Find E[(X-mu)^k] - Normal Random Variable

[motherh]

Hi,

I'm having a bit of a problem with a probability question. The question is

Let X be a normal random variable with mean \mu and variance \sigma^{2}. Find E[(X -\mu)^{k}] for all k = 1,2,...

I'm not really sure what to do and need some help to confirm how to approach the question. I've tried messing around with moment generating functions, using the binomial theorem to expand E[(X -\mu)^{k}] and also have tried computing the expectation directly. Are any of these methods the correct way to progress?

Any help is much appreciated, thanks!

 

[economicsnerd]

Letting m_k= \mathbb E[(X-\mu)^k], we know m_1=0 by definition of \mu. Next, maybe try to use integration by parts (and the exact functional form of the p.d.f. \phi) to express m_k (k=2,3,...) in terms of m_1,...,m_{k-1}.

 

[motherh]

Hi, thanks for your help so far! I'm a little bit confused though. What do I integrate by parts exactly? And I'm a little bit thrown by "the exact functional form of the p.d.f. ϕ" - what do you mean? Apologies, I use ϕ as part of my notation for moment generating functions.

 

[motherh]

Anyone? I'm still really confused. Do I need a binomial expansion of (X-mu)^k in my answer somewhere?

 

[economicsnerd]

Hey. I was using the notation \phi(z) = \frac1{\sqrt{2\pi}}e^{-\frac{z^2}2}, the probability density function of \mathcal N(0,1), so that the the p.d.f. of a \mathcal N(\mu,\sigma^2) random variable is f(x)=\phi\left(\frac{x-\mu}{\sigma}\right).

Then, we can express the desired numbers as m_k = \int_{-\infty}^\infty (x-\mu)^k f(x)\text{ d}x.

 

[economicsnerd]

To clean up notation, let's notice that m_k = \sigma^k n_k, where n_k = \mathbb E \left[ \left( \frac{x-\mu}{\sigma} \right)^k \right]. So we can just work with the system n_k = \int_{-\infty}^\infty z^k\phi(z)\text{ d}z, which I would suggest attacking with IBP.

 

[motherh]

Ah, thank you so much! I got the relation m_(k+2) = sigma^2 * (k+1) * m_k, does that sound right?

With m_1 = 0 it means m_k = 0 for odd k which does seem to make sense.