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Let $X$ and $Y$ be two random variable with density $f_{X,Y}(x,y)=\begin{cases} \frac{1}{y}, & \text{for } 0<x<y<1, \\[8pt] 0, & \text{otherwise}.\end{cases}$.

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<y<1 $. Then $f_{X|Y}(x|y)=\begin{cases} \frac{\frac{1}{y}}{1}, & \text{for } 0<x<y<1, \\[8pt] 0, & \text {otherwise.} \end{cases}$. Is that correct? I am very unsure about that.

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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, $$ f_{X|Y}(x,y) = \frac{f_{X,Y}(x,y)}{f_Y(y)} $$

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  • $\begingroup$ Is $f_{X|Y}(x|y)$ already the conditional distribution? Or is it only the conditional density and I have to integrate $f_{X|Y}(x|y)$? $\endgroup$ – tommy_m Apr 23 at 18:25
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    $\begingroup$ @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. $\endgroup$ – gt6989b Apr 23 at 18:41

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