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Let $\mu,\alpha_n:\mathbb R^+\to \mathbb R$ continuous function with $\mu$ bounded function. Let $N^{(n)}$ the trajectory of a Poisson process with intensity $(\alpha_n \mu)(t)$. Let $0=T_0^{(n)}<T_1^{(n)}<..$ jumps of $N^{(n)}$.

Let $M_n(t)=\sum_{i=1}^{N_t^{(n)}} \frac {1} {\alpha_n (T_i^{(n)})}$

Show $M_n(t)$ is an unbiased estimator of $M(t)=\int_0^t \mu (s) ds$.

My idea: I calculate $E(M_n(t)|N^n=u)$ and I find: $E(M_n(t)|N^n=u)=u \frac {1} {\int_0^t (\alpha_n \mu)(s)ds} M(t)$

$E(E(M_n(t)|N^n=u))=E(M_n(t))$. But $E(M_n(t))-M(t)=u \frac {1} {\int_0^t (\alpha_n \mu)(s)ds} M(t)-M(t)$ is not equal to 0.

Can you help me?

Davide Giraudo
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user298629
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  • I would appreciate you posting your work to show that $$E(M_n(t)|N^n_t = u) = \frac{u}{\int_0^t \alpha_n \mu (s) ,ds} M(t)$$. – Tom Dec 14 '15 at 14:59

1 Answers1

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You're close, it seems. The issue is that you've written $$ E[E[M_n(t) | N^n_t = u] ] = E[M_n(t)] $$ which is not true. What is true is that $$ E[E[M_n(t) | N^n_t] ] = E[M_n(t)] $$ With that in mind, you found that $$ E[M_n(t)|N^n_t = u] = \frac{u}{\int_0^t \alpha \mu (s) \, ds} M(t) $$ which implies that $$ E[M_n(t)|N^n_t] = \frac{N^n_t}{\int_0^t \alpha \mu (s) \, ds} M(t) $$ Therefore, $$ E[E[M_n(t) | N^n_t] ]= \frac{E[N^n_t]}{\int_0^t \alpha \mu (s) \, ds} M(t) = M(t). $$

Tom
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