Bhardwaj A. et al. Physical parameters of Cepheid and RR Lyrae stars

17 2021 10.1017/xxxxx

Proceedings IAU Symposium

# Predicting Physical Parameters of Cepheid and RR Lyrae variables in an
Instant with Machine Learning

A. Bhardwaj1 E. P. Bellinger2 S. M. Kanbur3 & M. Marconi1 1INAF-Osservatorio
Astronomico di Capodimonte, Salita Moiariello 16, 80131, Naples, Italy
email<EMAIL_ADDRESS>
2Max Planck Institute for Astrophysics, Karl-Schwarzschild-Straße 1, 85748,
Garching, Germany
3Department of Physics, State University of New York, Oswego, NY 13126, USA

###### Abstract

We present a machine learning method to estimate the physical parameters of
classical pulsating stars such as RR Lyrae and Cepheid variables based on an
automated comparison of their theoretical and observed light curve parameters
at multiple wavelengths. We train artificial neural networks (ANNs) on
theoretical pulsation models to predict the fundamental parameters (mass,
radius, luminosity, and effective temperature) of Cepheid and RR Lyrae stars
based on their period and light-curve parameters. The fundamental parameters
of these stars can be estimated up to 60 percent more accurately when the
light-curve parameters are taken into consideration. This method was applied
to the observations of hundreds of Cepheids and thousands of RR Lyrae in the
Magellanic Clouds to produce catalogs of estimated masses, radii,
luminosities, and other parameters of these stars.

###### keywords:

stars: variable, Cepheid, RR Lyrae - galaxies: Magellanic Clouds - cosmology:
distance scale

††editors: A. Mahabal, C. Fluke & J. Mclver, eds.

##  1 Introduction

Classical Cepheids and RR Lyrae variable stars are well known distance
indicators and are also excellent tracers of young and old age stellar
populations, respectively (Subramanian et al., 2017; Beaton et al., 2018;
Bhardwaj, 2020, 2022). Modern pulsation codes can reproduce observed pulsation
periods, light and radial velocity curves, and peak-to-peak amplitude
variations for these classical pulsating stars at all wavelengths (Bono,
Marconi & Stellingwerf, 2000; Marconi et al., 2013, 2015). In the era of large
variability surveys, a quantitative comparison of the predicted and observed
pulsation properties of Cepheid and RR Lyrae variables provides stringent
constraints on their intrinsic evolutionary parameters, and subsequently,
provides new challenges for the stellar evolution and pulsation theory
(Bhardwaj et al., 2015, 2017; Das et al., 2018). We aim to employ modern
automated methods for comparing observed and predicted light curve structure,
which is defined by its amplitude and phase parameters, of Cepheid and RR
Lyrae variables and generate catalogs of physical parameters of observed
variables in the Galaxy and the Magellanic Clouds.

##  2 Analysis and results

Our theoretical model grid includes 390 models with composition (Y = 0.25 and
Z = 0.008) representative of Cepheids in the Large Magellanic Cloud (LMC)
computed by Marconi et al. (2013), and a total of 270 models representative of
RR Lyrae metallicities ranging from Z = 0.0001 to Z = 0.02 and helium
abundances ranging from Y = 0.245 to Y = 0.27 (Marconi et al., 2015). The
pulsation models provide multiband theoretical light curves for a broad range
of stellar masses, luminosity levels, and chemical composition. The
observational $V-$ and $I-$band light curves were taken from the Optical
Gravitational Lensing Experiment survey (Soszyński et al., 2018). We use ANNs
trained using scikit-learn on the grid of theoretical models to predict
physical parameters of observed stars (see Bellinger et al., 2020, for more
details). The light curve structure importance was independently verified
using random forests.

We evaluate how well the ANN can predict physical parameters based on the
theoretical models and using: 1) period (linear model); 2) period and light
curve structure (machine learning method). The two-fold cross validation
method was used for model assessment and a quantification of improvement was
provided using the standard deviation of the errors and the coefficient of
variation ($R^{2})$ \- a higher $R^{2}$ is better. In Figure 1, a significant
increase in $R^{2}$ and a large reduction in the standard deviation of
residuals is clearly seen when using the machine learning method. Similar
results were also obtained for masses, radii, temperatures, absolute
magnitudes, colors, and Wesenheit magnitudes (Bellinger et al., 2020).

|
---|---
Figure 1: Predicted versus actual luminosities for estimates based on the
period (left) and period and light curve parameters (right). The trends in
errors (left) vanish when the light curve structure is included (right).

##  3 Conclusions

While the period–mean density relation suggests that the observable period is
the most important quantity in constraining the global stellar parameters of
classical pulsating stars, we showed that the light curve structure plays a
statistically significant part in determining these parameters. Once a smooth
grid of pulsation models covering the entire parameter space is available, the
catalogs of physical parameters can be instantly generated for all observed
Cepheid and RR Lyrae stars with unprecedented precision.

## References

  * Beaton et al. (2018) Beaton R. L. et al., 2018, Space Science Reviews, 214, 113
  * Bellinger et al. (2020) Bellinger E. P. et al., 2020, MNRAS, 491, 4752
  * Bhardwaj (2020) Bhardwaj A., 2020, Journal of Astrophysics and Astronomy, 41, 23
  * Bhardwaj (2022) Bhardwaj A., 2022, Universe, 8, 122
  * Bhardwaj et al. (2017) Bhardwaj A. et al., 2017, MNRAS, 466, 2805
  * Bhardwaj et al. (2015) Bhardwaj A. et al., 2015, MNRAS, 447, 3342
  * Bono, Marconi & Stellingwerf (2000) Bono G. et al., 2000, A&A, 360, 245
  * Das et al. (2018) Das S. et al., 2018, MNRAS, 481, 2000
  * Marconi et al. (2015) Marconi M. et al., 2015, ApJ, 808, 50
  * Marconi et al. (2013) Marconi M. et al., 2013, MNRAS, 428, 2185
  * Soszyński et al. (2018) Soszyński I. et al., 2018, Acta Astron., 68, 89
  * Subramanian et al. (2017) Subramanian S. et al., 2017, Space Science Reviews, 212, 1817


we get that for all of those intervals, any subsequence has a convergent
subsequence, any locally Lipschitz function on a compact set is Lipschitz.

For $J_{1}$ we take a convergent subsequence $\\{f_{1,k}\\}_{k\in\mathbb{N}}$,
and for every $i\geq 1$ we define the subsequence
$\\{f_{i+1,k}\\}_{k\in\mathbb{N}}$ to be a subsequence of
$\\{f_{i,k}\\}_{k\in\mathbb{N}}$ that is convergent on $J_{i+1}$, and thus it
converges on $J_{i+1}$ to a continuous function, and agrees with the previous
subsequences on $J_{1},\dots,J_{i}$ respectively.

We get that the subsequence $\left\\{f_{k,k}\right\\}_{k\in\mathbb{N}}$ is
convergent on $\mathcal{X}$, and converges to a continuous function. This
completes the proof for Case (1) of the lemma. We do the same thing for Case
(2), but apply Lemma B.3 on each interval, to get a convergent subsequence
where the limit is differentiable and satisfies the ODE presented by Lemma
B.3. This completes the proof for Case (2) of the lemma. ∎

######  Remark B.5.

Lemmas B.2-B.4 are also true if we apply them to a subsequence
$\left\\{f_{k_{j}}\right\\}$ and each of the conditions need only be true for
every $k_{j}$.

##  Appendix C Alternative Proof for Two Bidders

We will show here how the get a direct proof for the Theorem 3.1 (and by
extension Theorem 3.2 as well).

Proof overview. Consider the functions $Q^{*}_{k}$ and $F^{*}_{k}$ we get from
Theorem A.1 for each $k$. In Section C.1 we will prove that the sequences
$\left\\{Q^{*}_{k}\right\\}$ and $\left\\{F^{*}_{k}\right\\}$ are convergent,
and converge to $Q^{*}$ and $F^{*}$, respectively. This is the most
technically-challenging part of the proof, and is made possible by our toolbox
from Section B. In Section C.2 we will prove that $(Q^{*},F^{*})$ is an
equilibrium: For one side, we will show that any distribution of the bidders
can be approximated well enough using $\mathbf{r}_{k}$ for sufficiently large
$k$. For the other side, we use Claim 3.7, as it only utilize the indifference
of the seller over $[0,\overline{v}]$. We conclude that any other distribution
cannot gain a lower expected revenue when the seller uses $Q^{*}$.

###  C.1 Proof of Convergence

We define sequences $\left\\{Q^{*}_{k}\right\\}$ and
$\left\\{F^{*}_{k}\right\\}$, such that $\left(Q^{*}_{k},F^{*}_{k}\right)$ is
an equilibrium for the discrete setting with parameter $k$ that we get from
Theorem A.1. We show in this section that $\left\\{F^{*}_{k}\right\\}$
converges to $F^{*}$ (Claim C.1) and $\left\\{Q^{*}_{k}\right\\}$ converges to
$Q^{*}$ (Claim C.2). For the convergence of $\left\\{F^{*}_{k}\right\\}$ we
use Lemma B.4; for the convergence of $\left\\{Q^{*}_{k}\right\\}$ we use both
Lemma B.4 and Lemma B.2 as it provides guarantees uniform converges.

######  Claim C.1.

The sequence $\left\\{F^{*}_{k}\right\\}_{k\in\mathbb{N}}$ is uniformly
convergent to $F^{*}$ where

$F^{*}(v)=\begin{cases}0&v\in[0,\underline{v}]\\\
1-\frac{\underline{v}}{v}&v\in[\underline{v},\overline{v})\\\
1&v=\overline{v}\end{cases},$

Where $\underline{v}$ is the solution to
$\underline{v}\left(1+\log\overline{v}-\log\underline{v}\right)=\mu$.

###### Proof.

From Theorem A.1, for each $k>1$ and $1\leq i\leq k-1$ the function
$F^{*}_{k}(v)$ satisfies:

$F^{*}_{k}(r_{k,i+1})=\sqrt{\frac{\overline{v}}{k}\cdot\frac{2}{r_{k,i}+\frac{\overline{v}}{k}}(F^{*}_{k}(r_{k,i})-F^{*}_{k}(r_{k,i})^{2})+F^{*}_{k}(r_{k,i})^{2}};$

So we define a function
$\psi(\delta,r,y):[0,\overline{v}]\times(0,\overline{v}]\times[0,1]\to[0,1]$
by

$\psi(\delta,r,y)=\sqrt{\frac{2\delta}{r+\delta}(y-y^{2})+y^{2}}.$

For each $k>1$ and $1<i\leq k-1$ the function $F^{*}_{k}(v)$ satisfies:

$F^{*}_{k}(r_{k,i+1})=\psi\left(\frac{\overline{v}}{k},r_{k,i},F^{*}_{k}(r_{k,i})\right).$
(22)

For this proof, we would like to change the value of $F^{*}_{k}$ on
$r_{k,k+1}=\overline{v}$ from $1$ to
$\psi\left(\frac{\overline{v}}{k},r_{k,k},F^{*}_{k}(r_{k,k})\right)$, so
property $\eqref{eq:proof:clm:two_bidders:F_k_convergent:psi_induction}$ is
true for $k$ as well. For every $k$, we define a new function $\tilde{F}_{k}$
by $\tilde{F}_{k}(r)=F^{*}_{k}(r)$ for every $r\in[0,\overline{v})$ and
$\tilde{F}(\overline{v})=\psi\left(\frac{\overline{v}}{k},r_{k,k},\tilde{F}(r_{k,k})\right)$.
Now, for every $k>1$ and $1<i\leq k$

$\tilde{F}_{k}(r_{k,i+1})=\psi\left(\frac{\overline{v}}{k},r_{k,i},\tilde{F}_{k}(r_{k,i})\right).$

This will allow us get locally uniform convergence on a compact interval
$[0,\overline{v}]$, which means uniform convergence.

In order to prove that the sequence $\left\\{\tilde{F}_{k}\right\\}$ is
uniformly convergent, we need to analyze $\psi$. We can see is that for every
$y\in[0,1]$ and every $r\in(0,\overline{v}]$

$\psi(0,r,y)=y,$ (23)

$\psi$ is continuous, and properties (23) and (22) fit into Definition B.1’s
conditions (20) and (19) respectively, which mean that
$\left\\{\tilde{F}_{k}\right\\}$ is recursive discrete. In order to prove
convergence, we need to analyze $\psi$ further.

$\frac{\partial}{\partial\delta}\psi(\delta,r,y)=\frac{(y-y^{2})\left(\frac{2}{r+\delta}-\frac{2\delta}{(r+\delta)^{2}}\right)}{2\sqrt{\frac{2\delta}{r+\delta}(y-y^{2})+y^{2}}}=\frac{y(1-y)\left(\frac{1}{r+\delta}-\frac{\delta}{(r+\delta)^{2}}\right)}{\psi(\delta,r,y)}.$

We can see that it is non negative when $\delta\in[0,\overline{v}]$,
$r\in(0,\overline{v}]$, and $y\in(0,1]$. At $y=0$ this equation is not
defined, but $\psi(\delta,r,0)=0$ for every $\delta$ and $r$, which means that
the derivative at $y=0$ is always 0. The derivative at $\delta=0$ and $r,y>0$
is $\frac{1-y}{r}$ which implies that the derivative is not continuous at
$\delta=0$ and $y=0$. However, for any $r>0$ we can get an upper bound by
using $\psi(\delta,r,y)\geq y$

$\frac{\partial}{\partial\delta}\psi(\delta,r,y)\leq\left(1-y\right)\left(\frac{1}{r+\delta}-\frac{\delta}{(r+\delta)^{2}}\right)\leq\frac{1-y}{r}.$

This upper bound implies that $\psi$ is locally Lipschitz.

Since $\left\\{\tilde{F}_{k}\right\\}$ is recursive discrete, and the
corresponding $\psi$ is locally Lipschitz, we can apply Lemma B.4.1 (for
$\mathcal{X}=(0,\overline{v}]$, and $\mathcal{Y}=[0,1]$) and we get that any
subsequence of $\left\\{\tilde{F}_{k}\right\\}$ has a a locally uniform
convergent subsequence on $(0,\overline{v}]$, and each limit is continuous.

We take an arbitrary locally uniform convergent subsequence
$\left\\{\tilde{F}_{k_{j}}\right\\}$, and denote its limit by $\tilde{F}$. For
every $j$ we know that
$\int_{0}^{\overline{v}}(1-\tilde{F}_{k_{j}}(v))dv=\mu$, which implies that
$\int_{0}^{\overline{v}}(1-\tilde{F}(v))dv=\mu$ as well. We denote by
$\underline{v}=\inf_{r\in(0,\overline{v}]}\left\\{\tilde{F}(r)>0\right\\}$.
This is the lowest value that satisfies
$\frac{\partial}{\partial\delta}\psi(\delta,r,\tilde{F}(r))$ is continuous on
$r\in(\underline{v},\overline{v}]$.

Since $\left\\{\tilde{F}_{k_{j}}\right\\}$ is convergent to $\tilde{F}$, and
all of those are non decreasing functions, then on any compact sub-interval of
$(\underline{v},\overline{v}]$ there is a positive minimum value $a$ of
$\tilde{F}$ over that sun-interval, which implies that for any large enough
$j$, $F_{k_{j}}$ is at least $\frac{a}{2}$ on the same compact sub-interval.
Together with the continuous partial derivative, and Remark B.5, we can use
Lemma B.4.2 with the intervals $r\in(\underline{v},\overline{v}]$ and
$y\in(0,1]$, which implies that for any $r\in(\underline{v},\overline{v}]$

$\displaystyle\frac{d}{dr}\tilde{F}(r)$
$\displaystyle=\frac{\partial\psi(\delta,r,y)}{\partial\delta}\left(0,r,\tilde{F}(r)\right)$
$\displaystyle=\frac{1-\tilde{F}(r)}{r}.$

This implies that

$\tilde{F}(r)=1-\frac{c}{r}\quad r\in(\underline{v},\overline{v}]$

for some constant $c$. The function $\tilde{F}$ is defined from
$(0,\overline{v}]$ to $(0,1]$, continuous, and equal to 0 outside of
$(\underline{v},\overline{v}]$, which suggest that

$0=\lim_{r\to\underline{v}^{+}}\tilde{F}(r)=1-\frac{c}{\underline{v}}.$

Therefore $\underline{v}=c$ and we get that

$\tilde{F}(r)=\begin{cases}0&r\in(0,\underline{v}]\\\
1-\frac{\underline{v}}{r}&r\in[\underline{v},\overline{v}]\end{cases}.$

To get $\underline{v}$ we use the equality
$\int_{0}^{\overline{v}}(1-\tilde{F}(r))dv=\mu$, and we get that
$\underline{v}$ is the unique solution to

$\underline{v}\left(1+\log\overline{v}-\log\underline{v}\right)=\mu.$

Since the functions $\tilde{F}_{k}$ are all positive and non decreasing, we
get that $\lim_{k\to\infty}\tilde{F}_{k}(0)=0$ since it must be non negative
and $\tilde{F}_{k}$ is non decreasing; Thus, we can define $\tilde{F}(0)=0$.
The convergence on $[0,\overline{v}]$ is uniform since it is locally uniform
convergence to a continuous function on a compact set.

We have taken an arbitrary limit $\tilde{F}$ and proved that it is a specific
function. Since any subsequence has a uniformly convergent subsequence, the
sequence $\\{\tilde{F}_{k}\\}_{k\in\mathbb{N}}$ is uniformly convergent to
$\tilde{F}$.

Since $\\{\tilde{F}_{k}\\}$ and $\\{F^{*}_{k}\\}$ are identical on
$[0,\overline{v})$, we get uniform convergence there for $\\{F^{*}_{k}\\}$.
Since $F^{*}_{k}(\overline{v})=1$ for every $k$, we get convergence there as
well. And we conclude that $\\{F^{*}_{k}\\}$ convergence uniformly to $F^{*}$
that is defined in the claim’s description. ∎

######  Claim C.2.

The sequence $\left\\{Q^{*}_{k}\right\\}_{k\in\mathbb{N}}$ is convergent to
$Q^{*}$ where

$Q^{*}(r)=\begin{cases}\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{1}{\log\overline{v}-\log\underline{v}}\right)&r\in[0,\underline{v}]\\\
\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{r}{r-\underline{v}}\right)\cdot\left(\frac{\log
r-\log\underline{v}}{\log\overline{v}-\log\underline{v}}\right)&r\in(\underline{v},\overline{v}]\end{cases},$

Where $\underline{v}$ is the solution to
$\underline{v}\left(1+\log\overline{v}-\log\underline{v}\right)=\mu$.

###### Proof.

From Theorem A.1, for each $k>1$ and $2\leq i\leq k-1$ the function
$Q^{*}_{k}(v)$ satisfies:

$Q^{*}_{k}(r_{k,i-1})=Q^{*}_{k}(r_{k,i})+\left[Q^{*}_{k}(r_{k,i})(1-F^{*}_{k}(r_{k,i}))-Q^{*}_{k}(0)(1-F^{*}_{k}(0))\right]\cdot\frac{\overline{v}}{kr_{k,i}F^{*}_{k}(r_{k,i})};$

So we define a function $\psi(\delta,r,y,z,a)$ by

$\psi(\delta,r,y,z,a)=y+\left[y(1-z)-a)\right]\cdot\frac{\delta}{rz},$

and now for each $k>1$ and $2\leq i\leq k-1$ the function $F^{*}_{k}(v)$
satisfies:

$Q^{*}_{k}(r_{k,i-1})=\psi\left(\frac{\overline{v}}{k},r_{k,i},Q^{*}_{k}(r_{k,i}),F^{*}_{k}(r_{k,i}),Q^{*}_{k}(0)(1-F^{*}_{k}(0))\right).$
(24)

This fits property (21), since $\psi$ also satisfies property (20) from the
same definition, and from Claim C.1 $\\{F^{*}_{k}\\}$ is uniformly convergent
to a continuous function $F^{*}$ on $[0,\overline{v})$. We get that the
sequence $\left\\{Q^{*}_{k}\right\\}$ is recursive discrete- on
$\mathcal{I}=[0,\overline{v}],\mathcal{X}=(\underline{v},\overline{v}),\mathcal{Y}=[0,1],\mathcal{Z}=(0,1]$,
and $\mathcal{C}=[0,1]$. We defined $\mathcal{X}$ to be
$(\underline{v},\overline{v})$, because $F^{*}$ is not continuous at
$\overline{v}$, and because $F^{*}(v)=0$ for every $v\in[0,\underline{v}]$, so
$\psi(\cdot,\underline{v},\cdot,F^{*}(v),\cdot)$ is not defined.

Next, we differentiate by $\delta$,

$\displaystyle\frac{\partial}{\partial\delta}\psi(\delta,r,y,z,a)=\frac{y(1-z)-a}{rz},$

which is continuous when $r,z>0$.

For every compact sub-interval $I\subset(\underline{v},\overline{v})$, we can
find a minimum $\underline{z}$ of $F^{*}$ over $I$. Therefore
$\left[\nicefrac{{\underline{z}}}{{2}},1\right]$ is a compact sub-interval of
$\mathcal{Y}$ that contains $F^{*}_{k}(I)$ for every large enough $k$.

According to Lemma B.4.2 any subsequence of $\left\\{Q^{*}_{k}\right\\}$ has a
convergent subsequence. We take such convergent subsequence
$\left\\{Q^{*}_{k_{j}}\right\\}_{j\in\mathbb{N}}$, an it is convergent to a
continuous function $Q^{*}$ on $\mathcal{X}=(\underline{v},\overline{v})$, and
it satisfies

$\displaystyle\frac{dQ^{*}(r)}{dr}$
$\displaystyle=-\frac{\partial\psi\left(\delta,r,y,z,a\right)}{\partial\delta}\left(0,r,Q^{*}(r),F^{*}(r),a\right)$
$\displaystyle=-\frac{Q^{*}(r)(1-F^{*}(r))-a}{r\cdot F^{*}(r)}$
$\displaystyle=\frac{a-Q^{*}(r)\cdot\frac{\underline{v}}{r}}{r-\underline{v}},$

where $a$ is some limit of the sequence
$\left\\{Q^{*}_{k_{j}}(0)(1-F^{*}_{k_{j}}(0))\right\\}_{j\in\mathbb{N}}$. The
family of solutions for this ODE is

$Q^{*}(r)=\frac{r\left(a\log r+c\right)}{r-\underline{v}}\qquad
r\in(\underline{v},\overline{v}),$

For some constant $c$. From Theorem A.1, for every $k$

$Q^{*}_{k}(0)(1-F^{*}_{k}(0))=(k-1)(1-Q^{*}_{k}(r_{k,k-1}))F^{*}_{k}(r_{k,k})+(1-F^{*}_{k}(r_{k,k})).$

Since $Q^{*}_{k}(0)(1-F^{*}_{k}(0))\in[0,1]$ for every $k$, and
$\lim_{k\to\infty}F^{*}_{k}(r_{k,k})=1-\nicefrac{{\underline{v}}}{{\overline{v}}}>0$.
Therefore, we get that $\lim_{k\to\infty}Q^{*}_{k}(r_{k,k-1})=1$.

We can see that $\psi$ is Lipschitz (in $\delta$) for
$r\in[v^{\prime},\overline{v})$ for every $v^{\prime}>\underline{v}$, so from
Lemma B.2, $\\{Q^{*}_{k_{j}}\\}_{j\in\mathbb{N}}$ has a uniformly convergent
subsequence on $[v^{\prime},\overline{v})$ (and not just locally uniform).
Combined with $\lim_{k\to\infty}Q^{*}_{k}(r_{k,k-1})=1$, we get that
$\lim_{r\to\overline{v}}Q^{*}(r)=1$. We can use this to find $c$,

$\displaystyle\lim_{r\to\overline{v}}Q^{*}(r)=1$ $\displaystyle\implies$
$\displaystyle\frac{\overline{v}\left(a\log\overline{v}+c\right)}{\overline{v}-\underline{v}}=1$
$\displaystyle\implies$ $\displaystyle
c=1-\nicefrac{{\underline{v}}}{{\overline{v}}}-a\log\overline{v}.$

So far, we have that

$Q^{*}(r)=\frac{r\left(a(\log
r-\log\overline{v})+1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)}{r-\underline{v}}\qquad
r\in(\underline{v},\overline{v}).$

As a limit of non decreasing function in $\mathcal{Y}=[0,1]$, we get that
$Q^{*}$ is non decreasing in $[0,1]$. Therefore, the limit
$\lim_{r\to\underline{v}^{+}}Q^{*}(r)$ exists, and it is in $[0,1]$. This
implies that

$\lim_{r\to\underline{v}^{+}}r\left(a(\log
r-\log\overline{v})+1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)=0.$

Which means that

$a=\frac{1-\nicefrac{{\underline{v}}}{{\overline{v}}}}{\log\overline{v}-\log\underline{v}},$

and thus, for every $r\in(\underline{v},\overline{v})$

$\displaystyle Q^{*}(r)$
$\displaystyle=\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{r}{r-\underline{v}}\right)\cdot\left(\frac{\log
r-\log\overline{v}}{\log\overline{v}-\log\underline{v}}+1\right)$
$\displaystyle=\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{r}{r-\underline{v}}\right)\cdot\left(\frac{\log
r-\log\underline{v}}{\log\overline{v}-\log\underline{v}}\right).$

Since $Q^{*}_{k}(\overline{v})=1$ for every $k$, then by extending $Q^{*}$ to
$\overline{v}$, and defining

$Q^{*}(\overline{v})=\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{\overline{v}}{\overline{v}-\underline{v}}\right)\cdot\left(\frac{\log\overline{v}-\log\underline{v}}{\log\overline{v}-\log\underline{v}}\right)=1,$

we get convergence on $\overline{v}$ as well. $a$ was defined as some
arbitrary limit of
$\left\\{Q^{*}_{k_{j}}(0)(1-F^{*}_{k_{j}}(0)\right\\}_{j\in\mathbb{N}}$, since
we have proved that
$a=\frac{1-\nicefrac{{\underline{v}}}{{\overline{v}}}}{\log\overline{v}-\log\underline{v}}$,
which is only dependent on $\overline{v}$ and $\underline{v}$, then every
limit is identical, and the subsequence is convergent. Since
$\lim_{k\to\infty}F^{*}_{k}(0)=0$, we get

$\lim_{j\to\infty}Q^{*}_{k_{j}}(0)=\lim_{j\to\infty}Q^{*}_{k_{j}}(0)(1-F^{*}_{k_{j}}(0)=a.$

Likewise, we can see that $lim_{r\to\underline{v}^{+}}Q^{*}(r)=a$. Since
$Q^{*}_{k}$ is non decreasing for every $k$, we conclude that for every
$r\in[0,\underline{v}]$

$\lim_{j\to\infty}Q^{*}_{k_{j}}(r)=a,$

which allows us to extend the definition of $Q^{*}$ to the entire interval
$[0,\overline{v}]$.

We have defined $Q^{*}$ as an arbitrary limit of $\left\\{Q^{*}_{k}\right\\}$,
and proved it is a specific function. Since every subsequence has a convergent
subsequence, we conclude that $\lim_{k\to\infty}Q^{*}_{k}=Q^{*}$ where

$Q^{*}(r)=\begin{cases}\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{1}{\log\overline{v}-\log\underline{v}}\right)&r\in[0,\underline{v}]\\\
\left(1-\nicefrac{{\underline{v}}}{{\overline{v}}}\right)\cdot\left(\frac{r}{r-\underline{v}}\right)\cdot\left(\frac{\log
r-\log\underline{v}}{\log\overline{v}-\log\underline{v}}\right)&r\in(\underline{v},\overline{v}]\end{cases}.$

∎

###  C.2 Proof of Equilibrium in the Limit

While we have proved convergence of the equilibrium points of the discrete
case to $Q^{*}$ and $F^{*}$, it is not trivial that $(Q^{*},F^{*})$ is in
itself an equilibrium of the general case.

We can however prove Claim 3.6 directly using the convergence and correctness
in the discrete case.

Consider random variables $R_{k}\sim Q^{*}_{k}$ for every $k$, and $R\sim
Q^{*}$, where $Q^{*}_{k}$, and $Q^{*}$ are described by Theorem A.1 and Claim
C.2. Since $Q^{*}_{k}\to Q^{*}$, then $R_{k}\to R$ in distribution. In order
to prove Claim 3.6, we need to prove that for any $F\in\mathcal{F}$,

$\mathbb{E}\left[\operatorname*{Rev}_{F}(R)\right]\geq\mathbb{E}\left[\operatorname*{Rev}_{F^{*}}(R)\right].$

######  Claim C.3.

For every distribution function $F$ over $[0,\overline{v}]$,

$\lim_{k\to\infty}\mathbb{E}\left[\operatorname*{Rev}_{F}(R_{k})\right]=\mathbb{E}\left[\operatorname*{Rev}_{F}(R)\right].$

###### Proof.

For a given distribution function $F$, the function $\operatorname*{Rev}_{F}$
is continuous at $0$ and at every point where $F$ is continuous. Since $F$ is
monotone, then the set of all discontinuities is countable. $Q^{*}$ only point
of discontinuity is 0, and therefore $\Pr_{r\sim Q^{*}}(F\text{ is not
continuous at }r)=0$. Now, we can use Portmanteau Theorem which completes the
proof. ∎

Before we prove that $F^{*}$ is the adversary’s best response, we need to
prove a rather intuitive claim.

######  Claim C.4.

For $F^{*}_{k}$ as defined in the discrete case

$\lim_{k\to\infty}\mathbb{E}\left[\operatorname*{Rev}_{F_{k}^{*}}(R_{k})\right]=\mathbb{E}\left[\operatorname*{Rev}_{F^{*}}(R)\right].$

###### Proof.

Since $F^{*}_{k}\to F^{*}$ uniformly, then for any $\varepsilon>0$, for large
enough $k$

$\sup_{v\in[0,\overline{v}]}\left\lvert
F^{*}(v)-F^{*}_{k}(v)\right\rvert<\varepsilon.$

Therefore, for any $r\in[0,\overline{v}]$:

$\displaystyle\left\lvert\operatorname*{Rev}_{F^{*}}(r)-\operatorname*{Rev}_{F^{*}_{k}}(r)\right\rvert$
$\displaystyle=\left\lvert
r\left(F^{*2}_{k}(r)-F^{*2}(r)\right)+\int_{r}^{\overline{v}}\left(1-F^{*}(v)\right)^{2}-\left(1-F^{*}_{k}(v)\right)^{2}dv\right\rvert$
$\displaystyle\leq 2\overline{v}\left\lvert
F^{*}_{k}(r)-F^{*}(r)\right\rvert+2\int_{r}^{\overline{v}}\left\lvert
F^{*}_{k}(v)-F^{*}(v)\right\rvert dv$ $\displaystyle\leq
4\overline{v}\varepsilon.$

From Claim C.3, we get that:

$\lim_{k\to\infty}E\left[\operatorname*{Rev}_{F^{*}}(R_{k})\right]=E\left[\operatorname*{Rev}_{F^{*}}(R)\right].$

Therefore, for large enough $k$:

$\displaystyle\left\lvert
E\left[\operatorname*{Rev}_{F^{*}}(R)\right]-E\left[\operatorname*{Rev}_{F_{k}^{*}}(R_{k})\right]\right\rvert\leq$
$\displaystyle\left\lvert
E\left[\operatorname*{Rev}_{F^{*}}(R)\right]-E\left[\operatorname*{Rev}_{F^{*}}(R_{k})\right]\right\rvert$
$\displaystyle+\left\lvert
E\left[\operatorname*{Rev}_{F^{*}}(R_{k})\right]-E\left[\operatorname*{Rev}_{F_{k}^{*}}(R_{k})\right]\right\rvert$
$\displaystyle\leq$ $\displaystyle\varepsilon+4\overline{v}\varepsilon.$

∎

Now, we can check more general distributions, and prove that for every $F$,

$\mathbb{E}\left[\operatorname*{Rev}_{F^{*}}(R)\right]\geq\mathbb{E}\left[\operatorname*{Rev}_{F}(R)\right].$

######  Claim C.5.

For every $F\in\mathcal{F}$ there exists a sequence $\\{F_{k}\\}$ where
$F_{k}\in\mathcal{F}_{k}$ for every $k$, and for every $\varepsilon>0$, for
sufficiently large $k$,

$\left\lvert\mathbb{E}\left[\operatorname*{Rev}_{F}(R_{k})\right]-\mathbb{E}\left[\operatorname*{Rev}_{F_{k}}(R_{k})\right]\right\rvert<\varepsilon.$

###### Proof.

Fix $F\in\mathcal{F}$ and $\varepsilon\in(0,1)$. For any $k$ we define
$F_{k}\in\mathcal{F}_{k}$ by
$F_{k}(r_{k,i})=\frac{k}{\overline{v}}\cdot\int_{r_{k,i}}^{r_{k,i+1}}F(v)dv$
for every $1\leq i\leq k$. We immediately get that $F(r_{k,i})\leq
F_{k}(v)\leq F(r_{k,i+1})$ for every $1\leq i\leq k$ and
$v\in[r_{k,i},r_{k,i+1})$, and thus

$\left\lvert F(v)-F_{k}(v)\right\rvert\leq F(r_{k,i+1})-F(r_{k,i}).$

Denote by $I_{\varepsilon,k}=\left\\{i\in\\{1,\dots,k\\}\mid
F\left(r_{k,i+1}\right)-F(r_{k,i})>\varepsilon\right\\}$. It is easy to see
that $\left\lvert I_{\varepsilon,k}\right\rvert<\frac{1}{\varepsilon}$.

For every reserve price $r$ and $k>\frac{\overline{v}}{\varepsilon^{2}}$, we
can get an upper bound to the difference in the revenue between $F$ and
$F_{k}$,

$\displaystyle\left\lvert\operatorname*{Rev}_{F}(r)-\operatorname*{Rev}_{F_{k}}(r)\right\rvert$
$\displaystyle=\left\lvert
r\left(F^{2}_{k}(r)-F^{2}(r)\right)+\int_{r}^{\overline{v}}\left(1-F(v)\right)^{2}-\left(1-F_{k}(v)\right)^{2}dv\right\rvert$
$\displaystyle\leq 2r\left\lvert
F_{k}(r)-F(r)\right\rvert+2\int_{r}^{\overline{v}}\left\lvert
F_{k}(v)-F(v)\right\rvert dv$ $\displaystyle\leq 2r\left\lvert
F_{k}(r)-F(r)\right\rvert+2\overline{v}\varepsilon+2\sum_{r_{k,i}\in
I_{\varepsilon,k}}\int_{r_{k,i}}^{r_{k,i+1}}\left\lvert
F_{k}(v)-F(v)\right\rvert dv$ $\displaystyle\leq 2r\left\lvert
F_{k}(r)-F(r)\right\rvert+2\overline{v}\varepsilon+\frac{2\overline{v}}{k\varepsilon}$
$\displaystyle\leq 2r\left\lvert
F_{k}(r)-F(r)\right\rvert+4\overline{v}\varepsilon.$

Since $Q^{*}_{k}$ is uniformly convergent to $Q^{*}$ on
$[\underline{v}(1+4\varepsilon),\overline{v}]$ (Lemma B.2 on
$[\underline{v}(1+4\varepsilon),\overline{v})$ and adding another point
$\overline{v}$ to the domain), which is Lipschitz, we can get an upper bound
for $\Pr(R_{k}=r_{k,i})$, such that for large enough $k$ and any
$r_{k,i}\in[\underline{v}(1+4\varepsilon),\overline{v}]$

$\Pr(R_{k}=r_{k,i})<\varepsilon^{2}.$

Also

$\sum_{r_{k,i}\in\mathbf{r}_{k}\cap[0,\underline{v}\left(1+4\varepsilon\right)]}\Pr(R_{k}=r_{k,i})=F_{k}^{*}\left(\underline{v}\left(1+4\varepsilon\right)\right)\leq
1-\frac{1}{1+4\varepsilon}+\varepsilon<5\varepsilon.$

Therefore

$\displaystyle\mathbb{E}\left[\left\lvert
F_{k}(R_{k})-F(R_{k})\right\rvert\right]$
$\displaystyle=\sum_{i=1}^{k}\Pr(R_{k}=r_{k,i})\left\lvert
F_{k}(r_{k,i})-F(r_{k,i})\right\rvert$
$\displaystyle<5\varepsilon+\sum_{i=1}^{k}\Pr(R_{k}=r_{k,i})\varepsilon+\varepsilon^{2}\sum_{i\in
I_{\varepsilon,k}}\left\lvert F_{k}(r_{k,i})-F(r_{k,i})\right\rvert$
$\displaystyle<5\varepsilon+\varepsilon+\varepsilon^{2}\cdot\frac{1}{\varepsilon}$
$\displaystyle=7\varepsilon.$

Putting it all together, for large enough $k$

$\displaystyle\left\lvert\mathbb{E}\left[\operatorname*{Rev}_{F}(R_{k})\right]-\mathbb{E}\left[\operatorname*{Rev}_{F_{k}}(R_{k})\right]\right\rvert$
$\displaystyle\leq\mathbb{E}\left[\left\lvert\operatorname*{Rev}_{F}(R_{k})-\operatorname*{Rev}_{F_{k}}(R_{k})\right\rvert\right]$
$\displaystyle<4\overline{v}\varepsilon+2\overline{v}\mathbb{E}\left[\left\lvert
F_{k}(R_{k})-F(R_{k})\right\rvert\right]$
$\displaystyle<18\overline{v}\varepsilon.$

∎

######  Corollary C.6.

For any $F\in\mathcal{F}$ there is a sequence $\\{F_{k}\\}$ where
$F_{k}\in\mathcal{F}_{k}$ for every $k$ such that

$\lim_{k\to\infty}\mathbb{E}[\operatorname*{Rev}_{F_{k}}(R_{k})]=\mathbb{E}[\operatorname*{Rev}_{F}(R)].$

This is easy to see from Claims C.3 and C.5.

######  Corollary C.7.

For any $F\in\mathcal{F}$,

$\mathbb{E}[\operatorname*{Rev}_{F}(R)]\geq\mathbb{E}[\operatorname*{Rev}_{F^{*}}(R)].$

###### Proof.

Fix $F\in\mathcal{F}$. We use the sequence $\\{F_{k}\\}$ from Corollary C.6.
Since $F^{*}_{k}$ is the adversary’s optimal response to $Q_{k}^{*}$ then
$\mathbb{E}[\operatorname*{Rev}_{F_{k}}(R_{k})]\geq\mathbb{E}[\operatorname*{Rev}_{F^{*}_{k}}(R_{k})$
for any $k$. And thus

$\mathbb{E}[\operatorname*{Rev}_{F}(R)]=\lim_{k\to\infty}\mathbb{E}[\operatorname*{Rev}_{F_{k}}(R_{k})]\geq\lim_{k\to\infty}\mathbb{E}[\operatorname*{Rev}_{F_{k}^{*}}(R_{k})]=\mathbb{E}[\operatorname*{Rev}_{F^{*}}(R)].$

∎

This Completes the direct proof for Claim 3.6. The rest of the proof is
identical to Section 3.4 (from Claim 3.7).

##  Appendix D Missing Proofs from Section 3

### Proof of Claim 3.4

###### Proof.

For every $1\leq i\leq k-1$

$\displaystyle\frac{\partial}{\partial
q_{i}}\mathcal{L}_{k}(\mathbf{x},\mathbf{q})$
$\displaystyle=r_{k,i}(1-x_{i}^{2})+\frac{\overline{v}}{k}\left(1-x_{i}\right)^{2}-r_{k,i+1}(1-x_{i+1}^{2})$
$\displaystyle=r_{k,i+1}x_{i+1}^{2}-\frac{2\overline{v}}{k}\cdot
x_{i}+(r_{k,i+1}-2r_{i})x_{i}^{2}.$

Since $x_{i}\geq 0$ for all $1\leq i\leq k-1$

$\displaystyle\frac{\partial}{\partial
q_{i}}\operatorname*{Rev}(\mathbf{x},\mathbf{q})$ $\displaystyle=0$
$\displaystyle\iff x_{i+1}$
$\displaystyle=\sqrt{\frac{\frac{2\overline{v}}{k}\cdot
x_{i}+\left(r_{i}-\frac{\overline{v}}{k}\right)x_{i}^{2}}{r_{k,i}+\frac{\overline{v}}{k}}}$
$\displaystyle=\sqrt{\frac{\overline{v}}{k}\cdot\frac{2}{r_{k,i}+\nicefrac{{\overline{v}}}{{k}}}(x_{i}-x_{i}^{2})+x_{i}^{2}}.$

∎

### Proof of Claim 3.5

###### Proof.

Differentiating $\mathcal{L}_{k}(\mathbf{q},\mathbf{x})$ by $x_{i}$ is

$\frac{\partial}{\partial
x_{i}}\mathcal{L}_{k}(\mathbf{q},\mathbf{x})=-2(q_{i}-q_{i-1})r_{k,i}x_{i}-2\nicefrac{{\overline{v}}}{{k}}\cdot
q_{i}\left(1-x_{i}\right)+\lambda\cdot\nicefrac{{\overline{v}}}{{k}}.$

Since $r_{k,i},x_{i}>0$ for every $2\leq i\leq k$, we can easily get

$\frac{\partial}{\partial x_{i}}\mathcal{L}_{k}(\mathbf{q},\mathbf{x})=0\iff
q_{i-1}=q_{i}-\frac{\nicefrac{{\lambda}}{{2}}-q_{i}\left(1-x_{i}\right)}{r_{k,i}x_{i}}\cdot\nicefrac{{\overline{v}}}{{k}}.$

Since $r_{k,1}=0$, we get

$\frac{\partial}{\partial
x_{1}}\mathcal{L}_{k}(\mathbf{q},\mathbf{x})=-2\nicefrac{{\overline{v}}}{{k}}\cdot
q_{1}\left(1-x_{1}\right)+\lambda\cdot\nicefrac{{\overline{v}}}{{k}},$

which means that

$\frac{\partial}{\partial x_{1}}\mathcal{L}_{k}(\mathbf{q},\mathbf{x})=0\iff
q_{1}=\frac{\lambda}{2(1-x_{1})}.$

∎

##  Appendix E Missing Proofs from Section 4

### Proof of Claim 4.2

###### Proof.

For a fixed price $r$ the expected revenue is given by

$\displaystyle\operatorname*{Rev}_{F}(r)$ $\displaystyle=E\left[r\cdot
1_{v^{\left(2\right)}\leq
r<v^{\left(1\right)}}\right]+E\left[v^{\left(2\right)}\cdot
1_{r<v^{\left(2\right)}}\right]$
$\displaystyle=r\left(1-F^{n}\left(r\right)\right)+\int_{r}^{\overline{v}}\left(1-nF^{n-1}\left(v\right)+\left(n-1\right)F^{n}\left(v\right)\right)dv.$

When the seller randomizes a reserve price using distribution
$Q\in\mathcal{Q}$ that is continuous over $(0,\overline{v}]$ and
differentiable almost everywhere, the expected revenue would be

$\displaystyle\operatorname*{Rev}_{F}(Q)=$
$\displaystyle\int_{-\infty}^{\overline{v}}\left(r\left(1-F^{n}\left(r\right)\right)+\int_{r}^{\overline{v}}\left(1-nF^{n-1}\left(v\right)+\left(n-1\right)F^{n}\left(v\right)\right)dv\right)dQ\left(r\right).$
$\displaystyle=~{}$ $\displaystyle
Q(0)\int_{0}^{\overline{v}}\left(1-nF^{n-1}\left(v\right)+\left(n-1\right)F^{n}\left(v\right)\right)dv$
$\displaystyle+\int_{0}^{\overline{v}}Q^{\prime}(r)\left(r\left(1-F^{n}\left(r\right)\right)+\int_{r}^{\overline{v}}\left(1-nF^{n-1}\left(v\right)+\left(n-1\right)F^{n}\left(v\right)\right)dv\right)dr$
$\displaystyle=$
$\displaystyle\int_{0}^{\overline{v}}\left(Q^{\prime}\left(v\right)v\left(1-F^{n}\left(v\right)\right)+Q\left(v\right)\left(1-nF^{n-1}\left(v\right)+\left(n-1\right)F^{n}\left(v\right)\right)\right)dv.$

∎

###  E.1 Details for Proof of Claim 4.9

#### Second derivative is positive.

We substitute
$Q^{*}(v)=\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}\right)$,
and $z=1-\nicefrac{{a}}{{v}}$, in Eq (18), we get

$\displaystyle\left(\left(n-1\right)Q^{*}\left(v\right)-v\frac{dQ^{*}\left(v\right)}{dv}\right)z-\left(n-2\right)Q^{*}\left(v\right)$
$\displaystyle=$
$\displaystyle\left(n-1\right)\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}\right)\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle+v\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v\left(v-a\right)\left(n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}\right)\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle-\left(n-2\right)\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}\right)$
$\displaystyle=$
$\displaystyle\left(n-1\right)\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-2}\left(\frac{n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}\right)$
$\displaystyle+\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v\left(n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}\right)$
$\displaystyle-\left(n-2\right)\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}\right).$

We divide by
$\frac{\left(1-\nicefrac{{a}}{{\overline{v}}}\right)^{n-1}\left(\frac{v}{v-a}\right)^{n-1}}{n-1-\frac{1}{n-1}+\log\nicefrac{{\overline{v}}}{{\underline{v}}}}>0$
and get

$\displaystyle\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)$
$\displaystyle+\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v}$
$\displaystyle-\left(n-2\right)\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)$
$\displaystyle=$
$\displaystyle\left(\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)-\left(n-2\right)\right)\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)$
$\displaystyle+\left(n-1\right)\nicefrac{{a}}{{v}}\cdot\log\nicefrac{{v}}{{\underline{v}}}-\left(1-\nicefrac{{\underline{v}}}{{v}}\right)$
$\displaystyle=$
$\displaystyle\left(1-\left(n-1\right)\nicefrac{{a}}{{v}}\right)\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)+\left(n-1\right)\nicefrac{{a}}{{v}}\cdot\log\nicefrac{{v}}{{\underline{v}}}-1+\nicefrac{{\underline{v}}}{{v}}$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)-\left(n-1\right)\nicefrac{{a}}{{v}}\cdot\left(n-1-\frac{1}{n-1}\right)-1+\nicefrac{{\underline{v}}}{{v}}$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)-\nicefrac{{a}}{{v}}\cdot\left(\left(n-1\right)^{2}-1\right)-1+\nicefrac{{\underline{v}}}{{v}}$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)-\left(1-\nicefrac{{a}}{{v}}\right)-\nicefrac{{a}}{{v}}\cdot\left(n-1\right)^{2}+\nicefrac{{\underline{v}}}{{v}}.$

We substitute $a=\frac{\underline{v}}{(n-1)^{2}}$ (Eq (15)) and we get

$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)-\left(1-\frac{\underline{v}}{v\left(n-1\right)^{2}}\right)>\log\nicefrac{{v}}{{\underline{v}}}-\left(1-\nicefrac{{\underline{v}}}{{v}}\right)\geq
0.$

#### Details for $h_{v}\left(1-\nicefrac{{a}}{{v}}\right)\leq h_{v}(0)$.

First we express $\tilde{h}_{v}\left(0\right)$ by

$\displaystyle\tilde{h}_{v}\left(0\right)=$
$\displaystyle\left(\frac{v}{v-a}\right)^{n-1}\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)$
$\displaystyle-\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v-a}\right)-n.$

Next we express $\tilde{h}_{v}\left(1-\nicefrac{{a}}{{v}}\right)$ by

$\displaystyle\tilde{h}_{v}\left(1-\nicefrac{{a}}{{v}}\right)=$
$\displaystyle\left(\frac{v}{v-a}\right)^{n-1}\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)\left(1-n\left(1-\nicefrac{{a}}{{v}}\right)^{n-1}+\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)^{n}\right)$
$\displaystyle-\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v-a}\right)\left(1-\left(1-\nicefrac{{a}}{{v}}\right)^{n}\right)-n\cdot\nicefrac{{a}}{{v}}.$

Now we can calculate $h_{v}\left(1-\nicefrac{{a}}{{v}}\right)$.

$\displaystyle\tilde{h}_{v}\left(0\right)-\tilde{h}_{v}\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle=$
$\displaystyle\left(\frac{v}{v-a}\right)^{n-1}\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)\left(n\left(1-\nicefrac{{a}}{{v}}\right)^{n-1}-\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)^{n}\right)$
$\displaystyle-\left(\frac{v}{v-a}\right)^{n-1}\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v-a}\right)\left(1-\nicefrac{{a}}{{v}}\right)^{n}-n\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)\left(n-\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)\right)$
$\displaystyle-\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}+\underline{v}-v}{v-a}\right)\left(\frac{v-a}{v}\right)-n\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)\left(n-\left(n-1\right)\left(1-\nicefrac{{a}}{{v}}\right)\right)$
$\displaystyle-\left(\frac{\left(n-1\right)a\log\nicefrac{{v}}{{\underline{v}}}}{v}-\left(1-\nicefrac{{\underline{v}}}{{v}}\right)\right)-n\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle=$
$\displaystyle\left(n-1-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}\right)\left(1+\left(n-1\right)\nicefrac{{a}}{{v}}\right)$
$\displaystyle-\left(n-1\right)\nicefrac{{a}}{{v}}\cdot\log\nicefrac{{v}}{{\underline{v}}}+\left(1-\nicefrac{{\underline{v}}}{{v}}\right)-n\left(1-\nicefrac{{a}}{{v}}\right)$
$\displaystyle=$
$\displaystyle-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}+\left(1-\frac{1}{\left(n-1\right)^{2}}\right)\left(n-1\right)^{2}\nicefrac{{a}}{{v}}-\nicefrac{{\underline{v}}}{{v}}+n\cdot\nicefrac{{a}}{{v}}.$

We substitute $a=\frac{\underline{v}}{(n-1)^{2}}$ (Eq (15)) and we get

$\displaystyle-\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}+\left(1-\frac{1}{\left(n-1\right)^{2}}\right)\nicefrac{{\underline{v}}}{{v}}-\nicefrac{{\underline{v}}}{{v}}+\frac{n}{\left(n-1\right)^{2}}\nicefrac{{\underline{v}}}{{v}}$
$\displaystyle=$
$\displaystyle\frac{1}{n-1}+\log\nicefrac{{v}}{{\underline{v}}}+\frac{1}{n-1}\nicefrac{{\underline{v}}}{{v}}$
$\displaystyle=$
$\displaystyle\log\nicefrac{{v}}{{\underline{v}}}-\frac{1-\nicefrac{{\underline{v}}}{{v}}}{n-1}$
$\displaystyle\geq$ $\displaystyle 0.$

##  Appendix F Revisiting the Single bidder Case through Discretization

In this section we solve a ”simpler” setting (in a sense made precise below)
where there is only one bidder. This case is also solved by Carrasco et al.
[2018]. We show how we can use discretization to find the equilibrium, both
when the seller knows the bidder’s expected value and an upper bound on the
values (Section F.1), and when the seller knows the bidder’s expected value
and second moment (Section F.3). In order find the equilibrium for the latter,
we first solve a case where an upper bound is also known (Section F.2). For
this problem, we find that the seller has a certain freedom in the choice of
the pricing distribution, which is expressed by a family of distributions, all
of which yield the same expected revenue. We are able to identify a specific
member of this family, which does not rely on the upper bound on the
valuations to guarantee the same expected revenue. This solves the case where
the seller only knows the first two moments, and not an upper bound.

Our starting point is the expression for expected revenue. For one bidder with
a valuation sampled from a distribution $F$, the expected revenue of a seller
who sells the item for a price of $r$, is

$\operatorname*{Rev}_{F}(r)=r\left(1-F(r)\right).$

When the seller randomizes a price $r$, using distribution $Q$, the expected
revenue is

$\operatorname*{Rev}_{F}(Q)=E_{r\sim
R}[\operatorname*{Rev}_{F}(r)]=\int_{0}^{\infty}r\left(1-F(r)\right)dQ(r).$
(25)

Since the expected revenue function is linear in terms of both the seller’s
distribution and the adversary’s distribution, it is expected that at
equilibrium, both the seller and the adversary are indifferent over the
support (see discussion in Section 1 on indifference). This makes the single
bidder setting easier to analyze.

###  F.1 Single bidder: Upper Bound and First Moment Constraint

We assume the adversary can only choose distributions with a mean value of
$\mu$, and their support is contained in $[0,\overline{v}]$. Let
$\mathcal{F}_{\mu}^{\leq\overline{v}}$ be the set of distributions which have
expectation $\mu$, and support within $[0,\overline{v}]$. For
$F\in\mathcal{F}_{\mu}^{\leq\overline{v}}$ and $Q\in\mathcal{Q}$, the expected
revenue The problem

$\max_{Q\in\mathcal{Q}}\min_{F\in\mathcal{F}_{\mu}^{\leq\overline{v}}}\int_{0}^{\infty}r\left(1-F(r)\right)dQ(r).$
(26)

####  F.1.1 Discrete Case

For $k\geq 1$, we limit the possible prices and valuations to $k+1$ values
$\mathbf{r}_{k}=\left(r_{k,1},\dots,r_{k,k+1}\right)$, such that
$r_{k,k+1}=\overline{v}$. When the context is clear, we will write $r_{i}$
instead of $r_{k,i}$. Unlike the case for two bidders, in this case, using a
price of zero, will always yield an expected revenue of zero, so we assume
$r_{k,1}=\underline{v}>0$, for some value $0<\underline{v}<\mu$ that we will
determine when we solve for the general, non discrete case, to make sure is
would be a tight lower bound for the support of the distribution the adversary
would choose in that general case.

For every $k\geq 1$ and $1\leq i\leq k+1$, we assign
$r_{k,i}=\alpha^{i-1}\underline{v}$ where
$\alpha=\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)^{\nicefrac{{1}}{{k}}}$.101010A
simple analysis with a general vector of possible values shows that when the
ratio between consecutive prices is identical, we get simpler expressions.
This was not the case for two bidders, which is immediately clear since there
a price of zero is used by the seller. Given distributions $Q_{k},F_{k}$ over
$\mathbf{r}_{k}$, where $F_{k}$ has mean of $\mu$, we define the vectors
$\mathbf{q}_{k}=\left(q_{1},\dots,q_{k+1}\right)$ and
$\mathbf{x}_{k}=\left(x_{1},\dots,x_{k+1}\right)$ such that
$q_{i}=Q_{k}(r_{i})$ and $x_{i}=F_{k}(r_{i})$. We also define $q_{0}=0$.

The expected revenue of a mechanism where the seller randomizes a price using
$Q_{k}$ and the bidder’s valuation is sampled from $F_{k}$ is

$\operatorname*{Rev}(\mathbf{q}_{k},\mathbf{x}_{k})=\sum_{i=1}^{k}(q_{i}-q_{i-1})r_{i}(1-x_{i}).$

The mean value of $F_{k}$ is

$\displaystyle\int_{0}^{\overline{v}}(1-F(v))dv$
$\displaystyle=r_{1}+\sum_{i=1}^{k}(r_{i+1}-r_{i})(1-x_{i})$
$\displaystyle=\underline{v}+(\alpha-1)\sum_{i=1}^{k}r_{i}(1-x_{i}).$

Therefore the problem we solve for each $k$ is

$\displaystyle\max_{\mathbf{q}_{k}}\min_{\mathbf{x}_{k}}\ \ $
$\displaystyle\sum_{i=1}^{k}(q_{i}-q_{i-1})r_{i}(1-x_{i})$ (27) $\displaystyle
s.t\ \ $ $\displaystyle q_{1}\leq\dots q_{k}=q_{k+1}=1$ (28) $\displaystyle
x_{1}\leq\dots\leq x_{k+1}=1$ (29)
$\displaystyle(\alpha-1)\sum_{i=1}^{k}r_{i}(1-x_{i})=\mu-\underline{v}$ (30)

We note that we force $q_{k}=q_{k+1}=1$ since the seller could never seller
the item with a price of $r_{k+1}$. We allow values lower than zero in this
case, but we will make sure that when we take the limit to get the general
case, the image of the distribution function would be contained in $[0,1]$. We
do not necessarily get negative values, but we do not prove otherwise. The
Lagrangian we get for this problem is

$\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1})=\operatorname*{Rev}(\mathbf{q}_{k},\mathbf{x}_{k})-\lambda_{1}(\alpha-1)\sum_{i=1}^{k}r_{i}(1-x_{i}).$

We will ignore the inequalities constraints in
$\eqref{single:discrete:maxmin:q_nondec}$ and (29), so we could assume that if
a solution exists, then it would satisfy

$\displaystyle\frac{\partial\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1})}{\partial
q_{i}}$ $\displaystyle=0$ $\displaystyle 1\leq i\leq k-1$ (31)
$\displaystyle\frac{\partial\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1})}{\partial
x_{i}}$ $\displaystyle=0$ $\displaystyle 1\leq i\leq k.$ (32)

We will then see that the solution we find do satisfy those inequalities.

######  Claim F.1.

If $\mathbf{q}_{k},\mathbf{x}_{k}$ solves Problem (27)-(30), then

$x_{i}=1-\frac{\mu-\underline{v}}{kr_{i}(\alpha-1)}\qquad\forall 1\leq i\leq
k.$

###### Proof.

We look at the derivative by $q_{i}$ for every $1\leq i\leq k-1$,

$\displaystyle\frac{\partial\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1})}{\partial
q_{i}}=r_{i}(1-x_{i})-r_{i+1}(1-x_{i+1}).$

From equality (31) we get that $r_{i}(1-x_{i})=r_{i+1}(1-x_{i+1})$ for every
$1\leq i\leq k-1$, which means that for every $1\leq i\leq k$

$x_{i}=1-\frac{r_{1}}{r_{i}}(1-x_{1})=1-\frac{\underline{v}}{r_{i}}(1-x_{1}).$

Using the constraint (30) we get

$\displaystyle\mu-\underline{v}$
$\displaystyle=(\alpha-1)\sum_{i=1}^{k}r_{i}(1-x_{i})$
$\displaystyle=\underline{v}(\alpha-1)\sum_{i=1}^{k}(1-x_{1})$
$\displaystyle=k\underline{v}(\alpha-1)(1-x_{1}).$

And thus

$\displaystyle 1-x_{1}$
$\displaystyle=\frac{\mu-\underline{v}}{k\underline{v}(\alpha-1)}$
$\displaystyle\implies x_{i}$
$\displaystyle=1-\frac{\mu-\underline{v}}{kr_{i}(\alpha-1)}.$

We can notice that indeed $x_{1}\leq\dots\leq x_{k}\leq 1$. ∎

######  Claim F.2.

If $\mathbf{q}_{k},\mathbf{x}_{k}$ solves Problem (27)-(30), then for every
$1\leq i\leq k$

$q_{i}=\frac{i}{k}.$

###### Proof.

The derivative by $x_{i}$ for every $1\leq i\leq k$ is

$\frac{\partial\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1})}{\partial
x_{i}}=-(q_{i}-q_{i-1})r_{i}+\lambda_{1}r_{i}(\alpha-1).$

The derivative is zero only when

$q_{i}-q_{i-1}=\lambda_{1}(\alpha-1).$

Since $q_{0}=0$, we get that for every $1\leq i\leq k$

$\displaystyle q_{i}=i\lambda_{1}(\alpha-1).$

We know that $q_{k}=1$, which means

$k\lambda_{1}(\alpha-1)=1.$

So we can solve $\lambda_{1}=\frac{1}{k(\alpha-1)}$, and get And get

$q_{i}=\frac{i}{k}.$

We can also see that the inequalities $0\leq q_{1}\leq\dots\leq q_{k}$, are
satisfied. ∎

####  F.1.2 Mean and Upper Bound Solution

For every $k\geq 1$, there is a a solution $(\mathbf{q}_{k},\mathbf{x}_{k})$
that is described by Claims F.1, and F.2. Those correspond to functions
$Q^{*}_{k},F_{k}^{*}$ where $Q^{*}_{k}(r_{i})=q_{i}$, and
$F_{k}^{*}(r_{i})=x_{i}$ for every $1\leq i\leq k+1$. We defined the possible
values in the discrete case to be between in $[\underline{v},\overline{v}]$
where $\underline{v}>0$ is a parameter we can control, so we can set it in
such a way that the adversary’s best distribution’s support would be in
$[\underline{v},\overline{v}]$.

######  Claim F.3.

For arbitrary $\underline{v}>0$ the sequence
$\left\\{F^{*}_{k}\right\\}_{k\in\mathbb{N}}$ is pointwise convergent to
$F^{*}$,where

$F^{*}(v)=\begin{cases}0&v\in[0,\underline{v}),\\\
1-\frac{\mu-\underline{v}}{v\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}&v\in[\underline{v},\overline{v}),\\\
1&v\geq\overline{v}.\end{cases}$

###### Proof.

We know that $r_{k,1}\in[0,\overline{v}]$ for every $k$, so the sequence
$\\{r_{k,1}\\}_{k\in\mathbb{N}}$ has a convergent subsequence. We will assume
that the sequence itself is convergent, however all of the steps can be done
on every convergent subsequence, which would all yield the same limit, which
would prove that the sequence is indeed convergent.

Since
$\alpha=\left(\nicefrac{{\overline{v}}}{{\underline{v}}}^{\nicefrac{{1}}{{k}}}\right)$,
we get that

$\lim_{k\to\infty}k(\alpha-1)=\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right).$

From these limits, and from Claim F.1, we get

$\displaystyle
F^{*}(v)=1-\frac{\mu-\underline{v}}{v\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}.$

∎

######  Claim F.4.

For arbitrary $\underline{v}>0$ the sequence
$\left\\{Q^{*}_{k}\right\\}_{k\in\mathbb{N}}$ is pointwise convergent to
$Q^{*}$,where

$Q^{*}(v)=\begin{cases}0&r\in[0,\underline{v}),\\\ \frac{\log
r-\log\underline{v}}{\log\overline{v}-\log\underline{v}}&r\in[\underline{v},\overline{v}].\end{cases}$

We now need to set $\underline{v}$ such that neither the seller nor the
adversary would prefer having any mass in $[0,\underline{v})$. We can use the
fact that both the adversary and the seller are indifferent over
$[\underline{v},\overline{v}]$, when the other uses $Q^{*}$ or $F^{*}$
respectively.

######  Claim F.5.

If $\underline{v}$ solves
$\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}=\underline{v}$,
then $Q^{*}$, and $F^{*}$ defined in Claims F.3, F.4, are a solution to
Problem (26).

###### Proof.

When the seller uses distribution $Q$ to determine the price, and the bidder
samples valuation from distribution $F$, the expected revenue is given by

$Rev_{F}(Q)=\mathbb{E}_{r\sim
Q}\left[\operatorname*{Rev}_{F}(r)\right]=\int_{0}^{\infty}r\left(1-F(r)\right)dQ(r).$

If
$\underline{v}=\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}$,
then for $F=F^{*}$ we get a revenue of

$\displaystyle Rev_{F^{*}}(Q)$
$\displaystyle=\int_{0}^{\infty}r\left(1-F^{*}(r)\right)dQ(r)$
$\displaystyle=\int_{0}^{\underline{v}}rdQ(r)+\int_{\underline{v}}^{\overline{v}}\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}dQ(r)$
$\displaystyle\leq\underline{v}Q(\underline{v})+\underline{v}\left(Q(\overline{v})-Q(\underline{v})\right)$
$\displaystyle\leq\underline{v}.$

Likewise, if
$\underline{v}=\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}$,
then for $Q=Q^{*}$, and $F$ with a mean value of $\mu$, and a support
contained in $[0,\overline{v}]$, we get a revenue of

$\displaystyle Rev_{F}(Q^{*})$
$\displaystyle=\int_{0}^{\infty}r\left(1-F(r)\right)dQ^{*}(r)$
$\displaystyle=\int_{0}^{\infty}r\left(1-F(r)\right)(Q^{*})^{\prime}(r)dr$
$\displaystyle=\int_{\underline{v}}^{\overline{v}}\frac{\left(1-F(r)\right)}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}dr$
$\displaystyle=\frac{\mu-\int_{0}^{\underline{v}}(1-F(r))dr}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}$
$\displaystyle\geq\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}$
$\displaystyle=\underline{v}.$

∎

###  F.2 Single bidder: Upper Bound and Two Moment Constraints

In order to solve the case where only the mean value is known, and an upper
bound on the second moment, we first solve a simpler problem. In this problem,
we assume that the seller knows an upper bound $\mu_{2}$, on the second moment
of the bidder’s distribution, and also an upper bound $\overline{v}$ on the
valuations, and the mean value $\mu$ of the distribution. Without the
constraint on the second moment, we have solved this problem, and found that
for the equilibrium, the bidder sample her valuation using $F^{*}$ (Claim
F.5). We solve for the cases where $\mu_{2}$ is equal to the second moment of
$F^{*}$, which means that

$\displaystyle\mu_{2}$ $\displaystyle=\int_{0}^{\overline{v}}2v(1-F^{*}(v))dv$
$\displaystyle=\underline{v}^{2}+\int_{\underline{v}}^{\overline{v}}2\underline{v}dv$
$\displaystyle=2\overline{v}\underline{v}-\underline{v}^{2}.$

. With this $\mu_{2}$, we define
$\mathcal{F}_{\mu,\mu_{2}}^{\leq\overline{v}}$ to be the set of all
distributions over $[0,\overline{v}]$ with a mean value of $\mu$, and a second
moment of $\mu_{2}$. And the new problem is

$\max_{Q\in\mathcal{Q}}\min_{F\in\mathcal{F}_{\mu,\mu_{2}}^{\leq\overline{v}}}\int_{0}^{\infty}r\left(1-F(r)\right)dQ(r).$
(33)

Clearly, the solution for Problem (26) is also a solution for Problem (33).
However, with the new constraint, we can find a family of solutions for the
seller, that can grantee the same expected revenue. Afterwards, we will
identify a specific distribution in this family that solves that guarantee the
same revenue even when we remove the upper bound on the valuations.

####  F.2.1 Mean, Upper Bound, Second Moment: Discrete Case

For every $k\geq 1$, we solve for a discrete case with $k+1$ possibilities,
similarly to the case without the additional constraint on the second moment.
We also define $\underline{v}$ to be solution of
$\frac{\mu-\underline{v}}{\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)}=\underline{v}$,
like in Claim F.5. In order to make sure the adversary acts in the same way as
without the constraint, we allow the second moment to be higher than $\mu_{2}$
for each $k$, and the upper bound on the second moment we will use is the
second moment of $F^{*}_{k}$, where $F^{*}_{k}$ is the adversary’s previous
solution, according got Claim F.1. Which means

$\displaystyle\mu_{2,k}$
$\displaystyle=\int_{0}^{\overline{v}}2v(1-F^{*}_{k}(v))dv$
$\displaystyle=\underline{v}^{2}+\sum_{i=1}^{k}\int_{r_{i}}^{r_{i+1}}2v(1-x_{i}^{*})dv$
$\displaystyle=\underline{v}^{2}+\sum_{i=1}^{k}(1-x_{i}^{*})(r_{i+1}^{2}-r_{i}^{2})$
$\displaystyle=\underline{v}^{2}+\sum_{i=1}^{k}\frac{\mu-\underline{v}}{kr_{i}(\alpha-1)}(\alpha^{2}-1)r_{i}^{2}$
$\displaystyle=\underline{v}^{2}+\frac{(\mu-\underline{v})(\alpha+1)}{k}\sum_{i=1}^{k}r_{i}$
$\displaystyle=\underline{v}^{2}+\frac{(\mu-\underline{v})(\overline{v}-\underline{v})(\alpha+1)}{k(\alpha-1)}.$

The new problem is identical to 27-30, with the additional constraint

$(\alpha^{2}-1)\sum_{i=1}^{k}(1-x_{i})r_{i}^{2}\leq\mu_{2,k}-\underline{v}^{2}.$
(34)

That changes the Lagrangian for this problem to be

$\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1},\lambda_{2})=\operatorname*{Rev}(\mathbf{q}_{k},\mathbf{x}_{k})-\lambda_{1}(\alpha-1)\sum_{i=1}^{k}r_{i}(1-x_{i})-\lambda_{2}(\alpha^{2}-1)\sum_{i=1}^{k}(1-x_{i})r_{i}^{2}.$

The solution for the adversary remains completely identical. But for the
seller, we get a bit more freedom.

######  Claim F.6.

If $\mathbf{q}_{k},\mathbf{x}_{k}$ solves Problem (27)-(30), with constraint
(34) as well, then for every $1\leq i\leq k$

$q_{i}=i\lambda_{1}(\alpha-1)+\lambda_{2}(\alpha+1)(\alpha
r_{i}-\underline{v}),$

where
$\lambda_{1}=\frac{1-\lambda_{2}(\alpha+1)(\overline{v}-\underline{v})}{k(\alpha-1)}$.

###### Proof.

The derivative by $x_{i}$ for every $1\leq i\leq k$ is

$\frac{\partial\mathcal{L}(\mathbf{q}_{k},\mathbf{x}_{k},\lambda_{1},\lambda_{2})}{\partial
x_{i}}=-(q_{i}-q_{i-1})r_{i}+\lambda_{1}r_{i}(\alpha-1)+\lambda_{2}r_{i}^{2}(\alpha^{2}-1).$

The derivative is zero only when

$q_{i}-q_{i-1}=\lambda_{1}(\alpha-1)+\lambda_{2}r_{i}(\alpha^{2}-1).$

Since $q_{0}=0$, we get that for every $1\leq i\leq k$

$\displaystyle q_{i}$
$\displaystyle=\sum_{j=1}^{i}\lambda_{1}(\alpha-1)+\lambda_{2}r_{j}(\alpha^{2}-1)$
$\displaystyle=i\lambda_{1}(\alpha-1)+\lambda_{2}\underline{v}(\alpha^{2}-1)\sum_{j=1}^{i}\alpha^{j-1}$
$\displaystyle=i\lambda_{1}(\alpha-1)+\lambda_{2}(\alpha+1)(\alpha
r_{i}-\underline{v}).$

We know that $q_{k}=1$, which means

$k\lambda_{1}(\alpha-1)+\lambda_{2}(\alpha+1)(\overline{v}-\underline{v})=1.$

And we can solve $\lambda_{1}$,

$\lambda_{1}=\frac{1-\lambda_{2}(\alpha+1)(\overline{v}-\underline{v})}{k(\alpha-1)}.$

∎

####  F.2.2 Mean, Upper Bound, Second Moment: General Case

In the discrete case, we get freedom that depends on $\lambda_{2}$, which
yields a family of solutions. This translates to the general case as well, as
we take the limit.

From Claim F.6, for each $k$, the solution $Q^{*}_{k}$ for the discrete
problem satisfies

$Q_{k}^{*}(r_{i})=\frac{i}{k}(1-\lambda_{2}(\alpha+1)(\overline{v}-\underline{v}))+\lambda_{2}(\alpha+1)(\alpha
r_{i}-\underline{v}).$

And thus, $\left\\{Q_{k}^{*}\right\\}_{k\in\mathbb{N}}$ is pointwise
convergent to $Q^{*}$ where

$Q^{*}(r)=\frac{\log\nicefrac{{r}}{{\underline{v}}}}{\log\nicefrac{{\overline{v}}}{{\underline{v}}}}(1-2\lambda_{2}(\overline{v}-\underline{v}))+2\lambda_{2}(r-\underline{v}).$

Similarly, when $k\to\infty$

$\lambda_{1}=\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})}.$
(35)

And we get

$Q^{*}_{\lambda_{2}}(r)=\lambda_{1}\log\nicefrac{{r}}{{\underline{v}}}+2\lambda_{2}(r-\underline{v}).$

We just need to make sure that $Q^{*}$ is a distribution function, which means
we need to make sure that the formula we got is non-decreasing in
$[\underline{v},\overline{v}]$ and that its image over the same interval is
contained in $[0,1]$.

Since $Q^{*}(\underline{v})=0$, and $Q^{*}(\overline{v})=1$ regardless of
$\lambda_{2}$, we just need to check that it is non decreasing.

The derivative of $Q^{*}$ is

$\frac{dQ^{*}_{\lambda_{2}}(r)}{dr}=\frac{\lambda_{1}}{r}+2\lambda_{2}.$

So we demand

$\lambda_{1}\geq-2r\lambda_{2}\qquad\forall r\in[\underline{v},\overline{v}].$

######  Claim F.7.

When the seller randomizes a price using $Q^{*}_{\lambda_{2}}$, for
$\lambda_{1},\lambda_{2}$ satisfying

$\displaystyle\lambda_{1}$
$\displaystyle=\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})},$
$\displaystyle\lambda_{1}$ $\displaystyle\geq-2r\lambda_{2}\qquad\forall
r\in[\underline{v},\overline{v}],$

she guarantees an expected revenue of at least $\underline{v}$, when the
bidder samples her valuation from
$F\in\mathcal{F}_{\mu,\mu_{2}}^{\leq\overline{v}}$.

###### Proof.

The seller best response cannot have any mass below $\underline{v}$, since any
such distribution is dominated by a distribution that doesn’t. We can see that
since transferring the any mass below $\underline{v}$ to $\underline{v}$ would
not increase the expected revenue, while increasing the expected value and
decreasing the variance. This allows us to lower the values higher than $\mu$
to get to the right expected value, without increasing the revenue, while
still decreasing second moment. Therefore, we need only to check cases where
the support is fully contained in $[\underline{v},\overline{v}]$.

Let be $F\in\mathcal{F}_{\mu,\mu_{2}}^{\leq\overline{v}}$ such that the
support of $F$ is contained in $[\underline{v},\overline{v}]$.

$\displaystyle Rev_{F}(Q^{*})$
$\displaystyle=\int_{0}^{\infty}r\left(1-F(r)\right)(Q^{*})^{\prime}(r)dr$
$\displaystyle=\int_{\underline{v}}^{\overline{v}}\left(\lambda_{1}\left(1-F(r)\right)+2\lambda_{2}r\left(1-F(r)\right)\right)dr$
$\displaystyle=\lambda_{1}(\mu-\underline{v})+\lambda_{2}(\mu_{2}-\underline{v}^{2})$
$\displaystyle=\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})}(\mu-\underline{v})+\lambda_{2}(2\overline{v}\underline{v}-2\underline{v}^{2})$
$\displaystyle=\underline{v}-2\lambda_{2}\underline{v}(\overline{v}-\underline{v})+2\lambda_{2}\underline{v}(\overline{v}-\underline{v})$
$\displaystyle=\underline{v}.$

∎

###  F.3 Single bidder: Two Out of Three Constraints

In subsection F.2 we have introduced the second moment constraint, where
$\mu_{2}$ was set to be:

$\mu_{2}=2\overline{v}\underline{v}-\underline{v}^{2},$

where $\underline{v}$ is the unique solution to

$\mu=\underline{v}\left(1+\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)\right).$

In this section, we assume only knowledge of then mean value $\mu$, and an
upper bound on the second moment $\mu_{2}$.

We define $(\underline{v},\overline{v})$ to be the solution to:

$\displaystyle\mu_{2}$
$\displaystyle=2\overline{v}\underline{v}-\underline{v}^{2},$
$\displaystyle\mu$
$\displaystyle=\underline{v}\left(1+\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)\right).$

If the bidder could not have any values higher than $\overline{v}$, we would
have end up with the same solution as in subsection F.2, which is:

$\displaystyle Q^{*}_{\lambda_{2}}(r)$
$\displaystyle=\begin{cases}0&r\in[0,\underline{v}]\\\
\lambda_{1}\log\nicefrac{{r}}{{\underline{v}}}+2\lambda_{2}(r-\underline{v})&r\in(\underline{v},\overline{v}]\\\
1&r>\overline{v}\end{cases}$ $\displaystyle F^{*}(v)$
$\displaystyle=\begin{cases}0&v\in[0,\underline{v}]\\\
1-\nicefrac{{\underline{v}}}{{v}}&v\in(\underline{v},\overline{v})\\\
1&v\geq\overline{v}\end{cases}$ $\displaystyle\lambda_{1}$
$\displaystyle=\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})},$
$\displaystyle\lambda_{1}$ $\displaystyle\geq-2r\lambda_{2}\qquad\forall
r\in[\underline{v},\overline{v}].$

We now explore how we can set $\lambda_{2}$ such that the adversary would not
benefit from having mass above $\overline{v}$.

Fix $F\in\mathcal{F}_{\mu,\leq\mu_{2}}$ (that is $F$ only assumed to have an
expectation $\mu$, and a second moment that isn’t higher than $\mu_{2}$). We
also assume w.o.l.g that $F$ doesn’t have any mass lower than $\underline{v}$,
since if it does, than there exists $F_{2}\in\mathcal{F}_{\mu,\leq\mu_{2}}$
that doesn’t have any mass lower than $\underline{v}$, and yield no higher
expected revenue.

If $F$ has a second moment of exactly $\mu_{2}$, we get that the expected
revenue the seller guarantees herself when using $Q^{*}$ is:

$\displaystyle Rev_{F}(Q^{*})$
$\displaystyle=\int_{0}^{\infty}r\left(1-F(r)\right)(Q^{*})^{\prime}(r)dr$
$\displaystyle=\int_{\underline{v}}^{\overline{v}}\left(\lambda_{1}\left(1-F(r)\right)+2\lambda_{2}r\left(1-F(r)\right)\right)dr$
$\displaystyle=\lambda_{1}\left(\mu-\underline{v}-\int_{\overline{v}}^{\infty}(1-F(r))dr\right)+\lambda_{2}\left(\mu_{2}-\underline{v}^{2}-\int_{\overline{v}}^{\infty}2r(1-F(r))dr\right)$
$\displaystyle=\lambda_{1}(\mu-\underline{v})+\lambda_{2}(2\overline{v}\underline{v}-2\underline{v}^{2})-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr$
$\displaystyle=\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})}(\mu-\underline{v})+\lambda_{2}(2\overline{v}\underline{v}-2\underline{v}^{2})-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr$
$\displaystyle=\underline{v}-2\lambda_{2}\underline{v}(\overline{v}-\underline{v})+2\lambda_{2}\underline{v}(\overline{v}-\underline{v})-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr$
$\displaystyle=\underline{v}-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr.$

Therefore, we would require $\lambda_{1}\leq-2\lambda_{2}r$ for every
$r\geq\overline{v}$. We know that $\lambda_{1}\geq-2\lambda_{2}\overline{v}$,
and thus we can set

$\lambda_{1}=-2\lambda_{2}\overline{v}.$

In that case, we get

$\frac{1-2\lambda_{2}(\overline{v}-\underline{v})}{\log(\nicefrac{{\overline{v}}}{{\underline{v}}})}=-2\lambda_{2}\overline{v}.$

We can solve $\lambda_{2}$ and $\lambda_{1}$ to be

$\displaystyle\lambda_{2}$
$\displaystyle=\frac{1}{2((\overline{v}-\underline{v})-\overline{v}\log(\nicefrac{{\overline{v}}}{{\underline{v}}}))},$
(36) $\displaystyle\lambda_{1}$
$\displaystyle=-\frac{\overline{v}}{(\overline{v}-\underline{v})-\overline{v}\log(\nicefrac{{\overline{v}}}{{\underline{v}}})}.$
(37)

Note that $\lambda_{2}<0$, and thus the inequalities

$\displaystyle\lambda_{1}$ $\displaystyle\geq-2r\lambda_{2}\qquad\forall
r\in[\underline{v},\overline{v}],$ $\displaystyle\lambda_{1}$
$\displaystyle\leq-2r\lambda_{2}\qquad\forall r\geq\overline{v}$

are satisfied.

######  Claim F.8.

If the seller randomized the price by sampling from $Q^{*}$ defined below,
then she guarantees an expected revenue of at least $\underline{v}$ (defined
below) if the bidder uses a distribution $F$ with a mean value $\mu$, and a
second moment no larger than $\mu_{2}$.

$Q^{*}(r)=\begin{cases}0&r\in[0,\underline{v}]\\\
\lambda_{1}\log\nicefrac{{r}}{{\underline{v}}}+2\lambda_{2}(r-\underline{v})&r\in(\underline{v},\overline{v}]\\\
1&r>\overline{v}\end{cases}$

where

$\displaystyle\lambda_{2}$
$\displaystyle=\frac{1}{2((\overline{v}-\underline{v})-\overline{v}\log(\nicefrac{{\overline{v}}}{{\underline{v}}}))},$
$\displaystyle\lambda_{1}$
$\displaystyle=-\frac{\overline{v}}{(\overline{v}-\underline{v})-\overline{v}\log(\nicefrac{{\overline{v}}}{{\underline{v}}})},$

and $\underline{v},\overline{v}$ are the solution to

$\displaystyle\mu_{2}$
$\displaystyle=2\overline{v}\underline{v}-\underline{v}^{2},$
$\displaystyle\mu$
$\displaystyle=\underline{v}\left(1+\log\left(\nicefrac{{\overline{v}}}{{\underline{v}}}\right)\right).$

###### Proof.

As explained previously, we can assume that $F$ doesn’t have any mass below
$\underline{v}$. We note that $\lambda_{2}$ is always negative, and
$\lambda_{1}+2\lambda_{2}r$ is non-positive for every $r\geq\overline{v}$. The
expected revenue is

$\displaystyle Rev_{F}(Q^{*})$
$\displaystyle=\int_{0}^{\infty}r\left(1-F(r)\right)(Q^{*})^{\prime}(r)dr$
$\displaystyle=\int_{\underline{v}}^{\overline{v}}\left(\lambda_{1}\left(1-F(r)\right)+2\lambda_{2}r\left(1-F(r)\right)\right)dr$
$\displaystyle\geq\lambda_{1}\left(\mu-\underline{v}-\int_{\overline{v}}^{\infty}(1-F(r))dr\right)+\lambda_{2}\left(\mu_{2}-\underline{v}^{2}-\int_{\overline{v}}^{\infty}2r(1-F(r))dr\right)$
$\displaystyle=\lambda_{1}(\mu-\underline{v})+\lambda_{2}(2\overline{v}\underline{v}-2\underline{v}^{2})-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr$
$\displaystyle=\underline{v}-\int_{\overline{v}}^{\infty}(1-F(r))(\lambda_{1}+2\lambda_{2}r)dr$
$\displaystyle\geq\underline{v}.$

∎