I'm struggling to see how I would manipulate the PDF so that I could find the conditional probability?
X is an exponential RV with $\lambda>0$ and the pdf=$\lambda e^{-\lambda x}$
2026-03-26 03:12:16.1774494736
How do I find P(X > x + y | X > x), y > 0 when X is an exponential RV
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You have $$ \mathbb{P}[X > x+y | X>x] = \frac{\mathbb{P}[X > x+y, X>x]}{\mathbb{P}[X>x]} = \frac{\mathbb{P}[X > x+y]}{\mathbb{P}[X>x]} = \frac{1-F(x+y)}{1-F(x)}. $$ where $F$ is the cdf of the r.v. in question. Can you compute $F$ from your pdf and simplify the resulting expression? It looks quite nice in the end...