# Find the distribution

• Nov 23rd 2009, 10:03 AM
Twig
Find the distribution
Hi

Problem:

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

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

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

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

Fix t > 0
$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 $M(t)$ being even.

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

Thx
• Nov 23rd 2009, 10:20 AM
Moo
Hello,

$P(Y(t)=-1)=\tfrac 12 \cdot (1-e^{-2\lambda t})$

$P(Y(t)=1)=\tfrac 12 \cdot (1+e^{-2\lambda t})$

Isn't there something like a Bernoulli distribution ? :P
• Nov 23rd 2009, 10:44 AM
Twig
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 $e^{-2 \lambda t}$ .

Could this be correct? It seems reasonable that the expected value tends to zero as $t \to \infty$ , but also that the expected value is not zero when t is small, because it is more likely that $M(t)$ is even for small t.
• Nov 23rd 2009, 10:53 AM
Moo
I don't understand your problem actually.

Y(t) can only take two values : 1 and -1.
Let $p(t)=\tfrac 12 \cdot \left(1-e^{-2\lambda t}\right)$

We have $P(Y(t)=-1)=p(t)$ and $P(Y(t)=1)=1-p(t)$

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 $p(x,t)=\tfrac 12 \cdot \left(1-e^{-2\lambda t}\right) \cdot \bold{1}_{\{x=-1\}}+\tfrac 12 \cdot \left(1+e^{-2\lambda t}\right) \cdot \bold{1}_{\{x=1\}}$ if you really want to, but it's not really necessary...
• Nov 23rd 2009, 11:07 AM
Twig
Hi

You understand it correctly, and yes it should be a bernoulli distribution.

Can one say that $Y(t)$ has independant and stationary increments?