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Thread: Moment Generating Function question

  1. #1
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    Moment Generating Function question

    Find the moment generating funtion of Yn = Xn/n, where Xn has a gamma distribution with parameters alpha = n, and beta.

    $\displaystyle E(e^{tY_n}) = E(e^{t\frac{X_n}{n}})$ I'm stuck here. I know that the MGF of a gamma (in this case) is $\displaystyle (1-\beta t)^{-n}$.

    The question I have is what to do with the 1/n. I'm pretty sure it works out to be $\displaystyle (1-\frac{\beta t}{n})^{-n}$, but I'm not sure why.

    Would someone be able to explain this to me?

    Thanks in advance.
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    Quote Originally Posted by statmajor View Post
    Find the moment generating funtion of Yn = Xn/n, where Xn has a gamma distribution with parameters alpha = n, and beta.

    $\displaystyle E(e^{tY_n}) = E(e^{t\frac{X_n}{n}})$ I'm stuck here. I know that the MGF of a gamma (in this case) is $\displaystyle (1-\beta t)^{-n}$.

    The question I have is what to do with the 1/n. I'm pretty sure it works out to be $\displaystyle (1-\frac{\beta t}{n})^{-n}$, but I'm not sure why.

    Would someone be able to explain this to me?

    Thanks in advance.
    You know that $\displaystyle E(e^{tY_n}) = \frac{1}{\Gamma (n) \beta^n} \int_0^{+\infty} e^{tx/n} x^{n-1} e^{-x/\beta} \, dx$.

    Substitute $\displaystyle y = \frac{x}{n}$ and after some re-arranging the integral becomes $\displaystyle \frac{1}{\Gamma (n) \gamma^n} \int_0^{+\infty} e^{yt} y^{n-1} e^{-x/\gamma} \, dy$ where $\displaystyle \gamma = \frac{\beta}{n}$.

    This is recognised as the MGF for a random variable that has a gamma distribution with parameters $\displaystyle \alpha = n$ and $\displaystyle \gamma$, that is, $\displaystyle (1-\gamma t)^{-n}$. Now substitute back $\displaystyle \gamma = \frac{\beta}{n}$.



    Alternatively (and much easier), you could use the well known theorem that if X has a MGF $\displaystyle m_X(t)$ then $\displaystyle Y = aX + b$ has a MGF given by $\displaystyle m_Y(t) = e^{bt} m_X(at)$.
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