Finding the probability of $Y>1/4$ when conditioned on $X=x$.

Question

I am trying to find: $$\Pr(Y>1/4 \mid X = x) $$ where $X$ and $Y$ are two jointly continuous random variables with the following joint PDF: $$f_{X,Y} (x,y) = \begin{cases} 2xe^{x^2-y} & \text{if $0<x<1$ and $y>x^2$ } \\ 0 & \text{otherwise} \end{cases}$$

I am trying to use the formula $$\Pr(Y>1/4 \mid X=x) = \int_A f_{Y\mid X} (x,y) \,dy$$ where $$f_{Y\mid X} (x,y) = e^{x^2-y}$$

My main question is about the bounds of integration. I know that the $Y$ can vary between $1/4$ and $\infty$, but I do not know how to account for the fact that $y>x^2$ condition. Do I need to split this integral into pieces?

Thanks.

Answer 1

The way the conditional distribution of $Y$ given $X$ is found, is to integrate fixing the $x$ in $f$, so we are within the region $X = x$. However, note that if $P$ has marginal density zero at $x$, then the conditional density will also be zero.

That is, $$ P(Y \in A| X = x) = \frac{\int_A f(x,y)dy}{\int_{\mathbb R} f(x,y)dy} $$

for any subset $A \subset R$ and $x$ such that the denominator is non-zero. Otherwise the conditional probability is zero. In our case, $A = [\frac 14 , \infty)$.

Now, $\int_{\mathbb R} f(x,y)dy = 0$ for $x \leq 0$ or $x \geq 1$, so then the expression is zero. Else, $$\int_{\mathbb R} f(x,y) dy = \int_{x^2}^\infty 2xe^{x^2}e^{-y}dy = 2xe^{x^2}\int_{x^2}^\infty e^{-y}dy$$ is easily calculated.

Then, in similar fashion, $$ \int_{\frac 14}^\infty f(x,y)dy = 2xe^{x^2}\int_{\max\{x^2,\frac 14\}}^\infty e^{-y}dy $$

which is again very easily calculated, leading to the answer, following the cancellation of $2xe^{e^{x^2}}$ when we are dividing top by bottom.

Answer 2

Just use Bayes' Rule, and the Law of Total Probability

$$\begin{align}\mathsf P(Y>a\mid X=x)&=\int_a^\infty f_{\small Y\mid X}(y\mid x)~\mathrm d y\\[2ex]&=\int_a^\infty \dfrac{f_{\small X,Y}(x,y)}{f_{\small X}(x)}~\mathrm d y\\[2ex] &=\dfrac{\int_a^\infty f_{\small X,Y}(x,y)~\mathrm d y}{\int_\Bbb R f_{\small X,Y}(x,y)~\mathrm d y}\\[4ex]\text{so since}~f_{\small X,Y}(x,y)&=2x\mathrm e^{x^2}\mathrm e^{-y}\mathbf 1_{0\lt x\lt 1}\mathbf 1_{x^2\lt y}\\[4ex]\mathsf P(Y>1/4\mid X=x)&=\require{cancel}\dfrac{\cancel{2x\mathrm e^{x^2}}}{\cancel{2x\mathrm e^{x^2}}}\cdot\dfrac{\int_{\max\{1/4,x^2\}}^\infty \mathrm e^{-y}~\mathrm d y}{\int_{x^2}^\infty \mathrm e^{-y}~\mathrm d y}\cdot\mathbf 1_{0<x<1}\\[2ex]&=\dfrac{\int_{1/4}^\infty\mathrm e^{-y}~\mathrm d y}{\int_{x^2}^\infty \mathrm e^{-y}~\mathrm d y }\mathbf 1_{0<x<1/2}+\mathbf 1_{1/2\leqslant x\le 1}\\[2ex]&~~\vdots\end{align}$$