This question already has an answer here:
Let $Y \sim \exp(\delta)$ and $T \sim \exp(\lambda)$, and $Y$ and $T$ are independent. How do I get the density $f(x)$ where $X=Y-cT$, $c>0$? Thanks.
This question already has an answer here:
Let $Y \sim \exp(\delta)$ and $T \sim \exp(\lambda)$, and $Y$ and $T$ are independent. How do I get the density $f(x)$ where $X=Y-cT$, $c>0$? Thanks.
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$F(x)=P(X\leqslant x)=P(Y-cT\leqslant x)=\int\limits_{0}^{+\infty}P(Y\leqslant ct+x,T=t)dt$
and because X,Y independent:$F(x)=\int\limits_{0}^{+\infty}P(Y\leqslant ct+x)P(T=t)dt$
you find this,find the derivative and you should be fine. You could also write it that way:$P(Y\leqslant ct+x,T=t)=P(Y\leqslant ct+x|T=t)P(T=t)$.
This also works for discrete rvs where instead of $\int$ we have $\sum$.