My problem:
Suppose $Y$ is a real random variable with $Y \sim Exp(\lambda)$ and $Z$ is indipendent from $Y$, with density and $Z \geq 0$.
Why can I say $\mathbb{P}(Y-Z \geq t|Y>Z)=e^{-\lambda t}$?
My attempt:
I actually do not know how to proceed.
My problem:
Suppose $Y$ is a real random variable with $Y \sim Exp(\lambda)$ and $Z$ is indipendent from $Y$, with density and $Z \geq 0$.
Why can I say $\mathbb{P}(Y-Z \geq t|Y>Z)=e^{-\lambda t}$?
My attempt:
I actually do not know how to proceed.
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$P(Y >t+s)=e^{-\lambda(t+s)}=e^{-\lambda t} P(Y>s)$ for each $s$. Integrating w.r.t. the distribution $F_Z$ of $Z$ and using indepedence this becomes $P(Y>t+Z)=e^{-\lambda t} P(Y>Z)$. Divide by $P(Y>Z)$ to finish the proof.