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.

marked as duplicate by leonbloy, Davide Giraudo, Emily, rschwieb, Asaf Karagila Feb 19 '13 at 18:52

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.

  • Since $X$ and $Y$ are iid, you already know the density of $X$. Perhaps you meant that $Y$ and $T$ are iid? – joriki Feb 18 '13 at 13:44
  • @joriki But $Y$ and $T$ are given to have different parameters, and so couldn't be iid, though they could be i. I suppose we will have await clarification from the OP. – Dilip Sarwate Feb 18 '13 at 14:33
  • @Dilip: Well, strictly speaking we could have $\delta=\lambda$, but you're right, that wouldn't make much sense. – joriki Feb 18 '13 at 14:34
  • 1
    Note that if $X \sim Exp(\lambda)$ and $Y=cX$, then $Y \sim Exp (\frac{\lambda}{c})$, see also… – Alex Feb 18 '13 at 15:59
  • Apologies for the typo - what I meant was that Y and T are independent, but not identical. – Mike Feb 18 '13 at 18:38

$F(x)=P(X\leqslant x)=P(Y-cT\leqslant x)=\int\limits_{0}^{+\infty}P(Y\leqslant ct+x,T=t)dt$

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 $\sum$.

Not the answer you're looking for? Browse other questions tagged or ask your own question.