Determine the conditional probability $P\big( X>\frac{3}{4} \big| Y = \frac{1}{2}\big)$.

Question

Suppose that $X$ and $Y$ are random variables with joint density function

$$f_{X,Y}(x,y)=\begin{cases} 8xy \space \text{ for } 0<y<x<1\\ 0 \space\space\space\space\space\text{ otherwise.}\end{cases}$$

Determine the conditional probability $P\big( X>\frac{3}{4} \big| Y = \frac{1}{2}\big)$.

Is the answer $\frac{5}{12}$?

Answer 1

Guide:

Having a look at the density with fixed $y=\frac12$ we find the function:$$g_{X}\left(x\right)=f_{X,Y}\left(x,\frac{1}{2}\right)=\begin{cases} 4x & \text{for }\frac{1}{2}<x<1\\ 0 & \text{otherwise} \end{cases}$$

This function $g_{X}$ has in a certain sense the shape of the PDF of $X$ under condition $Y=\frac{1}{2}$.

However we cannot promote it to be this PDF because it is not a PDF.

Missing is that we do not have $\int g_{X}\left(x\right)dx=1$ as is necessary for a PDF.

This however can repaired taking $cg_{X}\left(x\right)$ as PDF where $c$ denotes a constant such that $\int cg_{X}\left(x\right)dx=1$.

Based on the equality $\int cg_{X}\left(x\right)dx=1$ we can easily find this $c$.

Answer 2

Hint

$$\mathbb P\{X\in A\mid Y=y\}=\int_A f_{X\mid Y=y}(x)\,\mathrm d x,$$ where $$f_{X\mid Y=y}(x)=\frac{f_{X,Y}(x,y)}{\int_{\mathbb R}f_{X,Y}(x,y)\,\mathrm d x}=\frac{f_{X,Y}(x,y)}{f_Y(y)}.$$