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Given a Poisson process $N(t),t\geq 0$ with rate $\lambda$ and another r.v. $T$ independent of $N(t)$ with mean $\mu$ and variance $\sigma^2$, I would like to compute the following quantities:

$$ \mathbb{Cov}(T,N(T)) \ \ \mbox{ and } \ \ \mathbb{Var}(N(T))$$

My guess is respectively: $\lambda \mu + \lambda \sigma^2$ and $\sigma^2\lambda$. But I am not sure it is correct nor how to justify some steps.

Anyone knows? Thank you very much!

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  • $\begingroup$ For the variance, use the conditional variance formula: $$\text{var}(N(T)) = E[\text{var}(N(T)\mid T] + \text{var}(E[N(T)\mid T).$$ This will show why your guessed answer is not correct (it is missing one term from the displayed expression above). $\endgroup$ – Dilip Sarwate Feb 28 '13 at 21:48
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Since $$ \mathbb{Cov}\left(T, N(T)\right) = \mathbb{E}\left(T N(T)\right) - \mathbb{E}(T) \mathbb{E}\left(N(T)\right) = \mathbb{E}\left(T \, \mathbb{E}\left(N(T)\mid T\right)\right) - \mathbb{E}(T) \, \mathbb{E}\left(\mathbb{E}\left(N(T)\mid T\right)\right) $$ But $\mathbb{E}\left(N(T) \mid T\right) = \lambda T$, thus $$ \mathbb{Cov}\left(T, N(T)\right) = \mathbb{E}\left(\lambda T^2\right) - \lambda \mathrm{E}(T)^2 = \lambda \mathbb{Var}(T) $$ Similarly: $$ \begin{eqnarray} \mathbb{Var}\left(N(T)\right) &=& \mathbb{E}\left( \mathbb{E}\left(N(T)^2|T\right) \right) - \mathbb{E}\left(\mathbb{E}\left( N(T) \mid T\right)\right)^2 \\ &=& \mathbb{E}\left( \lambda^2 T^2 + \lambda T\right) - \mathbb{E}\left(\lambda T\right)^2 \\ &=& \lambda^2 \mathbb{Var}(T) + \lambda \mathbb{E}(T) \end{eqnarray}$$

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  • $\begingroup$ Thanks a lot! Yes I see it, this is exactly what I did but something seemed strange to me.... Since we compose $N(T)$, it could be that we know the random variable $T$ but not $N$, what I mean is that $N(T) = N(T(\omega))(\tilde{\omega})$ where $\omega \in \Omega$, $\tilde{\omega} \in \tilde{\Omega}$ are the sample spaces of each variable. So when averaging out one, we should obtain a random variable? shouldn't we? $\endgroup$ – Dan Feb 28 '13 at 21:53
  • $\begingroup$ Yes, I think you are correct. The conditioning by $T$ indicates that we average over $\tilde\omega$ while keeping $\omega$ fixed, and then average over $\omega$. $\endgroup$ – Sasha Feb 28 '13 at 22:03

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