0
$\begingroup$

during an exercise on Martingales I came across the following request: let $X_n \sim \mathcal{E}(n)$ be a sequence of independent random variables, so the pdf is $F_{X_n}= 1 - \exp(-nt)$. Let $S_0 = 0, S_n = X_1+\ldots+X_n$, compute successively: $$ f_{S_1,S_2}(s,t) \text{ for } t\geq s \geq 0, f_{S_2}(t) \text{ for } t\geq 0 \text{ and } \mathbb{E}(S_1 | S_2) $$

I tried doing the following: $f_{x_1,x_2} = 2 e^{-x_1}\cdot e^{-2x_2} = 2 e^{-x_1-2x_2}$ . The transformation should be $s_1 = x_1, s_2=x_2+x_1\implies x_2 = s_2-s_1$. The determinant of the Jacobian is $2$, which leads me to : $$ f_{s_1,s_2}(s,t)= 4\cdot e^{-2t+s}$$ But then, how do I derive the marginal? When I integrate by $ds$ I get $+\infty$...did I do something wrong?

$\endgroup$
0
$\begingroup$

Your formula for $f_{s_1,s_2} (s,t)$ is valid only for $s\leq t$ since $s_1 \leq s_2$. $f_{s_1,s_2} (s,t)=0$ for $s>t$. Now you don't end up with a divergent integral, right?

$\endgroup$
  • $\begingroup$ I understand the reasoning and intuitively I can also see that it works, but when I switch to the integration, is this the correct way of doing it: $f_{s_2}(t) = 4\cdot e^{-2t} \int_0^t e^s ds = 4(e^t-e^{-2t})? $. Thanks for the answer! $\endgroup$ – user1868607 Jun 26 '18 at 9:06
  • $\begingroup$ It is $4(e^{-t} -e^{-2t})$. $\endgroup$ – Kavi Rama Murthy Jun 26 '18 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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