
Find the distribution
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)^{2n1}}{(2n1)!} = e^{\lambda t}\sinh(\lambda t) = \frac{1}{2}\left(1e^{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

Hello,
$\displaystyle P(Y(t)=1)=\tfrac 12 \cdot (1e^{2\lambda t})$
$\displaystyle P(Y(t)=1)=\tfrac 12 \cdot (1+e^{2\lambda t})$
Isn't there something like a Bernoulli distribution ? :P

How would I go about finding the distribution function from this. I looked at the bernoulli distribution, I didnīt really see how I could use it for my particular problem though. Help on this area would be appreciated.
Also, my attempt on the expected value gave me $\displaystyle e^{2 \lambda t} $ .
Could this be correct? It seems reasonable that the expected value tends to zero as $\displaystyle t \to \infty $ , but also that the expected value is not zero when t is small, because it is more likely that $\displaystyle M(t) $ is even for small t.

I don't understand your problem actually.
Y(t) can only take two values : 1 and 1.
Let $\displaystyle p(t)=\tfrac 12 \cdot \left(1e^{2\lambda t}\right)$
We have $\displaystyle P(Y(t)=1)=p(t)$ and $\displaystyle P(Y(t)=1)=1p(t)$
And this is exactly a Bernoulli distribution over {1,1} with parameter p(t)
Why do you absolutely want a distribution function ? You can write it $\displaystyle p(x,t)=\tfrac 12 \cdot \left(1e^{2\lambda t}\right) \cdot \bold{1}_{\{x=1\}}+\tfrac 12 \cdot \left(1+e^{2\lambda t}\right) \cdot \bold{1}_{\{x=1\}}$ if you really want to, but it's not really necessary...

Hi
You understand it correctly, and yes it should be a bernoulli distribution.
Can one say that $\displaystyle Y(t)$ has independant and stationary increments?