z$ | If $h(K_{1},...,K_{n})\in A$ the function equals 1 and 0 otherwise
$t(n)$ |  $t(n)\to\infty$ as $n\to\infty$, see 9 for definition
$s(n)$ |  $s(n)\to 0$ as $n\to\infty$(see 9)
$\gamma_{K},\ K=X,Y,Z$ | Tail index for distribution $P_{K}$
$\alpha_{K},\ K=X,Y,Z$ | $\alpha_{K}=1/\gamma_{K}$
$||K||_{p}$, | For random variable $K$, $||K||_{p}=(\mathbf{E}|K|^{p})^{1/p}$ being $L_{p}$ norm

######  Definition 1 ($\widehat{p}(x)$ and $\widehat{q}(x)$).

Suppose $x>0$ and sample size is $n$, then we respectively define
$\widehat{p}(x)$ and $\widehat{q}(x)$ as

$\widehat{p}(x)=\frac{1}{n}\Sigma_{i=1}^{n}\mathbf{1}_{Z_{i}\geq x}$ (6)

and

$\widehat{q}(x)=\frac{1}{n}\Sigma_{i=1}^{n}\delta_{i}\mathbf{1}_{Z_{i}\geq x}$
(7)

It is not difficult to see that $\widehat{p}(x)$ is estimator for $P_{Z}(x)$
and $\widehat{q}(x)$ is an estimator for $Prob(Z\leq x\cap X\leq Y)$. 21
demonstrates the motivation for us to estimate this probability. Similar with
[13], the second estimator $\widehat{\rho}$ is applied to estimate
$\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$.

######  Definition 2.

With the notation in table 1, we define estimator $\widehat{\rho}$ as

$\widehat{\rho}=\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}\geq
s(n)}$ (8)

Here $t(n)\to\infty$ and $s(n)\to 0$ satisfying

$\begin{cases}t(n)=n^{\beta},\ s(n)=n^{-c},\ \beta<\frac{\gamma_{0}}{2},\
\frac{\beta}{\gamma_{0}}<c<1/2,\ \text{If A3 is satisfied}\\\
t(n)=\log^{\beta}n,\ s(n)=n^{-c},\ \beta>0,\ 0<c<1/2\ \text{If A3 is not
satisfied}\end{cases}$ (9)

We apply estimator defined in [1], $\zeta$, for estimating tail index of the
censored data $\gamma_{Z}$. According to 1, it is reasonable for expecting
$\zeta$ to converge to $\frac{\gamma_{X}\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$.

######  Definition 3 (Estimator for tail index $\gamma_{Z}$).

With the notation in table 1, we define the estimator for tail index
$\gamma_{Z}$ as

$\widehat{\zeta}=\frac{1}{\widehat{p}(t(n))}\int_{t(n)}^{\infty}\frac{\widehat{p}(y)}{y}dy\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}=\frac{\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{n\widehat{p}(t(n))}\left(\Sigma_{i=1}^{n}\log\left(\frac{Z_{i}}{t(n)}\right)\mathbf{1}_{Z_{i}\geq
t(n)}\right)$ (10)

Here we use convention that $\infty\times 0=0$. Since
$\gamma_{X}=\frac{\gamma_{X}\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}/\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{y}}$,
estimator $\frac{\widehat{\zeta}}{\widehat{\rho}}$ is a candidate for
estimating tail index of $X$. We will use a truncated version of this
estimator. The key results of this paper is presented in theorem 11 and 12.

######  Theorem 1.

Suppose A1, A2, A3 and choose $H_{n}=\frac{1}{\log\log n}$,$m\geq 4$, then we
have

$||\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq
H_{n}}-\gamma_{X}||_{m/2}=O\left(\frac{1}{\log^{\kappa}n}\right)$ (11)

######  Theorem 2.

Suppose condition A1-A3 and choose $H_{n}$ as in theorem 11, then we have

$\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq
H_{n}}\to_{a.s.}\gamma_{X}$ (12)

##  2 $L_{p}$ convergence of truncated statistics

We first provide several crucial lemma that will be frequently used in the
following proof.

######  Lemma 1.

Suppose $X$ and $Y$ satisfy A1 and A2, then $Z=min(X,Y)$ has tail index
$\gamma_{Z}=\frac{\gamma_{X}\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$, and
$P_{Z}(x)=L_{Z}(x)x^{-1/\gamma_{Z}}$ with $L_{Z}$ satisfies A2

###### Proof.

Because of independents, we have, for arbitrary large $x$,

$P_{Z}(x)=L_{X}(x)L_{Y}(x)x^{-\alpha_{X}-\alpha_{Y}}$ (13)

Therefore,
$\gamma_{Z}=1/(\alpha_{X}+\alpha_{Y})=\frac{\gamma_{X}\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$
and the first part is proved. For the second part, notice that

$|\frac{xL_{Z}^{{}^{\prime}}(x)}{L_{Z}(x)}|\leq|\frac{xL_{X}^{{}^{\prime}}}{L_{X}}|+|\frac{xL_{Y}^{{}^{\prime}}}{L_{Y}}|=O\left(\frac{1}{\log^{\kappa}n}\right)$
(14)

and the result is proved. ∎

The next one is introduced to provide a bound for the slow varying function.

######  Lemma 2.

Suppose $L_{K},\ K=X,Y$ satisfies condition A2, then for $\forall\epsilon>0$
being given, for sufficiently large $x$, we have

$x^{-\epsilon}\leq L_{K}(x)\leq x^{\epsilon},\ K=X,Y,Z$ (15)

###### Proof.

This is equivalent as $|\log L_{K}(x)|\leq\epsilon\log x$ for large $x$.
According to lemma 1, $L_{Z}(x)$ also satisfies A2. We suppose $\kappa<1$ and
if $\kappa\geq 1$, the derivative is of $o\left(\frac{1}{\log^{0.5}x}\right)$
for large $x$. Because of A2, there exists a constant $C>0$ and $x_{0}>0$ such
that for arbitrary $x>x_{0}$,

$|\log L_{K}(x)|\leq|\log
L_{K}(x_{0})|+\frac{C}{1-\kappa}(\log^{1-\kappa}x-\log^{1-\kappa}x_{0})\leq\epsilon\log
x$ (16)

for large $x$, and the result is proved. ∎

The third one involves a frequently used inequality.

######  Lemma 3.

Suppose $K_{i},\ i=1,2,...,n$ being i.i.d random variables and define
$I_{i}=\mathbf{1}_{K_{i}\in A_{n}}$, here suppose $A_{n}$ is a Borel set with
positive measure as a function of sample size $n$. Define
$\widehat{r}=\frac{\Sigma_{i=1}^{n}I_{i}}{n}$ and $r=\mathbf{E}I_{i}$, and
suppose $m\geq 2$ being a constant, then there exists a constant $C_{m}$ such
that

$||\widehat{r}-r||_{m}\leq\frac{C_{m}(r(1-r))^{1/m}}{\sqrt{n}}$ (17)

###### Proof.

Define $f_{k}$ as

$f_{k}=\begin{cases}\Sigma_{i=1}^{k}(I_{i}-r),\ k\leq n\\\
\Sigma_{i=1}^{n}(I_{i}-r),\ k>n\end{cases}$ (18)

Then, $f_{k}$ is a martingale. Since $m\geq 2$, from Minkowski inequality and
Burkholder inequality [17], we have

$\displaystyle n||\widehat{r}-r||_{m}\leq
C_{m}||\sqrt{\Sigma_{i=1}^{n}(I_{i}-r)^{2}}||_{m}$ (19)
$\displaystyle=C_{m}\sqrt{||\Sigma_{i=1}^{n}(I_{i}-r)^{2}||_{m/2}}$
$\displaystyle\leq C_{m}\sqrt{\Sigma_{i=1}^{n}||(I_{i}-r)^{2}||_{m/2}}$

Since $m\geq 2$, we have $\mathbf{E}|I_{i}-r|^{m}\leq r(1-r)$ and the result
is proved. ∎

Now, we start proving the $L_{p}$ convergence of estimator $\widehat{\rho}$.

######  Theorem 3.

Suppose A1 and A2 and $m\geq 2$, then we have

$||\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}||_{m}=O\left(\frac{1}{s(n)\sqrt{n}}\right)+O\left(\frac{1}{P_{Z}(t(n))\sqrt{n}}\right)+O\left(\frac{1}{\log^{\kappa}t(n)}\right)$
(20)

###### Proof.

According to [18], we have that $\delta_{i}$ has the same distribution as
$\mathbf{1}_{U_{i}\leq\lambda(Z_{i})}$, here $U_{i}$ is uniform $[0,1]$ random
variable being independent with $Z_{i}$ and

$\lambda(x)=Prob(X\leq
Y|Z=x)=\frac{f_{X}(x)P_{Y}(x)}{f_{X}(x)P_{Y}(x)+f_{Y}(x)P_{X}(x)}=\frac{1/\gamma_{X}-\frac{xL_{X}^{{}^{\prime}}(x)}{L_{X}(x)}}{1/\gamma_{X}+1/\gamma_{Y}-\frac{xL_{X}^{{}^{\prime}}(x)}{L_{X}(x)}-\frac{xL_{Y}^{{}^{\prime}}(x)}{L_{Y}(x)}}$
(21)

Because of condition A2, for sufficiently large $x$, we have

$|\lambda(x)-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|\leq\frac{2}{(\gamma_{X}+\gamma_{Y})^{2}}\left(\gamma_{X}\gamma_{Y}^{2}|\frac{xL_{Y}^{{}^{\prime}}}{L_{Y}}|+\gamma^{2}_{X}\gamma_{Y}|\frac{xL_{X}^{{}^{\prime}}}{L_{X}}|\right)$
(22)

Thus, there exists a constant $C$ such that for sufficiently large $x$,
$|\lambda(x)-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|\leq\frac{C}{\log^{\kappa}(x)}$
and correspondingly, from mean value theorem, we have

$|\mathbf{E}\widehat{q}(t(n))-\frac{\gamma_{Y}P_{Z}(t(n))}{\gamma_{X}+\gamma_{Y}}|\leq\int_{t(n)}^{\infty}|\lambda(x)-\frac{\gamma_{X}}{\gamma_{X}+\gamma_{Y}}|f_{Z}(x)dx\leq\frac{C}{\log^{\kappa}t(n)}P_{Z}(t(n))$
(23)

From Minkowski inequality, we have

$\displaystyle||\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}||_{m}$ (24)
$\displaystyle\leq||\left(\frac{\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}\right)\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}+\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-1||_{m}$

If A3 is satisfied, then for sufficiently large $n$, from lemma 1 and 9, for
$\forall\epsilon>0$

$P_{Z}(t(n))\geq L_{Z}(t(n))t^{-1/\gamma_{0}}(n)\geq
n^{-\beta(1/\gamma_{0}+\epsilon)}$ (25)

choose small $\epsilon$ we have $c>\epsilon+\frac{\beta}{\gamma_{0}}$ and thus
for large $n$, we have $P_{Z}(t(n))/2>s(n)$.

If A3 is not satisfied, from 9, similar with 25, we have
$P_{Z}(t(n))\geq\log^{-\beta(\epsilon+1/\gamma_{0})}n>2s(n)$ for large $n$.

Thus from Chebyshev’s inequality and lemma 17, we have

$\displaystyle||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-1||_{m}\leq||\left(\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}-1\right)\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}+Prob\left(\widehat{p}(t(n))<s(n)\right)^{1/m}$ (26)
$\displaystyle\leq\frac{||\widehat{p}(t(n))-P_{Z}(t(n))||_{m}}{s(n)}+\frac{2||\widehat{p}(t(n))-P_{Z}(t(n))||_{m}}{P_{Z}(t(n))}$
$\displaystyle=O\left(\frac{1}{s(n)\sqrt{n}}\right)+O\left(\frac{1}{P_{Z}(t(n))\sqrt{n}}\right)$

This directly implies that
$||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}=O\left(1\right)$

For the first term, from Cauchy inequality, there exists a constant $C$ such
that we have

$\displaystyle||\left(\frac{\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}\right)\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}$ (27)
$\displaystyle\leq||\left(\frac{\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}\right)||_{2m}||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{2m}$ $\displaystyle\leq
C\left(\frac{1}{P_{Z}(t(n))}||\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))||_{2m}+O\left(\frac{1}{\log^{\kappa}t(n)}\right)\right)$

From lemma 17, we get the result. ∎

Notice that, if we assume A3, form 9, we know that the convergence rate is of
$O\left(\frac{1}{\log^{\kappa}n}\right)$, otherwise the convergence rate is of
$O\left(1/\log^{\kappa}\log n\right)$.

In the next part, we will concentrate on estimating $\gamma_{Z}$. According to
[18], since $\gamma_{Z}=\frac{\gamma_{X}\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$,
if we can find a suitable estimator $\widehat{\zeta}$ of $\gamma_{Z}$, since
$\gamma_{X}=\gamma_{Z}/\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$, it is
reasonable to consider $\widehat{\zeta}/\widehat{\rho}$. We will prove $L_{p}$
convergence of its truncated version below. First we give a lemma.

######  Lemma 4.

Suppose A1, A2, then as $t\to\infty$, we have

$\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{P_{Z}(y)}{y}dy=\gamma_{Z}+O\left(\frac{1}{\log^{\kappa}t}\right)$
(28)

###### Proof.

Since
$\gamma_{Z}=\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{L_{Z}(t)}{y^{(1+\alpha_{Z})}}dy$,
from Fubini-Tonelli theorem and A2, for large $t$, there exists constant $C$
such that

$\displaystyle|\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|=\frac{1}{P_{Z}(t)}|\int_{t}^{\infty}\frac{L_{Z}(y)-L_{Z}(t)}{y^{\alpha_{Z}+1}}dy|$
(29)
$\displaystyle\leq\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{dy}{y^{\alpha_{Z}+1}}\int_{t}^{y}|L_{Z}^{{}^{\prime}}(z)|dz$
$\displaystyle=\frac{\gamma_{Z}}{P_{Z}(t)}\int_{t}^{\infty}|L_{Z}^{{}^{\prime}}(z)|z^{-\alpha_{Z}}dz$
$\displaystyle\leq\frac{\gamma_{Z}C}{P_{Z}(t)}\int_{t}^{\infty}\frac{L_{Z}(z)}{\log^{\kappa}z}z^{-\alpha_{Z}-1}dz$

From mean value theorem, suppose $\alpha_{Z}=1/\gamma_{Z}$

$\displaystyle\int_{t}^{\infty}\frac{L_{Z}(z)}{\log^{\kappa}z}z^{-1/\gamma_{Z}-1}dz=\Sigma_{n=0}^{\infty}\int_{t\log^{n}t}^{t\log^{n+1}t}\frac{L_{Z}(z)}{z^{\alpha_{Z}+1}\log^{\kappa}z}dz$
(30)
$\displaystyle=\Sigma_{n=0}^{\infty}\frac{\gamma_{Z}L_{Z}(\eta_{n})}{\log^{\kappa}\eta_{n}}\frac{1}{t^{\alpha_{Z}}\log^{n\alpha_{Z}}t}\left(1-\frac{1}{\log^{\alpha_{Z}}t}\right)$
$\displaystyle\leq\Sigma_{n=0}^{\infty}\frac{\gamma_{Z}L_{Z}(\eta_{n})}{\left(\log
t+n\log\log t\right)^{\kappa}}\frac{1}{t^{\alpha_{Z}}\log^{n\alpha_{Z}}t}$

Notice that, from assumption A2, if $\kappa\neq 1$, then

$\displaystyle|\log L_{Z}(\eta_{n})-\log
L_{Z}(t)|\leq\int_{t}^{\eta_{n}}|\frac{L_{Z}^{{}^{\prime}}(z)}{L_{Z}(z)}|dz\leq\frac{C}{1-\kappa}(\log^{1-\kappa}\eta_{n}-\log^{1-\kappa}t)$
(31) $\displaystyle\leq\frac{C}{|1-\kappa|}|(\log t+(n+1)\log\log
t)^{1-\kappa}-\log^{1-\kappa}t|$

And if $\kappa=1$, then similarly we have $|\log L_{Z}(\eta_{n})-\log
L_{Z}(t)|\leq$, here $C$ is a constant. We continue proof with 3 different
cases.

Case 1: $0<\kappa<1$. In this case, for a given constant $D$ and sufficiently
large $t$, equation 31 is less than

$\frac{C}{1-\kappa}\left(\frac{\log t}{\log^{\kappa}t}+\frac{((n+1)\log\log
t)}{((n+1)\log\log
t)^{\kappa}}-\log^{1-\kappa}t\right)\leq\frac{(n+1)^{1-\kappa}}{(1-\kappa)D}\log\log
t$ (32)

This implies that $\frac{L_{Z}(\eta_{n})}{L_{Z}(t)}\leq(\log
t)^{\frac{(n+1)^{1-\kappa}}{D(1-\kappa)}}$, combine with 29 and 30, we have

$|\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|\leq
C\gamma_{z}^{2}\left(\frac{L_{Z}(\eta_{0})}{L_{Z}(t)\log^{\kappa}t}+\Sigma_{n=1}^{\infty}\frac{1}{(\log
t+n\log\log t)^{\kappa}(\log
t)^{n\alpha_{Z}-\frac{(n+1)^{1-\kappa}}{D(1-\kappa)}}}\right)$ (33)

Since

$\log L_{Z}(\eta_{0})-\log
L_{Z}(t)\leq\frac{1}{1-\kappa}\left(\log^{1-\kappa}t\left(1+\frac{\log\log
t}{\log
t}\right)^{1-\kappa}-\log^{1-\kappa}t\right)\leq\frac{1}{1-\kappa}\frac{\log\log
t}{\log^{\kappa}t}=o(1)$ (34)

Thus, for large $t$, $\frac{L_{Z}(\eta_{0})}{L_{Z}(t)}<2$. Also, for $n\geq
1$, we have $n+1\leq 2n$ and

$\frac{n\alpha_{Z}}{2}>\frac{(n+1)^{1-\kappa}}{D(1-\kappa)}\Leftarrow
Dn^{\kappa}\alpha_{Z}>\frac{2^{2-\kappa}}{1-\kappa}\Leftarrow
D\alpha_{Z}>\frac{2^{2-\kappa}}{1-\kappa}$ (35)

choose $D$ satisfies this condition then

$\Sigma_{n=1}^{\infty}\frac{1}{(\log
t)^{n\alpha_{Z}-\frac{(n+1)^{1-\kappa}}{D(1-\kappa)}}}\leq\Sigma_{n=1}^{\infty}\frac{1}{4^{n\alpha_{Z}-\frac{(n+1)^{1-\kappa}}{D(1-\kappa)}}}\leq\Sigma_{n=1}^{\infty}\frac{1}{2^{n\alpha_{Z}}}<\infty$
(36)

And we prove the result.

Case 2: $\kappa=1$. If $\kappa=1$, from 31, we have
$\frac{L_{Z}(\eta_{n})}{L_{Z}(t)}\leq\left(1+\frac{(n+1)\log\log t}{\log
t}\right)^{C}$, correspondingly, combine with 29 and 30, for large $t$, we
have

$\displaystyle|\frac{1}{P_{Z}(t)}\int_{t}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|\leq\frac{C\gamma_{Z}^{2}}{\log
t}\Sigma_{n=0}^{\infty}\frac{1}{\log^{n\alpha_{Z}}t}\left(1+\frac{(n+1)\log\log
t}{\log t}\right)^{C}$ (37) $\displaystyle\leq\frac{C\gamma_{Z}^{2}}{\log
t}\left(2^{C}+\Sigma_{n=1}^{\infty}\frac{(n+2)^{C}}{2^{n\alpha_{Z}}}\right)$

Since $\Sigma_{n=1}^{\infty}\frac{(n+2)^{C}}{2^{n\alpha_{Z}}}<\infty$, the
result is proved.

Case 3: $\kappa>1$. If $\kappa>1$, from 31, we have
$\frac{L_{Z}(\eta_{n})}{L_{Z}(t)}\leq\exp(2C/(\kappa-1))$, combine with 30 and
we prove the result. ∎

Now, we start to prove the $L_{p}$ convergence of statistics
$\widehat{\zeta}$.

######  Theorem 4.

Suppose A1,A2 and $m\geq 2$ then we have

$||\widehat{\zeta}-\gamma_{Z}||_{m}=O\left(\frac{1}{s(n)\sqrt{n}}+\frac{1}{P_{Z}(t(n))\sqrt{n}}+\frac{1}{\log^{\kappa}t(n)}\right)$
(38)

###### Proof.

From definition of $\widehat{\zeta}$ (see 10) and Minkowski inequality,

$\displaystyle||\widehat{\zeta}-\gamma_{Z}||_{m}\leq||\left(\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{\widehat{p}(y)}{y}dy-\gamma_{Z}\right)\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}+$ (39)
$\displaystyle\gamma_{Z}||\left(\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}-1\right)\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{m}+\gamma_{Z}Prob\left(\widehat{p}(t(n))<s(n)\right)^{1/m}$

For the second and the third term, from 26 we know that these term is of order
$O\left(\frac{1}{s(n)\sqrt{n}}+\frac{1}{P_{Z}(t(n))\sqrt{n}}\right)=o(1)$. For
the first term, from Cauchy inequality and Minkowski inequality, it is less
than

$||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{2m}\left(\frac{1}{P_{Z}(t(n))}||\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy||_{2m}+|\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|\right)$
(40)

From 26,
$||\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{2m}=O\left(1\right)$, from integral version Minkowski inequality and
lemma 17, we have

$\displaystyle||\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y))}{y}dy||_{2m}\leq\int_{t(n)}^{\infty}\frac{||\widehat{p}(y)-P_{Z}(y)||_{2m}}{y}dy$
(41)
$\displaystyle\leq\int_{t(n)}^{\infty}\frac{C_{2m}P_{Z}^{1/2m}(y)}{\sqrt{n}y}dy$
$\displaystyle=\int_{t(n)}^{\infty}\frac{C_{2m}L_{Z}^{1/2m}(y)}{\sqrt{n}y^{1+\alpha_{Z}/2m}}dy$

From lemma 15, choose $\epsilon=\alpha_{Z}/2$, for sufficiently large $y$,
$L_{Z}(y)^{1/2m}\leq y^{\alpha_{Z}/4m}$ and thus the integration is less than
$\frac{4mC_{2m}}{\alpha_{Z}\sqrt{n}}t(n)^{-\alpha_{Z}/4m}=o(1/\sqrt{n})$.
Thus,

$\frac{1}{P_{Z}(t(n))}||\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy||_{2m}=O\left(\frac{1}{P_{Z}(t(n))\sqrt{n}}\right)$
(42)

For the second term in 40, use lemma 28 and we prove the result. ∎

Similarly, if in addition we assume A3, from 9 we know that the convergence
rate is of $O\left(\frac{1}{\log^{\kappa}n}\right)$ and otherwise the
convergence rate becomes $O\left(\frac{1}{\log^{\kappa}\log n}\right)$.

Finally, we apply discussions above to prove theorem 11.

###### Proof for theorem 11.

We choose $\mu=\nu=m/2$ in theorem 1 of [15], according to 20 and 38, since
exists constant $C$ such that for sufficiently large $n$,

$\displaystyle\mathbf{E}|\widehat{\zeta}-\gamma_{Z}|^{2\nu}=||\widehat{\zeta}-\gamma_{Z}||_{m}^{m}\leq\frac{C^{m}}{\log^{m\kappa}n}$
(43) $\displaystyle
w_{n}=\mathbf{E}|\widehat{\rho}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|^{2\mu}\leq\frac{C^{m}}{\log^{m\kappa}n}$

Choose
$\phi_{n}(m)=\frac{2^{m-1}}{\left(\gamma_{Y}/(\gamma_{X}+\gamma_{Y})\right)^{m}}\frac{C^{m}(\gamma_{Z}^{m}+\left(\frac{\gamma_{X}}{\gamma_{X}+\gamma_{Y}}\right)^{m})}{\log^{m\kappa}n}=O\left(\frac{1}{\log^{m\kappa}n}\right)$
and choose $\beta=m/4$, $\beta\geq 1$, then we have

$\mathbf{E}|\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq
H_{n}}-\gamma_{X}|^{2\beta}\leq V_{n}(\beta)$ (44)

Here

$V_{n}(\beta)=O\left(\phi_{n}(2\beta)+\frac{\phi_{n}^{1/2}(m)w_{n}^{1/4}}{H_{n}^{\beta}}+\frac{\phi_{n}^{1/2}(m)w_{n}^{1/2}}{H_{n}^{2\beta}}+w_{n}\right)=O\left(\frac{1}{\log^{m\kappa/2}n}\right)$
(45)

Combine 44 and 45, we prove the result. ∎

In particular, this directly proves the $L_{m/2}$ convergence of the
statistics.

##  3 Almost sure convergence of tail index estimator

In this section, we try to prove the almost sure convergence of the tail index
estimator under assumption A1-A3. We first introduce two lemma.

######  Theorem 5.

Suppose A1-A3, and $t(n),s(n)$ are chosen as in 9, then we have

$\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}\to_{a.s.}\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$ (46)

###### Proof.

From Borel-Cantelli lemma [19], it suffices to show that, for
$\forall\epsilon>0$,

$\Sigma_{n=1}^{\infty}Prob\left(|\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|>3\epsilon\right)<\infty$ (47)

Since

$\displaystyle
Prob\left(|\frac{\widehat{q}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|>3\epsilon\right)$ (48)
$\displaystyle\leq
Prob\left(|\left(\frac{\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\right)\left(\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}-1\right)\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}|\geq\epsilon\right)$
$\displaystyle+Prob\left(|\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|\geq\epsilon\right)$
$\displaystyle+Prob\left(|\frac{\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}|\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}\geq\epsilon\right)$

We will separately discuss these 3 terms below.

For the first term, notice that for $\forall k>1$, from mean value inequality
and Minkowski inequality, we have

$\displaystyle
Prob\left(|\left(\frac{\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\right)\left(\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}-1\right)|\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}\geq\epsilon\right)$ (49)
$\displaystyle\leq\frac{1}{\epsilon^{k}}||\left(\frac{\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\right)\left(\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}-1\right)\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{k}^{k}$
$\displaystyle\leq\frac{1}{2^{k}\epsilon^{k}}||\left(\frac{\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\right)^{2}+\left(\frac{\widehat{p}(t(n))-\mathbf{E}\widehat{p}(t(n))}{\widehat{p}(t(n))}\right)^{2}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}||_{k}^{k}$
$\displaystyle\leq\frac{1}{2^{k}\epsilon^{k}}\left(||\frac{\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}||_{2k}^{2}+\frac{1}{s(n)^{2}}||\widehat{p}(t(n))-\mathbf{E}\widehat{p}(t(n))||_{2k}^{2}\right)^{k}$

From lemma 17,
$||\widehat{q}(t(n))-\mathbf{E}\widehat{q}(t(n))||_{2k}^{2}=O\left(\frac{P_{Z}^{1/k}(t(n))}{n}\right)$
and
$||\widehat{p}(t(n))-\mathbf{E}\widehat{p}(t(n))||_{2k}^{2}=O\left(\frac{P_{Z}^{1/k}(t(n))}{n}\right)$,
choose $k>\max\left(\frac{1}{1/2-\beta/\gamma_{0}},\frac{1}{1-2c}\right)$ then
the convergence of summation of first term is proved.

For the second term, notice that it is smaller than

$Prob\left(|\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}|\geq\epsilon-|\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|\right)$
(50)

From 23, for sufficiently large $n$,
$|\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|\leq\epsilon/2$,
and
$\frac{\mathbf{E}\widehat{q}(t(n))}{P_{Z}(t(n))}\leq\frac{2\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}$.
Since for $\forall k>1$

$\displaystyle
Prob\left(|\frac{P_{Z}(t(n))}{\widehat{p}(t(n))}\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}-1|\geq\frac{(\gamma_{X}+\gamma_{Y})\epsilon}{4\gamma_{Y}}\right)$ (51)
$\displaystyle\leq\left(\frac{4\gamma_{Y}}{\epsilon(\gamma_{X}+\gamma_{Y})}\right)^{k}\left(\frac{1}{s(n)}||\widehat{p}(t(n))-\mathbf{E}\widehat{p}(t(n))||_{k}+Prob(\widehat{p}(t(n))<s(n))\right)^{k}$

From 26 and similar to 49, choose sufficiently large $k$ and we know that
summation of this term converges.

For the third term, similar with 49 and we prove can prove the convergence of
summation. Since 47 is true, almost sure convergence is proved as well. ∎

######  Theorem 6.

Suppose A1-A3 and suppose $s(n)$ and $t(n)$ are chosen as in 9, then we have

$\widehat{\zeta}\to_{a.s.}\gamma_{Z}$ (52)

definition of $\widehat{\zeta}$ is in 10.

###### Proof.

From Borel-Cantelli [19] lemma, it suffices to show that, for
$\forall\epsilon>0$,

$\Sigma_{n=1}^{\infty}Prob\left(|\widehat{\zeta}-\gamma_{Z}|\geq
4\epsilon\right)<\infty$ (53)

Since the above term is less than

$|\widehat{\zeta}-\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{\widehat{p}(y)}{y}dy|+|\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy|+|\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|$
(54)

According to lemma 28, for sufficiently large $n$,
$|\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy-\gamma_{Z}|<2\epsilon$.
Thus, there exists a constant $n_{0}$ such that

$\displaystyle\Sigma_{n=n_{0}}^{\infty}Prob\left(|\widehat{\zeta}-\gamma_{Z}|\geq
4\epsilon\right)$ (55)
$\displaystyle\leq\Sigma_{n=n_{0}}^{\infty}Prob\left(|\widehat{\zeta}-\frac{1}{P_{Z}(t(n))}\int_{t(n)}^{\infty}\frac{\widehat{p}(y)}{y}dy|\geq\epsilon\right)+Prob\left(\frac{1}{P_{Z}(t(n))}|\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy|\geq\epsilon\right)$

For the second term, notice for arbitrary $k>1$, from 42

$\displaystyle
Prob\left(\frac{1}{P_{Z}(t(n))}|\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy|\geq\epsilon\right)\leq\frac{1}{\epsilon^{k}}\left(\frac{1}{P_{Z}(t(n))}||\int_{t(n)}^{\infty}\frac{\widehat{p}-P_{Z}(y)}{y}dy||_{k}\right)^{k}$
(56) $\displaystyle=O\left(\frac{1}{P_{Z}^{k}(t(n))n^{k/2}}\right)$

Choose $k>\frac{2}{\frac{1}{2}-\frac{\beta}{\gamma_{0}}}$ we prove the
convergence of summation for the second term.

The first term of 55 is less than

$\displaystyle
Prob\left(\frac{\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy}{P_{Z}(t(n))}|\frac{P_{Z}(t(n))\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{\widehat{p}(t(n))}-1|\geq\frac{\epsilon}{2}\right)$ (57)
$\displaystyle+Prob\left(\frac{1}{P_{Z}(t(n))}|\left(\frac{P_{Z}(t(n))\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{\widehat{p}(t(n))}-1\right)\left(\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy\right)|\geq\frac{\epsilon}{2}\right)$

According to lemma 28, For large $n$,
$\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy/P_{Z}(t(n))\leq 2\gamma_{Z}$, so for
sufficiently large $n$,

$\displaystyle
Prob\left(\frac{\int_{t(n)}^{\infty}\frac{P_{Z}(y)}{y}dy}{P_{Z}(t(n))}|\frac{P_{Z}(t(n))\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{\widehat{p}(t(n))}-1|\geq\frac{\epsilon}{2}\right)$ (58)
$\displaystyle\leq
Prob\left(\frac{|P_{Z}(t(n))-\widehat{p}(t(n))|}{s(n)}\geq\frac{\epsilon}{4\gamma_{Z}}\right)+Prob\left(\widehat{p}(t(n))<s(n)\right)$

From 51 we know the convergence of summation on this term.

Also, from mean value inequality,

$\displaystyle
Prob\left(\frac{1}{P_{Z}(t(n))}|\left(\frac{P_{Z}(t(n))\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{\widehat{p}(t(n))}-1\right)\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy|\geq\frac{\epsilon}{2}\right)$
(59) $\displaystyle\leq
Prob\left(\frac{1}{P_{Z}^{2}(t(n))}|\int_{t(n)}^{\infty}\frac{\widehat{p}(y)-P_{Z}(y)}{y}dy|^{2}\geq\frac{\epsilon}{2}\right)$
$\displaystyle+Prob\left(\left(\frac{P_{Z}(t(n))\mathbf{1}_{\widehat{p}(t(n))\geq
s(n)}}{\widehat{p}(t(n))}-1\right)^{2}\geq\epsilon/2\right)$

From 56 and 51 we know the convergence of summation. Thus, 53 is proved and
the almost sure convergence is also proved. ∎

Finally, we prove the almost sure convergence of
$\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq H_{n}}$.

###### Proof for theorem 12.

According to theorem 46 and 6, since $\gamma_{X},\gamma_{Y}>0$, we have

$\frac{\widehat{\zeta}}{\widehat{\rho}}\to_{a.s.}\gamma_{X}$ (60)

Since $H_{n}\to 0$ as $n\to\infty$, according to theorem 46, there exists a
$n_{0}>0$ such that

$\displaystyle\Sigma_{n=n_{0}}^{\infty}\mathbf{1}_{\widehat{\rho}<H_{n}}\leq\Sigma_{n=n_{0}}^{\infty}\mathbf{1}_{|\widehat{\rho}-\frac{\gamma_{Y}}{\gamma_{X}+\gamma_{Y}}|>\frac{\gamma_{Y}}{2(\gamma_{X}+\gamma_{Y})}}<\infty$
(61)

which means that $\mathbf{1}_{\widehat{\rho}\geq H_{n}}\to_{a.s.}1$. Thus, the
product of these two terms converges to $\gamma_{X}$. ∎

##  4 Simulations and numerical examples

In this section, we suppose $X_{i}$ and $Y_{i},\ i=1,2,...n$ obey log gamma
distribution, whose density is

$f_{K}(x)=C_{f_{K}}x^{-\alpha_{K}-1}\log^{\beta_{K}-1}x,\ K=X,Y,\ x\geq 1,\
\alpha_{K}>0$ (62)

We first prove that this distribution satisfies assumption A2.

######  Theorem 7.

Distribution with density 62 satisfies condition A2

###### Proof.

Slow varying part of distribution 62 is

$L_{K}(x)=C_{f_{K}}x^{\alpha_{K}}\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\
dy$ (63)

Since

$\alpha_{K}\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\
dy=x^{-\alpha_{K}}\log^{\beta_{K}-1}x+(\beta_{K}-1)\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-2}y\
dy$ (64)

We have

$\displaystyle|\frac{xL_{K}^{{}^{\prime}}(x)}{L_{K}(x)}|=\frac{|(\alpha_{K}\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\
dy)-x^{-\alpha_{K}}\log^{\beta_{K}-1}x|}{\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\
dy}$ (65)
$\displaystyle=\frac{|\beta_{K}-1|\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-2}y\
dy}{\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\ dy}$

Suppose $h_{K}(x)=\int_{x}^{\infty}y^{-\alpha_{K}-1}\log^{\beta_{K}-1}y\ dy$,
since $h_{K}$ is decreasing and $h_{K}(\infty)=0$, then

$\frac{\int_{x}^{\infty}\frac{-h_{K}^{{}^{\prime}}(y)}{\log y}\
dy}{h_{K}(x)}\leq\frac{1}{\log x}$ (66)

and thus assumption A2 is satisfied with $\kappa=1$. ∎

Figure 1 to -1 demonstrates the performance of estimator
$\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq H_{n}}$
under different conditions. Parameters we choose for simulation is listed in
table 2, sample size is assumed to be 10000 for case 1-5 and 50000 for case 6.
We use relative error

$\delta=\frac{|\frac{\widehat{\zeta}}{\widehat{\rho}}\mathbf{1}_{\widehat{\rho}\geq
H_{n}}-\gamma_{X}|}{\gamma_{X}}$ (67)

to evaluate finite sample performance of our estimator. We perform 50 times
numerical experiments and the error bars in figure 1 to -1 show the maximum,
minimum and average relative error under different $t(n)$. Following
definition 9, $t(n)=n^{\beta}$, $\beta$ coincides with notation Beta in figure
1--1. As we can see,

1\. If tail index of censor time is less than the underlying data, performance
of tail index estimator will be inferior.

2\. Choosing suitable $t(n)$ is critical for making tail index estimator
reliable. Choosing too small or too big $t(n)$ leads to increase of relative
error.

3\. For suitable $t(n)$, tail index estimator has good performance even when
censor rate is high.

Table 2: Parameter for simulation, definition of $\beta_{K},K=X,Y$ see 62 Case | $\gamma_{X}$ | $\gamma_{Y}$ | $\beta_{X}$ | $\beta_{Y}$ | $\gamma_{0}$ | Average censor rate
---|---|---|---|---|---|---
1 | 2.0 | 2.0 | 1.2 | 1.4 | 0.2 | 0.441
2 | 1.0 | 2.0 | 0.5 | 0.5 | 0.3 | 0.392
3 | 1.0 | 2.0 | 1.5 | 1.5 | 0.3 | 0.291
4 | 0.5 | 0.476 | 1.0 | 1.0 | 0.1 | 0.512
5 | 0.5 | 0.4 | 1.0 | 1.0 | 0.1 | 0.555
6 | 0.5 | 0.4 | 1.0 | 1.0 | 0.1 | 0.556

Figure 1: Numerical experiment case 1(left) and 2(right), sample size is
10000, parameters used in these cases coincide with table 2

Figure 0: Numerical experiment case 3(left) and 4(right), sample size is
10000, parameters used in these cases coincide with table 2

Figure -1: Numerical experiment case 5(left) and 6(right), sample size is
10000 for case 5 and 50000 for case 6, parameters used in these cases coincide
with table 2

##  5 Conclusion

In this paper, we focus on proving almost sure convergence and $L_{p}$
convergence of estimator provided by Grama and Spokoiny [1] under random
censoring and condition A1-A3. We also perform numerical experiments with data
satisfying log gamma distribution. Numerical results demonstrate the
usefulness of our tail index estimator when sample size is finite.

## References

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# Multi-Object Grasping – Generating Efficient Robotic Picking and
Transferring Policy

Adheesh Shenoy, Tianze Chen, and Yu Sun The authors are from the Robot
Perception and Action Lab (RPAL) of Computer Science and Engineering
Department, University of South Florida, Tampa, FL 33620, USA. Email: {shenoy,
tianzechen<EMAIL_ADDRESS>Adheesh Shenoy is an undergraduate student.

###### Abstract

Transferring multiple objects between bins is a common task for many
applications. In robotics, a standard approach is to pick up one object and
transfer it at a time. However, grasping and picking up multiple objects and
transferring them together at once is more efficient. This paper presents a
set of novel strategies for efficiently grasping multiple objects in a bin to
transfer them to another. The strategies enable a robotic hand to identify an
optimal ready hand configuration (pre-grasp) and calculate a flexion synergy
based on the desired quantity of objects to be grasped. This paper also
presents an approach that uses the Markov decision process (MDP) to model the
pick-transfer routines when the required quantity is larger than the
capability of a single grasp. Using the MDP model, the proposed approach can
generate an optimal pick-transfer routine that minimizes the number of
transfers, representing efficiency. The proposed approach has been evaluated
in both a simulation environment and on a real robotic system. The results
show the approach reduces the number of transfers by 59% and the number of
lifts by 58% compared to an optimal single object pick-transfer solution.

##  I Introduction

Transferring multiple objects from one bin to another can be considered a
menial task for humans. Using the sense of touch and experience, we can simply
grasp multiple objects from a pile and move them to a bin. We face these types
of tasks in multiple situations. When we cook, based on the recipe, we grasp
and transfer multiple cloves of garlic into a pot. In logistics, workers are
expected to transfer objects such as bulbs from a pile to fill 10-pack or
5-pack bins. In all these tasks, we grasp several objects at a time and
transfer them together because it is more efficient than grasping and
transferring an object one at a time.

However, we have not observed many works on multiple-object grasping (MOG) or
multiple-object bin-picking in robotics. A significant amount of works have
been focusing on single object bin-picking, pick-and-place, and grasping for
manipulations. In traditional single object picking/grasping, an object’s pose
would be estimated using a vision system [1, 2, 3, 4] to guide a robotic
hand/gripper. Since humans have demonstrated outstanding grasping skills,
several approaches extract grasping strategies and use them to reduce the
complexity of grasp planning [5, 6, 7, 8]. More recently, deep-learning-based
approaches use large labeled datasets and deep neural networks to directly
find good grasp points from dense 3D point clouds [9, 10, 11, 12]. A
comprehensive review can be found in [13].

Only a limited amount of work on grasping multiple objects has been carried
out for static grasp stability analysis. [14] discusses the enveloping grasp
of multiple objects under rolling contacts and [15] studied force closure of
multiple objects. It builds the theoretical basis for later work on active
force closure analysis for the manipulation of multiple objects in [15]. [16,
17, 18] try to achieve stably grasping of multiple objects through force-
closure-based strategies. In their studies, the target objects are already in
the air and traditional grasp quality measures were used to analyze the grasps
[19, 20, 21, 22]. Our recent work [23] has studied the tactile sensing aspect
of MOG and developed a deep learning approach to estimate the object quantity
in the grasp.

There are several technical challenges in MOG. Firstly, estimating the object
quantity and their pose in a bin is very challenging. Occlusion among objects
of similar color and texture makes a computer-vision-based approach prone to
error. For example, in the 2019-2020 IROS Robotic Grasping and Manipulation
Competition (RGMC) [24], all teams failed at picking ice cubes from an ice
bucket. Secondly, The displacement of objects within the bin in contact with
the hand voids previously estimated poses. When the hand makes contact with
the object pile in the bin, it displaces the objects. If we only rely on a
computer vision system, the eye of the hand-eye system can no longer view the
majority of the hand. Therefore, it can no longer update the estimated pose of
the objects to be grasped.

A robot will have to use tactile sensors and torque sensors when grasping
objects from a bin. Tactile sensing has widely been recognized as a critical
perception component in object grasping and manipulation. It has been used
with vision sensors to estimate the location of an object relative in a world
coordinate system [25, 26], embedded force sensors on a robotic hand [27], and
a 6-axis force/torques sensor on a robot’s wrist [28, 29, 30, 31, 32]. In
those works, tactile/force sensors are used to reduce the uncertainty in the
perception of the vision system or are only used for single object grasping.
For picking the desired quantity, the robot would need to predict how many
objects will remain in grasp after lifting the hand out of the bin. The robot
needs to make the prediction before lifting the hand so that it can adjust or
simply try again without lifting the hand if the predicted quantity is
different from the desired one.

When the desired quantity is small, MOG once could be sufficient. However, if
the desired quantity is large, one grasp and transfer would not produce it. We
would need to develop an approach that can produce a large quantity MOG policy
to ensure both the precision of the total outcome of several MOGs and
transfers, and the efficiency of the combined MOGs and transfers.

The contribution of this paper includes two novel procedures for transferring
a targeted quantity of objects from a pile into a bin using MOG. It also
introduces several new techniques that are designed for MOG, including a pre-
grasp selection, end-grasp selection, maximum capability grasp selection, and
in-grasp object quantity estimation. Compared to single object transferring,
our approach reduces the total number of transfers between bins by 59% and the
number of lifts from the bin by 58% assuming that the single object grasping
algorithm has a 100% success rate.

##  II Problem Description and Approaches

Figure 1: A multi-object grasping and transferring example. A robot is asked
to transfer 10 tomatoes from the blue bin to the yellow bin.

In this paper, we focus on transferring a large quantity of uniformly shaped
and sized objects from one bin to another. An example setup is illustrated in
Figure 1. The objects in the original bin are randomly piled up. The pick-and-
transfer process for a large quantity can be solved in several different ways.
This paper compares three different approaches. The first approach is the
basic single-object grasping approach to pick and transfer one object at a
time. If the quantity is $N$, it would need at least $N$ times of picking and
transferring.

The second approach is a naive multi-object grasping approach that first
grasps as many objects as possible to quickly reach or get close to the
desired target quantity and then grasps the remaining number. If a robot can
grasp and hold $q$ objects at the most, the robot will first perform
$p=rounddown(\frac{m}{q})$ times of picking and placing $q$ objects and then
grasp the remaining $r=N-p*q$ number of objects. For example, if the demanded
quantity is $10$ and a robotic hand can grasp $4$ objects at the most, the
robot will grasp and transfer $4$ objects twice and then pick and transfer the
remaining $2$ objects. However, in reality, even if a robotic hand can grasp
$4$ objects at the most, it can rarely do it successfully and consistently. It
may need many re-grasps (open and close the hand in the pile) to achieve it.
On the other hand, because of the perception error, a robot may think it holds
$4$ objects and lifts the hand, but it may have grasped $1$ or $2$ or even $0$
objects in the hand. Therefore, the naive multi-object grasping approach may
not be the most efficient approach for those two reasons.

This paper introduces another approach that models the picking and
transferring process as a Markov decision process (MDP) because of the
stochastic feature in grasping multiple objects. As illustrated in Figure 1,
the states are the object quantity in the receiving bin and the actions are
grasping actions for a different number of objects. Since the grasping action
may pick up different quantities with different probabilities, it is an MDP.
The optimization goal is to reach the desired quantity while minimizing the
number of grasps and transfers between bins. The model may generate a policy
that requires the robot to perform a grasp action for any number of objects at
a particular step. We call this approach the _Markov-decision-process-based
multi-object grasping and transferring or MDP-MOGT_.

Figure 2: High quantity grasping and transferring process is modeled with a
MDP. $0$, $1$, … $n$ in the circles are states. They represents the number of
objects in the receiving bin. $g_{i}$’s are MOG actions. Each MOG action $i$
could end up having $0$, $1$, $2$, …, or $q$ objects in the grasp and transit
to state $i$, $i+1$, …, or $i+q$ with $p_{i0}$, $p_{i1}$, …, $p_{iq}$
probabilities respectively.

##  III Markov-decision-process-based multi-object grasping and transferring

###  III-A State space and rewards

The state is the object quantity in the receiving bin. If the receiving bin
has the desired quantity, we reach the goal state. The start state is zero
that represents an empty bin. At a given step, a robot takes an action of MOG
and transfers objects to the receiving bin if there are objects grasped. At
the end of each step after the robot has taken an action, the state of the
robot will change. As illustrated in Figure 2, the states are from $0$ to $n$,
where $n$ is the desired quantity.

The reward is associated with the new state. We define the reward as follows:

$r(s^{{}^{\prime}}|(s,a))=\begin{cases}-(n-s^{{}^{\prime}}),&\text{if
}s^{{}^{\prime}}<n\\\ 100,000&\text{if }s^{{}^{\prime}}=n\\\ -1000,&\text{if
}s^{{}^{\prime}}>n\\\ \end{cases}$ (1)

where $r(s^{{}^{\prime}}|(s,a))$ is the reward of taking action $a$ at state
$s$ resulting in the new state $s^{{}^{\prime}}$.

###  III-B MOGT action space

A MOGT action contains three sub-actions: MOG, lifting, and transferring. Any
sequence of finger flexion and extension could be a grasping action. However,
some sequences can rarely pick up any objects, while a small few can grasp
multiple objects with a high success rate. Therefore, first, we must explore
the grasping action space and identify several grasps that are suitable for
multiple object grasping.

To fully explore the grasp action space, we have developed a stochastic
flexing routine (SFR) in [23] and used it to perform a bias random walk grasp
from several pre-grasps. The pre-grasps set has a dense sample of feasible
hand configurations. In our experiment, we use uniform sampling to obtain
9,000 pre-grasps for a Barrett hand. Figure 3 shows the upper and lower bounds
of the spread angle and finger angle that form the pre-grasp set. For the set,
we choose $20^{\circ}$/step for the spread angle and $3^{\circ}$/step for the
finger base joint. These step sizes were chosen to reduces the sample size,
yet be extensive enough for our search.

Figure 3: (A) Spread is at $0^{\circ}$. (B) Spread is at $360^{\circ}$. (C)
Fingers on the left side are at $30^{\circ}$. (D) Fingers on the left are at
$90^{\circ}$.

####  III-B1 Selecting pre-grasps based on potentials

Not all pre-grasps lead to good MOGs. To pick the best ones, the robot would
perform SFR $10$ times from every pre-grasp in the pre-grasp set. We use their
grasping results to calculate the potential pre-grasp value (PPG).

$PPG(\mathbf{\theta},\bf{O})=\\{p_{0},p_{1},p_{2},...,p_{m}\\},$ (2)

where $\boldsymbol{\theta}$ is the robotic hand’s joint angle vector, $\bf{O}$
is the object’s geometry model, $p_{i}$’s are the probabilities of the pre-
grasp leading to a successful grasp of $i$ objects.

We compare the average $PPG$ of each spread to define the best spread for
grasping the target quantity of objects. We filter the pre-grasps within that
spread based on their success rate for grasping the target quantity of objects
and apply K-mean clustering on the remaining data to obtain clusters of hand
configurations.

We perform the SFR on the centroids of the clusters 100 times and compute the
$PPG$ for the centroids. Finally, we select the best pre-grasp from the
centroids based on its $PPG$ to give us the clustered-probability-based pre-
grasp (CPPG) for each targeted grasp

####  III-B2 Selecting pre-grasps based on expectations

We can also find the pre-grasp that can successfully transfer a large quantity
of objects at once based on the best expectation pre-grasp (BEPG). We define
the BEPG as the pre-grasp that, on average, yields the highest quantity of
objects. The strategy we propose to acquire the BEPG also relies on the pre-
grasp set mentioned in section III-B1.

We use the $PPG$ to calculate the average grasp potential (AGP) or grasp
number expectation for each spread within the pre-grasp set.

$AGP(\mathbf{\theta},\bf{O})=p_{1}+p_{2}*2+...+p_{i}*i+p_{m}*m,$ (3)

where the AGP is the weighted sum of the quantity of grasped objects and the
weights are their probabilities. Therefore, we select the spread with the best
average AGP. We filter the pre-grasps within that spread based on their AGP
and apply K-mean clustering to obtain clusters of hand configuration that have
a high potential for grasping large quantities of objects.

Similar to the strategy in section III-B1, we perform the SFR on the centroids
of the clusters 100 times and select the pre-grasp with the highest AGP value.
This pre-grasp would be the BEPG.

####  III-B3 Selecting pre-grasps based on volume

When transferring a large quantity of objects between bins, humans usually
grasp a handful of objects in each attempt. We try to expand our hands as
large as we can and attempt to grasp the maximum quantity in each transfer. We
call this pre-grasp, the maximum capability pre-grasp (MCPG). We believe that
during the preliminary transfers between bins, the robot hand should grasp as
many objects as possible. One approach for finding this pre-grasp is to
compute the volume of the in-grasp space of the hand. Therefore, we can use
the grasp with the largest volume to get the MCPG for a hand.

We use the strategy mentioned in [23] to compute the volume of every pre-grasp
in the pre-grasp set. We can use the pre-grasp with the largest volume as the
MCPG for the hand.

####  III-B4 Selecting finger flexion synergy

Using SFR, for a single pre-grasp we discovered several distinct end-grasp
types associated with the quantity of objects in the hand. We define the end-
grasp as the hand configuration when the grasping routine has been completed.
Based on our observation, some end-grasps have a higher chance of grasping a
particular quantity of objects compared to others. Therefore, to compare the
end-grasps we define a success rate for the grasp type with

$SRG(k,\bf{O})=\\{s_{0},s_{1},s_{2},...,s_{m}\\},$ (4)

where $k$ is the index of a grasp type, $s_{i}$ is the success rate that the
grasp type $k$ has in grasping $i$ objects. Therefore, a end-grasp type can be
selected based on its $SRG$ values.

Using the data collected, we apply K-mean clustering on the end-grasps to
obtain clusters of hand configurations that are likely to fit the desired
quantity of objects. We use the centroids with the highest neighbors as the
end-grasp. This end-grasp is used to compute the finger flexion synergy.

####  III-B5 Lifting based on prediction

Before the robot lifts the hand, it must sense the objects in the grasp and
predict how many objects will remain in the grasp after lifting. It is primary
for successful MOG and consequently, successful transfer of objects between
bins. It is part of the MOG action because the robot will continuously re-
grasp until the robot senses that the desired number has been reached and it
can lift the hand from the original bin.

Since modeling the physics of the objects within the bin is computationally
expensive, we decided to develop a data-driven deep learning approach to
estimate the quantity of objects within the grasp when the hand is inside the
bin.

We train five classifiers that estimate zero or non-zero objects, two or non-
two objects, three or non-three objects, and greater than or equal to 2
objects or lesser than 2 objects in the hand. The structure of the classifier
is shown in Figure 4. The input data dimension of the model is $113$. The
input data contains the pre-grasp and the current hand configuration ($14$),
the tactile sensor information ($96$), and the readings from the strain gauges
present in the couple joints of the hand ($3$). The activation function used
for the output layer is sigmoid with a single class as the output. The whole
model can be represented as

$n=f(\mathbf{h},\mathbf{t},\mathbf{s}),$ (5)

where $\mathbf{h}$ is a vector representing the hand configuration,
$\mathbf{t}$ is the vector representing the tactile sensor array, and
$\mathbf{s}$ is the vector containing the three strain gauge readings. The
output $n$ is the prediction of whether the grasp would contain the desired
quantity of objects. The shape for the tactile sensor array is rearranged to
account for the spatial information. We represent the tactile sensors in the
palm as a $4$ by $7$ matrix and the tactile sensors in each finger as a $3$ by
$8$ matrix. The result of the prediction models can be found in section IV-D

To reduce the number of false positives when performing the grasping routine,
we use the non-zero model along with either of the models to estimate when to
lift the hand. If the non-zero model estimates non-zero for three consecutive
time steps and the other model estimates True for one time-step we lift the
hand. We define this as the voting algorithm. We use the $\geq 2$ model when
grasping the most quantity of objects and the $1$, $2$, and $3$ object models
when we grasp their respective quantity of objects.

Figure 4: $hand\ configuration$ is a vector that contains the pre-grasp and
the current hand configuration. $torque$ is a vector that contains the strain
gauge readings from each finger. $palm$ represents the tactile sensors on the
palm of the hand and $finger1$ to $finger3$ represents the tactile sensor
array on each finger. $prediction$ is the classification result.

####  III-B6 Transfer sub-action

The last part of the MOGT action is the transfer sub-action. If the number of
objects in the grasp after lifting is desirable, we transfer them to the
destination bin. If it is not, we drop the objects in the origin bin and
repeat the grasping routine. For the sake of this paper, we assume that we can
detect the quantity of objects within the hand once it has been lifted from
the pile.

###  III-C State transition probability

After a MOGT action, a number of objects could be added to the receiving bin.
So the system will transit to a new state with a probability. The actions we
focus on are to grasp a maximum quantity of objects, $1$ object, $2$ objects,
and $3$ objects. We obtain the state transition probability distribution for
each action through data collection.

###  III-D Obtaining and Applying MDP-MOGT Policy

We utilize the value iteration method based on the Bellman equation in
equation 6 to compute the optimal policy when we are at each state of the MDP-
MOGT model. After we have got the policy we will follow the algorithm in
Algorithm 1.

$V_{t+1}^{*}(s)=\max\sum_{s^{{}^{\prime}}}p(s,a,s^{{}^{\prime}})[r(s,a,s^{{}^{\prime}})+\lambda
V_{t}^{*}(s)]$ (6)

Algorithm 1 MDP-MOGT algorithm

1: Initialize $target$, $current$ as 0, $objectsGrasped$ as 0

2: while $current\ \neq\ target$ do

3: Initialize $isLift$ as 0

4: choose pre-grasp Type following policy $\pi$

5: choose corresponding end-grasp Type

6: insert hand and start grasping

7: $isLift$ = output from voting algorithm

8: if $isLift\ =\ 1$ then

9: lift

10: else

11: reset hand to pre-grasp and regrasp

12: end if

13: lift and check number of objectGrasped

14: if $current+objectsGrasped<=target$ then

15: $current+=objectsGrasped$

16: transfer objects to destination bin

17: end if

18: end while

##  IV Experiments and Results

We have evaluated our proposed approach in simulation as well as the real
world using spheres. The robotic hand used in the evaluation is the Barrett
Hand. Based on our data collection, the hand is capable of grasping a maximum
of 5 objects, therefore, we use a target of $10$ objects, to showcase the
advantage of high quantity transferring using MOG.

The two metrics we use for evaluating the entire approach are the number of
transfers made between the two bins and the total number of lifts performed
from the pile. These two metrics highlight the efficiency of the transfer
approach.

###  IV-A System setup

The system for data collection was set up within simulation and the real
world. We used CoppeliaSim as the simulation software. The setup within the
simulation is shown in Figure 5. We attach the Barrett hand to the UR5e robot
arm and place the objects in a bin in front of the robot. The Barrett hand
model provided in CoppeliaSim only contains one tactile sensor on each tactile
sensing region. Therefore, we attached tactile sensors to the region to
replicate the Barrett hand within the real system. The setup for the real
system is also shown in Figure 5.

Figure 5: Setup of our system in CoppeliaSim (left) and real system (right)

We used spheres in both the simulation and the real system for our
experiments. The spheres in both the systems have a radius of $2cm$ and a
weight of $2.7g$.

###  IV-B Data collection

####  IV-B1 Data collection for CPPG and BEPG

To gather the data for computing the pre-grasps, we used the simulation and
the real system.

In the simulation, we repeat each of the $9000$ hand configurations from the
pre-grasp set $10$ times using SFR. For CPPG and BEPG, we use the data as
described in sections III-B1 and III-B2 respectively to get the best pre-
grasps. For computing CPPG, we focus on selecting the pre-grasps for grasping
$1$, $2$, and $3$ objects. For selecting the number of clusters within each
section, we visualize the inertia and distortion of the data and select the
number of clusters where the inertia and distortion are decreasing linearly.
Inertia can be defined as the sum of the squared distances of the data sample
to their closest centroids. Distortion can be defined as the average of the
squared distances from the centroid of each cluster.

For collecting data in the real system, we used the top 3 pre-grasps for the
spheres from the simulation data for both CPPG and BEPG. We then performed SFR
using those pre-grasps $50$ times using the real system to select the best
pre-grasp for each grasp type for the real system.

####  IV-B2 Data collection for the finger flexion synergy

For getting the end-grasp for the pre-grasps within the simulation, we used
the data collected within the simulation for the pre-grasps and use the
approach in section IV-B. We use inertia and distortion to select the clusters
here as well. For getting the end-grasp for the grasps in the real system, we
used the data collected within the real system for each of the real system
pre-grasps and perform the same approach as we do for getting the simulation
end-grasps.

####  IV-B3 Data collection for MDP-MOGT

For the MDP-MOGT approach, we need to possess the state transition
probabilities for each of the pre-grasps in the action space. Therefore, to
get the state transition probabilities for the pre-grasps within the
simulation, we used each pre-grasp and its corresponding end-grasp along with
the estimation model to grasp the spheres $50$ times. This data provides us
with the state transition probabilities for each pre-grasp when performing the
grasping routine with the model and the finger flexion synergy. We also
computed the state transition probabilities for each pre-grasp with SFR. We
performed the same procedure in the real system to collect data and compute
the state transition probabilities for the real system pre-grasps.

###  IV-C Results of grasping maximum quantity of objects

For grasping the maximum quantity of objects, we described two approaches in
section III-B2 and III-B3, BEPG, and MCPG. We performed each grasp type $10$
times using SFR on the spheres in the real system. The results can be found in
table I. The percentages within the table represent the percentage of times
the pre-grasp was able to grasp the corresponding number of spheres. Based on
the results, BEPG significantly outperforms MCPG in yielding more quantity of
objects on average.

TABLE I: best-expectation pre-grasp vs. maximum capability pre-grasp | Number
---
of spheres
| BEPG
---
| MCPG
---
0 | 6% | 80%
1 | 16% | 10%
2 | 16% | 10%
3 | 42% | 0%
4 | 20% | 0%

###  IV-D Results of estimation models

We train a total of $5$ estimation models. The models are for estimating if
the grasp contains at least 1 object (non-zero), $1$ object, $2$ objects, $3$
objects or at least $2$ objects. The precision and the RMSE for each model can
be found in table II. Precision represents the percentage of the correct
predictions among all predictions. RMSE represents the standard deviation of
the prediction error.

For training each model, we use early stopping on the validation loss to
prevent over-fitting. We also use the Adam optimizer with a learning rate of
$0.001$ and use binary cross-entropy as our loss function. We apply SMOTE [33]
on the training data to fix the imbalanced classes. Lastly, we perform
transfer learning using the real-system data to acquire the models for the
real system.

Based on the results, the precision for all the models reduces when performing
transfer learning, except the model trained to estimate $\geq 2$ objects. We
believe the reduction in the precision is because of the noise within the real
system data. The reduction can also be attributed to limited real system data.

TABLE II: Precision and RMSE of estimation models for sphere | Estimation
---
model
| Precision
---
(Simulation)
| Precision
---
(real)
| RMSE
---
(Simulation)
| RMSE
---
(Real)
non-zero | 95.1% | 96.09% | NA | NA
1 | 85.97% | 51.72% | 0.6 | 0.69
2 | 67.18% | 51.46% | 0.97 | 0.87
3 | 40.95% | 38.24% | 1.84 | 0.92
$\geq 2$ | 78.97% | 83.04% | NA | NA

###  IV-E Results of transfer approach

For transferring objects between the bins, we have evaluated two approaches,
the naive transfer approach, and the proposed MDP-MOGT approach. For MDP-MOGT,
we collected data to compute the state transition probability for each pre-
grasp and defined the problem as an MDP to acquire an optimum policy.

For the experiments on both approaches, we used the BEPG as the pre-grasp to
grasp the maximum quantity of objects and the CPPG for grasping the target
quantity of objects. The results for the experiments are shown in table III.

Based on the results, when using the SFR with the pre-grasp in the real
system, MDP-MOGT reduced the number of transfers by 6.38% and the number of
lifts by 9.26% when compared to the naive approach. Similar results were
observed when comparing the two transfer approaches using the grasping routine
with the model and the finger flexion synergy. This showcases the superiority
of the MDP-MOGT approach. Similarly, When comparing the MDP-MOGT results
between the SFR with the pre-grasp and the grasping routine using the pre-
grasps, finger flexion synergy, and the models, the routine with models
outperformed the SFR routine by 6.81% in the number of transfers and 14.28% in
the number of lifts in the real system. Similar results were observed in
simulations as well. This showcases the improvement made because of the models
and the finger flexion synergy.

TABLE III: Results for transfer approaches | Approach
---
| Grasping
---
routine
| Average
---
transfers
| Average
---
lifts
Naive | SFR (Real) | 4.7 | 5.4
MDP-MOGT | SFR (Real) | 4.4 | 4.9
Naive | Estimation models (Real) | 4.2 | 4.9
MDP-MOGT | Estimation models (Real) | 4.1 | 4.2
MDP-MOGT | SFR (Simulation) | 6.0 | 7.6
MDP-MOGT | Estimation models (Simulation) | 5.9 | 6.8

##  V Discussion & Conclusion

This paper proposed several new MOG techniques and two object transferring
approaches that take advantage of MOG. The MOG techniques include clustered-
probability-based pre-grasp, best expectation pre-grasp, maximum capability
pre-grasp, and a data-driven deep learning model to predict the quantity of
objects in a grasp after the hand lifts, when the hand is in the pile. The two
object transferring approaches introduced is the naive transfer approach and
the Markov-decision-process-based multi-object grasping transfer approach. We
have evaluated the proposed strategies for transferring spheres in a
simulation environment and the real system. The experiment results show that
our approach outperforms single object transferring and the naive transfer
approach. We also discovered that a large volume grasp does not translate to
grasping a higher quantity of objects. In the future, we will test our
approach on different shapes of objects and explore other deep neural network
architectures to increase the precision of the models.

## Acknowledgment

This material is based upon work supported by the National Science Foundation
under Grants Nos. 1812933 and 191004.

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