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Math Help - Poisson Process - Integrability Condition

  1. #1
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    Poisson Process - Integrability Condition

    Hello everyone.

    I always seem to have troubles when it comes to proving an integrability condition of a particular process... somehow I don't have the right method.
    I have the following stochastic process:

    M(t) = N(t) - kt

    Where N(t) is a poisson process of parameter kt.
    M is called a compensated poisson process.

    I want to prove that: E[ | M(t) | ] < +infinity (in other words, that it is finite).

    I have searched for a while, and all my attempts have been infructuous.
    Thank you for your help.

    Jack
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  2. #2
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    Quote Originally Posted by JackOLantern View Post
    Hello everyone.

    I always seem to have troubles when it comes to proving an integrability condition of a particular process... somehow I don't have the right method.
    I have the following stochastic process:

    M(t) = N(t) - kt

    Where N(t) is a poisson process of parameter kt.
    M is called a compensated poisson process.

    I want to prove that: E[ | M(t) | ] < +infinity (in other words, that it is finite).

    I have searched for a while, and all my attempts have been infructuous.
    Thank you for your help.

    Jack
    Hi,

    well, N(t) is just a Poisson random variable of parameter kt, so that E[N(t)]=kt<\infty. Then, since |M(t)|=|N(t)-kt|\leq N(t)+kt, you can conclude E[|M(t)|]\leq E[N(t)]+kt<\infty. And E[M(t)]=0 by the way.

    I guess you were looking for something more "sophisticated"... The difficult question is probably the next one in your problem.
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  3. #3
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    Hi Laurent,

    Thank you very much.
    Stupid question: since k > 0 and t >= 0, why is kt < infinity?
    The weird thing is that I found the second question easier than this one (the second question was obviously proving the martingale characterization).
    When it comes to integrability condition, I am always a bit unsure about how to tackle the thing.

    Jack
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  4. #4
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    Quote Originally Posted by JackOLantern View Post
    Hi Laurent,

    Thank you very much.
    Stupid question: since k > 0 and t >= 0, why is kt < infinity?
    The weird thing is that I found the second question easier than this one (the second question was obviously proving the martingale characterization).
    When it comes to integrability condition, I am always a bit unsure about how to tackle the thing.

    Jack
    In my answer, t is fixed (as well as k). Did you mean you want a uniform bound for all t\geq 0? (I doubt this would be possible, since E[|N(t)-kt|^2]=kt\to_{t\to\infty}\infty)
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  5. #5
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    Hello Laurent,

    Okay I got it. k and t are fixed so the product is obviously finite.
    I guess this is the correct reasoning when you try to prove that some expression is finite.
    Would you have an example in which E[ | some stochastic process | ] is infinite? (so that I can picture it in my mind).

    Thanks a bunch

    Jack
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  6. #6
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    The next question asked to show that (M(t))_{t\geq 0} is a martingale, which means :
    1) E[|M(t)|]<\infty
    2) for any 0<s<t,...

    The assumption 1) is the one that makes assumption 2) make sense. That's probably why the first question was to show 1).

    Quote Originally Posted by JackOLantern View Post
    Would you have an example in which E[ | some stochastic process | ] is infinite? (so that I can picture it in my mind).
    Don't be impressed by the "stochastic process" thing: since t is fixed, you only deal with real random variables, no more no less.

    So what you're asking for is just an example of distribution with infinite mean. For instance, take a Cauchy distribution (density \frac{1}{\pi(1+t^2)}). If you want a "stochastic process", take the constant process X_t=X for all t, where X is Cauchy distributed... A more natural example would be a so-called Cauchy process (or other stable processes), but it would be unnecessarily complicated to try to define it here.
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  7. #7
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    Hi Laurent,

    Thank you very much for your help, you've made it clearer to me.

    Jack
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