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Math Help - Find the Limiting Distribution of X_n/n With a given MGF

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    Find the Limiting Distribution of X_n/n With a given MGF

    I am having extreme difficulty solving the following problem;

    Let X_n have mgf given by

    M_xn (t) = (1 - 2 at - nat^2 + na^2t^3)^-n/2, t<1/a , a>0

    Find the limiting distribution for X_n?n.

    Any help on this one would be appreciated. Not even sure where to really begin on this....
    Last edited by mr fantastic; April 22nd 2009 at 01:24 PM. Reason: Restored question deleted by the OP
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    Hello,
    Quote Originally Posted by roresc View Post
    I am having extreme difficulty solving the following problem;

    Let X_n have mgf given by

    M_xn (t) = (1 - 2 at - nat^2 + na^2t^3)^-n/2, t<1/a , a>0

    Find the limiting distribution for X_n?n.

    Any help on this one would be appreciated. Not even sure where to really begin on this....
    So a method would be to find the mgf of \frac{X_n}{n} (let's call it Y_n) and then find its limit as n goes to infinity.

    \begin{aligned}<br />
M_{Y_n}(t)<br />
&=\mathbb{E}\left(e^{tY_n}\right)=\mathbb{E}\left(  e^{t \cdot \frac{X_n}{n}}\right)=\mathbb{E}\left(e^{X_n \cdot \frac tn}\right)=M_{X_n}\left(\tfrac tn\right) \\<br />
&=\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)^{-n/2}<br />
\end{aligned}


    That's some start


    Hint :
    Spoiler:
    take the logarithm of M_{Y_n}(t)


    If you want more (not necessarily with a method you would know) :
    Spoiler:

    \ln(M_{Y_n}(t))=-\frac n2 \ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)

    Since there's a coefficient n in front of the logarithm, it is sufficient to get a o\left(\frac 1n\right) (asymptotic equivalence)

    But we know that \ln(1+x)=x+o(x)

    So \ln\left(1-2a \frac tn-a \frac{t^2}{n}+a^2 \frac{t^3}{n^2}\right)=-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right) (the \frac{t^3}{n^2} can be omitted, because the power of n is smaller than 1/n)

    So \ln(M_{Y_n}(t))=-\frac n2 \left(-2a \frac tn-a \frac{t^2}{n}+o\left(\frac 1n\right)\right)=at+a\frac{t^2}{2}+o(1)

    Hence, as n goes to infinity, \ln(M_{Y_n}(t)) \to at+a \frac{t^2}{2}

    This gives you the limit for M_{Y_n}(t)
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