Conditional probability $P(Y>X^2|X+Y)$

Question

Random variables $X$ and $Y$ are independent with the same exponential distribution. Calculate $P(Y>X^2 | X + Y)$. I have no idea how to go about it.

Answer 1

We assume that the variables $X,Y$ are i.i.d following the law $h$. We can write down the conditional density as follows

$$F(X=x|X+Y=a)=F(X=x|Y=a-X)=\frac{f_{X,Y}(x,a-x)}{f_{X+Y}(a)}=\frac{h(x)h(a-x)}{\int_{-\infty}^{\infty}h(x)h(a-x)dx}$$

Note now that for $a>-1/4:$

$$P(Y>X^2|X+Y=a)=P(X^2+X-a<0|X+Y=a)=P(X\in\left(\frac{-1-\sqrt{1+4a}}{2},\frac{-1+\sqrt{1+4a}}{2}\right)|X+Y=a)\\=\frac{\int_{\frac{-1-\sqrt{1+4a}}{2}}^\frac{-1+\sqrt{1+4a}}{2}h(x)h(a-x)dx}{\int_{-\infty}^{\infty}h(x)h(a-x)dx}$$

and finally, noting that $Y>X^2$ is impossible if $a<-1/4$

$$P(Y>X^2|X+Y=a)=\begin{Bmatrix}0&,~~a\leq -1/4\\\frac{\int_{\frac{-1-\sqrt{1+4a}}{2}}^\frac{-1+\sqrt{1+4a}}{2}h(x)h(a-x)dx}{\int_{-\infty}^{\infty}h(x)h(a-x)dx}&,~~a>-1/4\end{Bmatrix}$$