## Ratio of expectation involving random unit vectors

Let $$u=(u_1,…,u_n), v=(v_1,…,v_n)$$ be two random vectors independently and uniformly distributed on the unit sphere in $$\mathbb{R}^n$$. Define two other random variables $$X=\sqrt{\sum_{i=1}^nu_i^2v_i^2}$$, $$Y=|u_1v_1|$$. Consider the following ratio of expectation: $$r_n(\alpha)=\frac{\mathbb{E}\{\exp[-\frac{\alpha^2-\alpha^2X^2+\alpha X}{2}]\}}{\mathbb{E}\{\exp[-(\alpha^2-\alpha^2Y^2+\alpha Y)]\}}$$ Does there exist a finite upper bound for $$r_n(\alpha)$$, independent of $$\alpha$$, for all $$\alpha\geq0$$?

## Ratio of expectation involving random unit vectors

Let $$u=(u_1,…,u_n), v=(v_1,…,v_n)$$ be two random vectors independently and uniformly distributed on the unit sphere in $$\mathbb{R}^n$$. Define two other random variables $$X=\sqrt{\sum_{i=1}^nu_i^2v_i^2}$$, $$Y=|u_1v_1|$$. Consider the following ratio of expectation: $$r_n(\alpha)=\frac{\mathbb{E}\{\exp[-\frac{\alpha^2-\alpha^2X^2+\alpha X}{2}]\}}{\mathbb{E}\{\exp[-(\alpha^2-\alpha^2Y^2+\alpha Y)]\}}$$ Does there exist a finite upper bound for $$r_n(\alpha)$$, independent of $$\alpha$$, for all $$\alpha\geq0$$?

## Root of the expectation of a random rational function

I am trying to figure out a formula for the unique $$\lambda>1$$ such that $$\mathbb{E}\bigg[\frac{X}{\lambda -X}\bigg]=1$$ where $$X$$ is a discrete random variable taking values in $$\{\frac{1}{n},…,\frac{n-1}{n},1\}$$, distributed w.r.t. some distribution $$\mu$$.

We can rewrite the expression above which yields $$\sum_{k=1}^n \frac{\mu(\frac{k}{n})\frac{k}{n}}{x-\frac{k}{n}} = 1.$$

I know that there are no closed solutions for the roots of such a function, since they are based on solving for zeros of a high degree polynomial. Still, I think I miss some obvious results here how to analyse such a function.

I’d appreciate any kind of help. Thank’s a lot!

## Calculating expectation and variance for having rolled 1 and 6 twice out of rolling a die 12 times

First i have calculated the probability to get each possible number {1,2,3,4,5,6} twice from 12 rolls(A).

$$Pr[A]=\frac{\binom{12}{2,2,2,2,2,2}}{6^{12}}.$$

Then there are 2 random variables:

X-number of times that 1 was received.

Y-number of times that 6 was received.

Before calculating $$E(X),Var(X),E(Y),Var(Y)$$ i’m uncertain of how i should calculate the probabilities of X and Y

## Obtaining a lower bound on the expectation using the Sudakov-Fernique inequality

In my work I wish to obtain a lower bound for the term below, independent of the vector $$x$$. Here the expectation is taken over $$h$$, a standard random Gaussian vector of length $$n$$. The vector $$x$$ is fixed. The minimum is taken over all $$\{i_1,\dots,i_L\} \in \{1,\dots,n\}$$. Can this be done using the Sudakov-Fernique inequality? $$\mathbb{E}_{h} \min _{i_{1}, \ldots, i_{L}}\left[\sum_{j\neq i_1,\dots,i_L}h_j\mathrm{sign}(x_j^*)\right].$$

## Conditional Expectation: Intergrating indicator function multiplied by the joint denisity

I am currently reading “Measure, Integral and Probability” by Capinski, Marek (see p179). It includes some motivation for the definition of the conditional expectation. For example, given two random variables $$X,Y$$ with joint density $$f_{(X,Y)}$$ (and so the marginal and conditional densities), we want to show that for any set $$A \subset \Omega, A=X^{-1}(B), B$$ Borel, that $$\int_A\mathbb{E}(Y|X)dP= \int_A \mathbb{E}(Y)dP.$$ This is one of the defining condition of an conditional expectation. The book shows the following calculation, \begin{align} \int_A\mathbb{E}(Y|X)dP &= \int_\Omega 1_B(X)\mathbb{E}(Y|X)dP\ &= \int_\Omega 1_B(X(\omega))\left(\int_\mathbb{R}yf_{Y|X}(y|X(\omega))dy\right)dP(\omega)\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{(Y|X)}(y|x)dy f_X(x)dx\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{X,Y}(x,y)dxdy\ &= \int_\Omega 1_A(X)YdP\ &= \int_A YdP. \end{align} What I don’t understand is the second to last equality immediately above, i.e. $$\int_\mathbb{R} y \int_\mathbb{R}1_B(x)f_{X,Y}(x,y)dxdy = \int_\Omega 1_A(X)YdP .$$ I think it is a typo since $$X\in \mathbb{R}$$ and $$A \subset \Omega$$ — however, I cant figure the correction either!

## Expectation of number of hubs in a random graph

Suppose $$\Gamma(V, E)$$ is a finite simple graph. Let’s call a vertex $$v \in V$$ a hub if $$deg(v)^2 > \Sigma_{w \in O(v)} deg(w)$$. Here $$deg$$ stands for the vertex degree, and $$O(v)$$ for the set of all vertices adjacent to $$v$$. Let’s define $$H(\Gamma)$$ as the number of all hubs in $$\Gamma$$.

Now, suppose $$G(n, p)$$ is an Erdos-Renyi random graph with $$n$$ vertices and edge probability $$p$$. Does there exist some sort of explicit formula for $$E(H(G(n, p)))$$ (as a function of $$n$$ and $$p$$)?

How did this question arise:

I have recently heard of a so called «friendship paradox» that states, that the number of your friends usually does not exceed the average number of friends your friends have. When I at first heard about it, I wondered, if that is just some specifics of human society, or is there a mathematical explanation behind it. First I tried to translate the statement of the «friendship paradox» to the mathematical language as

Any graph with $$n$$ vertices has $$o(n)$$ hubs,

but then I quickly found, that it it is blatantly false this way:

Suppose $$n > 2$$, let’s take the full graph on $$n$$ vertices $$K_n$$ and then remove one edge from it, then, the resulting graph will have $$n – 2$$ hubs, which is clearly not $$o(n)$$.

So, the «friendship paradox» can not be translated to deterministic graph theory using the notion of «hubs». So it can not be interpreted that way. So, I thought, that maybe something similar with random graphs should work.

Also:

Later, I found out, that there actually is a consistent mathematical interpretation of «friendship paradox» in terms of deterministic graph theory: Friendship paradox demonstration

However it does not solve my problem – which is finding the expectation of the number of hubs in a random graph. So, please do not mark my question as a duplicate of the aforementioned question.

## Is my interpretation of the expression on the expectation on a correct one?

Consider the following equation expression

$$E_{x\sim p_{data}(x)}[ f(x) ]$$

I am understanding it as

1) if X is a collection of a discrete random variable (X_1, X_2,……, X_n), and x is generated from X by a particular assignment of all random variables in the tuple

Then

$$E_{x\sim p_{data}(x)}[ f(x) ] = \sum\limits_{x} f(x) p_{data}(x)$$

2) if X is a collection of a continuous random variable (X_1, X_2,……, X_n), and x is generated from X by a particular assignment of all random variables in the tuple

Then

$$E_{x\sim p_{data}(x)}[ f(x) ] = \int\limits_{x_1}\int\limits_{x_2} \cdots \int\limits_{x_n} f(x) p_{data}(x) dx_n \cdots dx_1$$

Is my interpretation exact? If wrong, where am I going wrong?

## The expectation of partition times needed separate two elements in a set

I met a problem which can be formulated as set partition.

Given a set $$S=\{s_1,s_2,…,s_n\}$$ having $$n$$ elements, I want to separate two elements, say $$s_1,s_2$$, in $$S$$ by repeatedly using set partition operations. Each set partition operation randomly partitions a set, say $$A$$, into two non-empty subsets, $$B$$ and $$C$$, such that $$A=B\cup C$$ and $$B\cap C=\emptyset$$. I want to calculate or approximate the expectation of partition time, $$E(n)$$, to separate $$s_1$$ and $$s_2$$.

Let see two simple cases:

(1) $$n=2$$:

In this case, $$S=\{s_1,s_2\}$$. The only feasible partition will separate $$S=\{s_1,s_2\}$$ as $$\{s_1\}$$ and $$\{s_2\}$$. So, $$E(2) = 1$$.

(2) $$n=3$$:

In this case, $$S=\{s_1,s_2,s_3\}$$. There are two situations:

(a) if the first partition is $$\{s_1\},\{s_2,s_3\}$$ or $$\{s_2\},\{s_1,s_3\}$$, then 1 partition is ok!

(b) if the first partition is $$\{s_1,s_2\},\{s_3\}$$, then I need a second partition making $$\{s_1,s_2\}$$ into $$\{s_1\},\{s_2\}$$. So the partition time is 2.

The possibility of situation (a) is 2/3 and situation (b) 1/3. So the $$E(3)=1*(2/3)+2*(1/3)=4/3$$.

I tried using recursive formula but it seems to be a non-closed form. I also wonder whether or not $$E(n)$$ can be approximated by some other continuous functions?

## Bound for Expectation of Singular Value

In my case, $$X_{\boldsymbol{\delta}}\in\mathbb{R}^{d\times M}$$ is a function of Rademacher variables $$\boldsymbol{\delta}\in\{1,-1\}^M$$ with $$\delta_i$$ independent uniform random variables taking values in $$\{−1, +1\}$$. $$X_{\boldsymbol{\delta}}=[\sum_{i=1}^{I_1}\delta_{i}\mathbf{x}_{i},\sum_{i=I_1+1}^{I_2}\delta_{i}\mathbf{x}_{i},…,\sum_{i=I_{M-1}+1}^{I_M}\delta_i\mathbf{x}_{i}]$$ is a group-wise sum with known $$I_1,I_2,…,I_M$$ and non-singular $$X=(\mathbf{x}_1,\mathbf{x}_2,…,\mathbf{x}_N)\in\mathbb{R}^{d\times N}$$ where $$N>M\gg d$$.

Given that $$\sigma_i(X_{\boldsymbol{\delta}})$$ denotes $$i$$-th smallest singular value, how can I find the lower bound of the expectation $$\underset{\boldsymbol{\delta}}E\left[\sum_{i=1}^{k} \sigma_{i}^{2}\left(X_{\boldsymbol{\delta}}\right)\right]$$ assuming $$k?

Note: I can find an upper bound by Jensen’s inequality and concavity of sum of $$k$$ smallest eigenvalue, but I am curious about whether it is possible to get a lower bound.

I have also posted the question here.