Useful exercises Brownian Motion and Stochastic Calculus

Let $(N^{(n)}(\cdot ){)}_{n\in \mathbb{Z}}$ be an i.i.d (identically independently distributed) family of Poisson process. We define

$$Y(t):=\sum _{n=0}^{\infty }{4}^{-n}{N}^{(n)}(3^{n}t)$$

Compute $E(Y(t))$

Solution

Here the important part is to notice that the Poisson distribution defined by

$$\mathbb{P}(\Gamma =k)=e^{-\lambda }\frac{{\lambda }^k}{k!}$$

Is always positive for all $\lambda$, thus the summands in the infinite sum are all positive and we can apply the monotonous convergence theorem, saying that the expectation of an infinite sums can be written as the infinite sum of expectations due to linearity of $E$. Thus,

$$E(Y(t))=E\left(\sum _{n=0}^{\infty }{4}^{-n}{N}^{(n)}(3^{n}t)\right)=\sum _{n=0}^{\infty }{4}^{-n}E(N^{(n)}(3^{n}t))$$

Now we need to find the expectation value of an arbitrary Poisson distributed r.v. (random variable).

$$E[N^{\lambda }]=\sum _{k=0}^{\infty }\mathbb{P}(N=k)=\sum _{k=0}^{\infty }{e}^{-\lambda }\frac{{\lambda }^k}{k!}\cdot k$$ $$=\sum _{k=1}^{\infty }{e}^{-\lambda }\frac{{\lambda }^{k-1}}{(k-1)!}\lambda =\lambda e^{-\lambda }\sum _{k^{\prime }=0}^{\infty }\frac{{\lambda }^{k^{\prime }}}{k!}=\lambda e^{-\lambda }{e}^{\lambda }=\lambda$$

With this knowledge, the expectation of $Y$ simplifies to

$$E(Y(t))=\sum _{n=0}^{\infty }{4}^{-n}{3}^{n}t=\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^{n}t=4t$$

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