Hi there, I'm having trouble fully understanding the steps in differentiating the MGF of the Gamma distribution 3 times to get E(X^3). Could someone please outline the steps involved? I'd be most grateful. Thank you
It is often good idea to expand the expression for MGF to retrieve the moments about zero as differentiation of higher order could be a daunting task. First lets start by the definition of MGF of a rv \(X\), \begin{align*} M_X(t) &= \mathbb E\left [e^{Xt} \right ]\\ &= \mathbb E\left [ 1+Xt + \frac{(Xt)^2}{2!} + ...+ \frac{(Xt)^k}{k!}+\ldots \right ] \quad \text{By using the expansion of } e^x \\ &= 1+ t \mathbb E\left [X\right ] + \frac{t^2 \mathbb E\left [X^2 \right ]}{2!} + ...+ \frac{t^k \mathbb E\left [X^k\right ]}{k!}+\ldots \end{align*} So what do we see is that if we expand MGF of an rv \(X\) then the \(r^{th}\) raw moment of \(X\) i.e. \(\mathbb E\left [X^r\right ] \) is the coefficient of \(\dfrac{t^r}{r!}\) Now lets come to the original problem of finding the the third raw moment of a gamma distribution. The MGF of gamma rv \(X \sim Ga(\alpha, \lambda)\) \(M_X(s) = \left ( 1-\frac{t}{\lambda } \right )^{-\alpha}\) Expanding the above series we get, \begin{align*} M_X(t) &= 1 + \frac{\alpha}{\lambda}\frac{t}{1!} + \frac{\alpha (\alpha +1)}{\lambda ^2}\frac{t^2}{2!} +\frac{\alpha (\alpha +1)(\alpha+2)}{\lambda ^3}\frac{t^3}{3!} +\ldots + \frac{\alpha (\alpha +1)\ldots (\alpha + k-1)}{\lambda ^k}\frac{t^k}{k!}+ \ldots \end{align*} As can be seen in the above expression \(\mathbb{E}[X^3]=\dfrac{\alpha (\alpha +1)(\alpha+2)}{\lambda ^3}\)
