Let X and Y be two independent random variables. If I know for some function f that $E_Y[f(X,Y)]$ exists and $P_X[f(X,Y)>t|Y]<\delta$ holds (for any fixed instance of Y) then does it follow that $P_X[E_Y[f(X,Y)]>t]<\delta$?

Here is an attempt, but I'm not sure if it is correct or not:

Define the event $A="f(X,Y)<t"$. We know that $P_X(A)\ge 1-\delta$. Let event B be $B="E_Y[f(X,Y)]<t"$. Now, $A=>B$. Therefore, $P_X(A) <= P_X(B)$, and since $P_X(A)\ge 1-\delta$ we get that $P_X(B)\ge 1-\delta$, which is the same as $P_X(\neg B)<\delta$.

  • $\begingroup$ No; to see this, consider the case where $X$ and $Y$ are distributed independent Cauchy, and $f(X,Y) = X+Y$. The expectation does not exist (not the same as "is infinite".) For a discussion of this latter point (irrelevant to your question, I admit), see stats.stackexchange.com/questions/117376/…. $\endgroup$ – jbowman Nov 5 '14 at 0:22
  • 1
    $\begingroup$ Thanks. I should add that in my case the expectation does exist. $\endgroup$ – axk Nov 5 '14 at 0:24

I believe that this is not true. Let $X$ and $Y$ be uniform on $[0,1]$, let $\mu$ be the measure associated to $X$ and $\nu$ be the measure associated to $Y$. Fix $n \in \mathbb{N}$ and consider the function $$ \begin{align} f(x,y) = n(t+1) \sum_{k=0}^{n-1} {\bf 1} \left[\frac{k}{n} \leq x,y \leq \frac{k+1}{n} \right]. \end{align} $$ This function is $0$ everywhere except on the squares of size $1/n$ along the diagonal.

Then $\mathbb{P}(f(x,y) > t|Y) = 1/n $, $\nu$ a.s., but $\mathbb{E} \left[ f(X,Y) | X \right] = t+1$, $\mu$ a.s.

Here's why your answer doesn't work. You have $A = \left\{ (x,y) : f(x,y) < t \right\}$. Let $A_y = \left\{ x : f(x,y) < t \right\}$. Your hypothesis is that $\mu(A_y) \geq 1 - \delta$ a.s. with respect to $\nu$. Your event $B$ is $\left \{ x : \int f(x,y) d\nu(y) < t \right \}$. It is not true that for a particular $y$ the random set $A_y$ contain $B$. It is true that $B$ is contained in $\cup_y A_y$ but this is an uncountable union. So you can't say anything about $\mu(A)$ from knowing $\mu(A_y)$ for $y$ a.s with respect to $\nu$.


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.