CMP Chapter 3 Loss Distributions Q3.3

[mcallist]

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

 

[Bharti Singla]

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

 

[vgarg]

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 \(\mathbb{E}[X^3]=\dfrac{\alpha (\alpha +1)(\alpha+2)}{\lambda ^3}\)

 

[mcallist]

Thanks very much Bharti and Vgarg. I can understand it now. Cheers