# Find the conditional distribution of $X$ given that $Y=y$

Let $$X$$ and $$Y$$ be two random variable with density $$f_{X,Y}(x,y)=\begin{cases} \frac{1}{y}, & \text{for } 0.

To find the conditional distribution of $$X$$ given that $$Y=y$$ first I calculated: $$f_Y(y)=\int_0^y \frac{1}{y}dx = 1$$ for $$0. Then $$f_{X|Y}(x|y)=\begin{cases} \frac{\frac{1}{y}}{1}, & \text{for } 0. Is that correct? I am very unsure about that.

Yes, just like for 2 events $$A,B$$ you have $$\mathbb{P}[A|B] = \frac{\mathbb{P}[A \cap B]}{\mathbb{P}[B]}$$ so too more generally in terms of random variables, $$f_{X|Y}(x,y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$$
• Is $f_{X|Y}(x|y)$ already the conditional distribution? Or is it only the conditional density and I have to integrate $f_{X|Y}(x|y)$? – tommy_m Apr 23 at 18:25
• @tommy_m Not sure what you mean by conditional distribution. The usual notation denotes the pdf as $f(\cdot)$ and the cdf as $F(\cdot)$, both can characterize the distribution, and so can the mgf $M_X(t)$... To be specific, in this case $f_{X|Y}(x,y)$ denotes the conditional pdf of $X$ given $Y$, but if you need the conditional cdf, you would indeed need to integrate it. – gt6989b Apr 23 at 18:41