Moving an expectation inside a probability? Let $X$ and $Y$ be two independent random vectors such that $E[X^TY]>0$, and all components of $X$ are positive valued. 
Then, is it true that, $P_{X,Y}\{X^T Y > 0\} \le P_{Y}\{E_X[X]^T Y > 0\}$ ?
 A: For the simple case of vectors of length $1$, the question is

For independent random variables $X$ and $Y$ such that $P\{X > 0\} = 1$ and
  $E[XY] > 0$, is it true that
  $$P\{XY > 0\} \leq P\{E[X]Y > 0\}~??$$

and the answer is Yes, and in fact the inequality is an equality, with
both probabilities having value $P\{Y > 0\}$.
To see this, note that
$$\begin{align}
P\{XY > 0\} &= P\{X > 0, Y > 0\} + P\{X < 0, Y < 0\}\\
&= P\{X > 0, Y > 0\} & \scriptstyle{\text{because}~ P\{X<0\}=0}\\
&= P\{X > 0\}P\{Y > 0\}&\scriptstyle{\text{since $X$ and $Y$ are independent}}\\
&= P\{Y > 0\}&\scriptstyle{\text{since P\{X > 0 = 1\}}}
\end{align}$$
while $X$ being positive means that $E[X] > 0$, and so we have that
$$P\{E[X]Y > 0\} = P\left\{Y > \frac{0}{E[X]}\right\} = P\{Y > 0\}.$$
Note that the condition $E[XY]>0$ was not used anywhere in the proof.
I will leave it to you to figure out whether this proof can be extended to
random vectors with more than one component. Indeed, without any
information about the distributions of the $X_i$ and $Y_i$ other than
that $(X_1,X_2,\ldots, X_n)$ is independent of $(Y_1,Y_2,\ldots, Y_n)$,
 $$E\left[X^TY\right]=\sum_{i=1}^n E[X_iY_i]=\sum_{i=1}^n E[X_i]E[Y_i] > 0$$
where all the $E[X_i]$ are positive numbers, the quantity 
$$P\{X^TY > 0\} = P\{X_1Y_1+X_2Y_2+\cdots+X_nY_n > 0\}$$
seems to be difficult to estimate or bound.
A: The inequality for $n=1$ obviously holds, even without the independance assumption, because both events $\{XY>0\}$ and $\{E[X]Y>0\}$ equal the event $\{Y>0\}$.
For $n=2$, sometimes the inequality is true, sometimes it is wrong. Consider $n=2$ and $X_2=Y_2=1$, and denote by $F$ the cdf of $Y_1$. Then, using independence between $X_1$ and $Y_1$, one has $\Pr(X_1Y_1+X_2Y_2)=\Pr(Y_1>-1/X_1)=1-E[F(W)]$ where $W \sim -1/X_1$. On the other hand, $\Pr(E(X_1)Y_1+E(X_2)Y_2)=1-F(-1/E(X_1))$. Now take $X_1$ non-constant and such as $E(1/X_1)=1/E(X_1)$, for example $X_1\sim\text{Beta}'(2,2)$ (beta-prime distribution). Then the problem consists in comparing $E[F(W)]$ with $F(E[W])$. If $\Pr(Y<0)=0$ then both quantities are zero. Otherwise, by Jensen's inequality,  $F(E[W])< E[F(W)]$ when $F$ is strictly convex on $(-\infty,0)$, and the inequality is reversed when $F$ is strictly concave $(-\infty,0)$. It remains to show that there exist such examples of $F$ such that the additional assumption $E[X^TY]=E[X_1]E[Y_1]+1>0$, for example taking $E[Y_1]=0$. This should cause no difficulty to find an example of strictly convex $F$ on $(-\infty,0)$. However this is not possible to get an example of strictly concave $F$ on $(-\infty,0)$; in this case one could follow the same idea but taking $\Pr(X_1>\epsilon)=1$, then taking an example of $F$ strictly concave on $(\frac{-1}{\epsilon}, 0)$. 
I'm sorry my answer is not very clean. This is not a complete answer, I just hope these hints are helpful.
