# Thread: Moment Generating Function question

1. ## 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.

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

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

Would someone be able to explain this to me?

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

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

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

Would someone be able to explain this to me?

You know that $E(e^{tY_n}) = \frac{1}{\Gamma (n) \beta^n} \int_0^{+\infty} e^{tx/n} x^{n-1} e^{-x/\beta} \, dx$.
Substitute $y = \frac{x}{n}$ and after some re-arranging the integral becomes $\frac{1}{\Gamma (n) \gamma^n} \int_0^{+\infty} e^{yt} y^{n-1} e^{-x/\gamma} \, dy$ where $\gamma = \frac{\beta}{n}$.
This is recognised as the MGF for a random variable that has a gamma distribution with parameters $\alpha = n$ and $\gamma$, that is, $(1-\gamma t)^{-n}$. Now substitute back $\gamma = \frac{\beta}{n}$.
Alternatively (and much easier), you could use the well known theorem that if X has a MGF $m_X(t)$ then $Y = aX + b$ has a MGF given by $m_Y(t) = e^{bt} m_X(at)$.