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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
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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
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@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
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@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
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1Note that if $X \sim Exp(\lambda)$ and $Y=cX$, then $Y \sim Exp (\frac{\lambda}{c})$, see also math.stackexchange.com/questions/115022/… – Alex Feb 18 '13 at 15:59
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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 $\sum$.
