Results 1 to 5 of 5

Math Help - Find the distribution

  1. #1
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    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)<br />

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    Hi

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

    Can one say that Y(t) has independant and stationary increments?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. find distribution
    Posted in the Advanced Statistics Forum
    Replies: 7
    Last Post: June 27th 2010, 02:14 PM
  2. find asympto. distribution of MLE
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 24th 2009, 11:37 PM
  3. Find the distribution
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: November 2nd 2009, 02:41 PM
  4. find mle of this distribution
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 20th 2009, 02:39 PM
  5. [SOLVED] How To Find The Distribution
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 25th 2005, 03:33 AM

Search Tags


/mathhelpforum @mathhelpforum