$\mu _{r+1} = r \lambda \mu_{r-1} + \lambda\frac{d\mu_{r}}{d\lambda}$ where $\mu_r = E[( X - E[X] )^r]$ and where $X$ is in Poisson distribution.

Question

I was given the following question in my assignment, and i am totally stumped as to how to proceed.

If $X$ is a Possion variate with mean $\lambda$ , show that

$\displaystyle \mu _{r+1} = r \lambda \mu_{r-1} + \lambda\frac{d\mu_{r}}{d\lambda} \text{ for } r = 1, 2, \ldots$

where $\mu_{r} = E[( X - E[X] )^r]$

A couple things here confuse me immensely. How exactly can I differentiate with respect to $\lambda$? Because as I understand it, it is a discrete value, and I thought differentiation was only for continuous variables. Also, Since $X$ is also discrete, I had the same confusion.

Answer 1

I found the solution to this problem. Am posting it for anyone that wants a straight forward algebraic proof. I had a hard time finding this , and felt stupid once i realized how simple it is. Nevertheless, hope this helps someone:

We start with differenciating the central moment

$$\mu_n = E[ (X - \lambda)^n ]$$

$$\mu_n = \sum_{x=0}^{\infty} \frac{(x-\lambda)^n \lambda^xe^{-\lambda}}{x!} $$

$$\frac{d\mu_n}{d\lambda} = \sum_{x=0}^{\infty} \frac{-n(x-\lambda)^{n-1} \lambda^xe^{-\lambda}}{x!} + \frac{x(x-\lambda)^n \lambda^{x-1}e^{-\lambda}}{x!} + \frac{-(x-\lambda)^n \lambda^xe^{-\lambda}}{x!}$$

$$\frac{d\mu_n}{d\lambda} = -n\mu_{n-1} + \sum_{x=0}^{\infty} \frac{(x-\lambda)^n \lambda^{x}e^{-\lambda}}{x!}(\frac{x}{\lambda} - 1)$$

$$\frac{d\mu_n}{d\lambda} = -n\mu_{n-1} + \frac{\mu_{n+1}}\lambda{}$$

The rest is trivial from here.