Given the joint pdf $f_{X,Y}(x,y) = 2e^{-(x+y)}$, $0 \leq x \leq y$,
$ y\geq 0$. . Find $P(Y < 1| X = 1)$.
Attempt: $P(Y < 1| X = 1) \frac{P(Y<1, X = 1)}{P(X = 1)}$
Can someone please help me setting the problem? Thank you.
2026-04-01 21:56:50.1775080610
Given the joint pdf $f_{X,Y}(x,y) = 2e^{-(x+y)}$, $0 \leq x \leq y$, $ y\geq 0$. . Find $P(Y < 1| X = 1)$.
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In this case the result is immediate based on the support of the joint distribution. The support of $f_{X,Y}$ is $$0\le x\color{blue}{\le y}$$ which means that $Y$ takes (with positive probability) only values that at least equal as the value of $X$. So, if $X=1$ then $Y$ takes with positive probability only values $y$ such that $$1\color{blue}{\le y}$$ or equivalently $$P(Y\ge 1\mid X=1)=1$$ which implies that the required probability is equal to $0$.