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Thread: Find the distribution

  1. #1
    Senior Member Twig's Avatar
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    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)^{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
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  2. #2
    Moo
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    Hello,

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

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


    Isn't there something like a Bernoulli distribution ? :P
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  3. #3
    Senior Member Twig's Avatar
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    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.
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  4. #4
    Moo
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    I don't understand your problem actually.

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

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

    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 $\displaystyle p(x,t)=\tfrac 12 \cdot \left(1-e^{-2\lambda t}\right) \cdot \bold{1}_{\{x=-1\}}+\tfrac 12 \cdot \left(1+e^{-2\lambda t}\right) \cdot \bold{1}_{\{x=1\}}$ if you really want to, but it's not really necessary...
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  5. #5
    Senior Member Twig's Avatar
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    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?
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