Obtaining a probability inequality from another?

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 $P_X(\neg B)<\delta$.

• 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/…. – jbowman Nov 5 '14 at 0:22
• Thanks. I should add that in my case the expectation does exist. – 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 $\nu$.