# Conditional Probability given Joint Density Problem

X and Y are random variables with joint density function $$f_{xy}(x , y)= \begin{cases} 8xy, & 0 \lt y \lt x \lt 1 \\ 0,& \text{otherwise} \end{cases}$$ Find $P(X \gt \frac{3}4 \mid Y = \frac14)$

Working: I have calculated $$\ f_{x \mid y}\left(x\mid\frac14\right) = \frac{32}{15}x$$ and know that $$P\left(X \gt \frac{3}4 \mid Y = \frac14\right) =\int_{-\infty}^{\infty} \ f_{x \mid y}\left(x\mid\frac{1}4\right) \ dx$$

But what I don't understand is what to put as the limits of integration! Any help is appreciated!

• by $X>\frac33$ you mean $X>1$? – Jimmy R. Nov 18 '15 at 11:28
• @Stef No, sorry! I have edited it now! thanks – jessicarevelo Nov 18 '15 at 11:34
• To find the conditional PDF (i.e. finding constant $\frac{32}{15}$ I mean) you correctly practicized the limits $\frac14$ and $1$. Actually the PDF takes value $0$ beyond those limits. You should use them again, but now with $\frac34$ in stead of $\frac14$. – drhab Nov 18 '15 at 11:39

When you calculate a function (here $f_{X\mid Y=1/4}$) try always writting also its domain. Here the domain of $f_{X\mid Y=1/4}$ is: $$f_{X\mid Y=1/4}(x \mid 1/4)=\begin{cases}\frac{32}{15}x, & 1/4<x<1 \\0, & \text{else } \end{cases}$$ Now you see that $$P(X >3/4 \mid Y=1/4)=\int_{3/4}^{1}\frac{32}{15}x\ dx$$
If $g$ denotes the conditional PDF then it is prescribed by $x\mapsto\frac{32}{15}x$ if $\frac14<x<1$ and $x\mapsto0$ otherwise.
$$P(X>\frac34|Y=\frac14)=\int\chi_{(\frac34,\infty)}g(x)dx=\int_{\frac34}^1\frac{32}{15}xdx$$