1
$\begingroup$

I am trying to compute expectation of $X\mathbb I_{[X+Y\le a]}$ where $a$ is a fixed positive integer, $X$ is discrete uniform random variable taking values from $1$ to $a$, and $Y$ another random variable of unknown distribution, independent of $X$. I need it as part of a proof I am doing. My calculations are as follows: \begin{align} E(X\mathbb I_{[X+Y\le a]})&=E(E(X\mathbb I_{[X+Y\le a]}|X+Y\le a))&&(\text{By tower property})\\ &=E(E(X|X+Y\le a))\\ &=E(E(X|X\le a-Y))\\ &=E\left(\dfrac{a-Y+1}{2}\right)&&(\because X|X\le b\sim \mathrm{dunif}(1,2,\dots,b))\\ &=\dfrac{a+1}{2}-\dfrac12E(Y) \end{align} Now my problem is that, the random variable of which I want to calculate expectation is non-negative, but the expectation that I am getting is negative if $E(Y)>a+1$, which is possible. So what is the problem in my calculations?? Also, how do I sort that problem out?

$\endgroup$
3
  • $\begingroup$ Is $Y$ positive ? Negative ? What is the support ? $\endgroup$
    – ArnoV
    Feb 14 '21 at 11:10
  • $\begingroup$ Support of $Y$ is a subset of $\mathbb N$, only that is known. $\endgroup$
    – Martund
    Feb 14 '21 at 11:11
  • $\begingroup$ After a few mistakes of my own I've added the correct answer: you mistakenly forget that $a-Y$ can be negative. $\endgroup$
    – ArnoV
    Feb 14 '21 at 17:37
2
$\begingroup$

Focus on the computation of $\mathbf{E}\left[X\mid X \leq a - Y\right]$ You have 2 possibilities:

  • If $Y=y > a$ then $\mathbf{E}\left[X\mid X\leq a - y\right] = 0$ as $X$ has zero mass outside its support.
  • If $Y = y < a$ then $$\mathbf{E}\left[X\mid X\leq a - y\right] = \sum_{x=1}^ax\,p(x\mid a,y) = \frac{1}{a-y}\sum_{x=1}^{a-y}x = \dfrac{a - y +1}{2}$$ because the conditional probability simplifies to $$ p(x\mid a,y) = \dfrac{\mathbf{P}(X=x, X\leq a-y)}{\mathbf{P}(X \leq a - y)} = \frac{a}{a(a-y)}\mathbb{I}_{x \leq a - y} = \frac{1}{a-y}\mathbb{I}_{x \leq a - y}$$ Therefore the expectation over $Y$ becomes $$\sum_{y=1}^{a-1}p_Y(y)\frac{a-y+1}{2} = \frac{a+1}{2}\mathbf{P}(Y<a) - \frac{1}{2}\sum_{y=1}^{a-1}y\,p_Y(y)$$ This is positive because the second term is bounded as $$\sum_{y=1}^{a-1}y\,p_Y(y) \leq (a-1)\mathbf{P}(Y<a)$$ which yields that your quantity is bigger than $\mathbf{P}(Y<a)\dfrac{1 - a + 1 + a}{2} > 0 $
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.