Hi

Problem:

Let $\displaystyle M(t), t\geq 0 $ be a Poisson process with rate $\displaystyle \lambda $.

For $\displaystyle t \geq 0$ define $\displaystyle Y(t)=(-1)^{M(t)} $

Compute the distribution and the expected value of $\displaystyle Y(t) $ for $\displaystyle t > 0 $ .

I have calculated the probability of $\displaystyle M(t)$ being odd or even, which will control probability of $\displaystyle Y(t)$ being -1 or +1.

Fix t > 0

$\displaystyle P(M(t) \mbox{ is odd } ) = \displaystyle\sum_{n=1}^{\infty}e^{-\lambda t} \frac{(\lambda t)^{2n-1}}{(2n-1)!} = e^{-\lambda t}\sinh(\lambda t) = \frac{1}{2}\left(1-e^{-2 \lambda t}\right)

$

Similarily for probability of $\displaystyle M(t) $ being even.

How do I go about finding the distribution function, I assume I have found something like density function?

Thx