Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\{X(t), t\ge 0\}$ and $\{Y(t),t\ge 0\}$ be independent Poisson processes with parameters $\lambda_1$ and $\lambda_2$, respectively. Define $Z_1(t)=X(t)+Y(t)$, $Z_2(t)=X(t)-Y(t)$, $Z_3(t)=X(t)+k$, $k$ a positive integer. Determine which of the above processes are Poisson and find $\lambda$.

Any help is appreciated! This is not homework! Thanks.

share|cite|improve this question

1 Answer 1

up vote 4 down vote accepted

The last one is not Poisson since $Z_{3}(0) \neq 0$. Consider $t_1 \leq t_2 \leq t_3 \leq t_4$. Then look at $Z_{1}(t_2)-Z_{1}(t_1), Z_{1}(t_4)-Z_{1}(t_3)$ and see if they are independent. Do the same for $Z_{2}(t)$. Then you have to look at stationary increments.

share|cite|improve this answer
@PEV Do the same for $Z_2(t)$? Hmmm... Rather, check the sign of the variations of $Z_2$, I would say. – Did Mar 5 '11 at 17:19
@Didier Piau: Can you explain more on what you said? Thanks! – kira Mar 5 '11 at 19:30
@PEV: Can you also do some detailed explanation? Thanks! – kira Mar 5 '11 at 19:33
@kira Since the process $Z_2$ has independent increments, one must rely on another property to show that $Z_2$ fails being Poisson. For example, the fact that the random function $t\mapsto Z_2(t)$ is not almost surely nondecreasing. [Re $Z_3$, if ever this question was, after all, part of some homework of yours, you should check whether your professor imposed the condition $Z(0)=0$ for a process $(Z(t))_{t\ge0}$ to be Poisson, or not (most people do but not everybody does).] – Did Mar 6 '11 at 20:56
@kira: For the covariance, the only non-trivial part is the calculation of ${\rm E}[X(t) X(t+h)]$. For this purpose, write ${\rm E}[X(t)X(t + h)]$ as ${\rm E}[X(t)(X(t) + X(t + h) - X(t))]$, and note that $X(t)$ and $X(t + h) - X(t)$ are independent (further note that ${\rm E}[X(t)X(t)]$ can be found from ${\rm Var}[X(t)]$). – Shai Covo Mar 8 '11 at 0:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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