Given that $X,Y$ are two continuous Random variables with joint PDF given by $$f_{XY}(x, y)=\left\{\begin{array}{cl} \frac{1}{2} e^{-x}, & |y| \leq x, x>0 \\ 0 & \text { else } \end{array}\right.$$ Find $$f_{X \mid Y}(x \mid y)$$ My try:
I have figured out the marginal density function of $Y$ as :
$$f_{Y}(y)=\left\{\begin{array}{ll} \frac{1}{2} e^{-y}, & y>0 \\ \frac{1}{2} e^{y}, & y<0 \end{array}\right\}$$
Now we know that $$f_{X\mid Y}(x \mid y)=\frac{f_{XY}(x, y)}{f_{Y}(y)}$$
Now i am confused how to do the division?
You are almost there, simply express
$$f_Y(y)=\frac{e^{-|y|}}{2}$$
Thus
$$f_{X|Y}(x|y)=e^{|y|-x}$$
$x\ge|y|$
It's a truncated negative exponential