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Thread: MGF of sample mean of IID poisson RVs

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    MGF of sample mean of IID poisson RVs




    I am given the MGF of IID RVs as e^{\lambda(e^t-1)} and am supposed to find the MGF of their sample mean of n such RVs.
    I have am arriving at

    M_n(t) = e^{n\lambda(e^t - 1)+1}

    This doesn't appear to be a valid MGF and there is another part to the question where it asks for the
    \lim_{n \to \infty} M_n(t)
    I am confused where might i be making a mistake and if it is proper to have the MGF running to infinity.
    Thank You

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    Re: MGF of sample mean of IID poisson RVs

    What is the equation for MGF of sample mean of n RVs? Is it [Mx(t/n)]^n ?

    Because if I use the above, then I am getting e^{n\lambda(e^{t/n} - 1)} My attempt
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

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    Re: MGF of sample mean of IID poisson RVs

    Quote Originally Posted by nsus View Post
    I am given the MGF of IID RVs as e^{\lambda(e^t-1)} and am supposed to find the MGF of their sample mean of n such RVs.
    I have am arriving at

    M_n(t) = e^{n\lambda(e^t - 1)+1}

    This doesn't appear to be a valid MGF and there is another part to the question where it asks for the
    \lim_{n \to \infty} M_n(t)
    I am confused where might i be making a mistake and if it is proper to have the MGF running to infinity.
    Thank You
    Well, Buckeye's close, however, you should end up with:

    M_n(t) = ( Exp [ (Lambda/n) * ((Exp[t] - 1)) ] )^n.

    Taking the limit as n->Infinity will yield the MGF you provided above in your first sentence. I hope this helps.
    Last edited by Dragan; 02-13-2017 at 04:51 AM.

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    Re: MGF of sample mean of IID poisson RVs

    Thanks for your replies,

    I have arrived at a result similar to Buckeye and Dragan's result eludes me.
    I used M_Y(t)=\prod M_X(\frac{t}{n}) , where Y is the sample mean.
    But i still can't figure out how should an mgf look as n tends to infinity.
    Any pointers will be appreciated.
    Last edited by nsus; 02-13-2017 at 08:12 AM.

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    Re: MGF of sample mean of IID poisson RVs

    Quote Originally Posted by nsus View Post
    Thanks for your replies,

    I have arrived at a result similar to Buckeye and Dragan's result eludes me.
    I used M_Y(t)=\prod M_X(\frac{t}{n}) , where Y is the sample mean.
    But i still can't figure out how should an mgf look as n tends to infinity.
    Any pointers will be appreciated.
    Well, if you take the limit of the functions that either you or Buckeye provided as n-> Infinity, then the answer is Zero. Thus, if that is a satisfactory answer to you, then that is just fine with me.

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    Re: MGF of sample mean of IID poisson RVs


    Quote Originally Posted by Dragan View Post
    Well, if you take the limit of the functions that either you or Buckeye provided as n-> Infinity, then the answer is Zero. Thus, if that is a satisfactory answer to you, then that is just fine with me.
    Thank you so much for your time.
    I did a bit of reading and figured that an MGF of One would imply the RV to be equal to zero. I used the Taylor's series to solve for the limit and found that

    M_y(t/n) = e^{\lambda t} as n \to \infty.
    This seems to be an acceptable solution.
    Thanks for your time once again.

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