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9 votes

$E[(X+Y)^{a}] > E[(X)^{a}]$?

For all $x>0$, $y>0$ and $a>0$ you have $x+y>x$ so that: $$(x+y)^a>x^a.$$ Consequently, for any joint distribution of $(X,Y)$ where the support is on the strictly positive values for ...
Ben's user avatar
  • 127k
5 votes
Accepted

Probability algorithm on strings

Ok, so you have strings $Z_n$ that are exactly half 0s and half 1s and you have $X_n$ that have exactly one more 1. One way to generate these sorts of strings is to generate a $Z_n$ uniformly at ...
Thomas Lumley's user avatar
5 votes

Expected Value Chi Square distribution

Your (5) "I changed $\frac{(n−1)s^2}{\sigma^2}∼\chi^2_{n-1}$ to $\frac{(n−1)s^2}{\chi^2_{n-1}}∼\sigma^2$" does not make much sense as $\sigma^2$ does not have a distribution while your $(n-...
Henry's user avatar
  • 40.3k
4 votes
Accepted

Expectation under convex order

The implication does not hold. For example, let $M \equiv 0$, $N$ be the Radamacher random variable (i.e., $P(N = \pm 1) = \frac{1}{2}$). Then for any concave function $u$, it follows by Jensen's ...
Zhanxiong's user avatar
  • 20.4k
4 votes

Validating binary prediction model

There are a bunch of ways. Two reasonably popular ways are the Brier score and the log loss. The Brier score is square loss, the same as mean squared error: $\text{Brier}\left(e, p\right) = \dfrac{1}{...
Dave's user avatar
  • 64.3k
3 votes

Expected value of quotient of Poisson distributions

This conditional expectation can be directly evaluated by the formula $E[\xi|A] = \frac{E[\xi I_A]}{P(A)}$ with $P(A) > 0$. Since $X \sim \text{Poisson}(\lambda c)$ and $Y \sim \text{Poission}(\...
Zhanxiong's user avatar
  • 20.4k
3 votes

Expectation of Inverse Logit of Normal Random Variable

The logistic distribution function in the integral can be approximated by the distribution function of a Gaussian having the same mean & variance; the new integral can be evaluated exactly, as per ...
Scortchi - Reinstate Monica's user avatar
2 votes

Expected value of quotient of Poisson distributions

We can rewrite the expectation as: $$\mathrm{E}_{X+Y}\left[\left.\mathrm{E}_X\left. \left[{X\over X+Y}\right| X+Y\right]\right| X+Y>0\right]$$ We note that, as is well-known, the distribution of $X$...
jbowman's user avatar
  • 40k
2 votes
Accepted

Expectation under convex order by multiplying

Define $N$ and $M$ as in a similar question asked by you before, so that $E[u(N)] \leq E[u(M)]$ holds for any concave non-decreasing function. Let $X = N + 2 > 0$, then \begin{align*} & E[u(NX)]...
Zhanxiong's user avatar
  • 20.4k
1 vote

Distribution of outcomes of multiple binomial distributions

Let $S_i$, for $i=1,2,\ldots,50$, be the raw score for the $i$th subject, i.e. the number of tasks completed with the right hand. If $S_i \sim \mathrm{Bin}(30, 0.5)$, I'm assuming that you calculate ...
Doctor Milt's user avatar
  • 3,301
1 vote

Iterated expectation over different sets of variables

The conjecture is not true. An easy way to see that is whereas $E[E[Y|X, Z]|Z, Q]$ is a function of $(Z, Q)$, $E[Y | X, Q]$ is a function of $(X, Q)$, hence they cannot be identical. To give you a ...
Zhanxiong's user avatar
  • 20.4k
1 vote

How to show that $E|X-\mu| \leq E|X-Y|$?

To begin with, convince yourself that it is equivalent to prove $E[|X|] \leq E[|X - Y|]$ holds for $E[X] = 0$. Since the function $x \mapsto |x|$ is convex on $\mathbb{R}$, it follows by the ...
Zhanxiong's user avatar
  • 20.4k

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