Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Since integration is not my strong suit I need some feedback on this, please:

Let $Y$ be $\mathcal{N}(\mu,\sigma^2)$, the normal distrubution with parameters $\mu$ and $\sigma^2$. I know $\mu$ is the expectation value and $\sigma$ is the variance of $Y$.

I want to calculate the $n$-th central moments of $Y$.

The density function of $Y$ is $$f(x)=\frac{1}{\sigma\sqrt {2\pi}}e^{-\frac{1}{2}\left(\frac{y-\mu}{\sigma}\right)^2}$$

The $n$-th central moment of $Y$ is $$E[(Y-E(Y))^n]$$

The $n$-th moment of $Y$ is $$E(Y^n)=\psi^{(n)}(0)$$ where $\psi$ is the Moment-generating function $$\psi(t)=E(e^{tX})$$

So I started calculating:

$$\begin{align} E[(Y-E(Y))^n]&=\int_\mathbb{R}\left(f(x)-\int_\mathbb{R}f(x)dx\right)^n\,dx \\ &=\int_\mathbb{R}\sum_{k=0}^n\left[\binom{n}{k}(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx \\ &=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\int_\mathbb{R}f(x)dx\right)^{n-k}\right]\,dx\right) \\ &=\sum_{k=0}^n\binom{n}{k}\left(\int_\mathbb{R}\left[(f(x))^k\left(-\mu\right)^{n-k}\right]\,dx\right) \\ &=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}\int_\mathbb{R}(f(x))^k\,dx\right) \\ &=\sum_{k=0}^n\binom{n}{k}\left((-\mu)^{n-k}E\left(Y^k\right)\right) \\ \end{align}$$

Am I on the right track or completely misguided? If I have made no mistakes so far, I would be glad to get some inspiration because I am stuck here. Thanks!

share|improve this question
1  
Since $Y−E(Y)$ has mean $0$ and in this case is normally distributed $N(0,\sigma^2)$, the $n$-th central moment should not be affected by the original mean $\mu$. –  Henry Dec 19 '11 at 0:42
    
What you have so far is correct, but as @Henry points out, the central moments are invariant under a shift. So you may as well simplify things by taking $\mu=0$ from the start. In any case, you still need to find $E[Y^n]$ for the normal distribution with mean $0$. –  mjqxxxx Dec 19 '11 at 1:02
1  
Your question has a typo in the normal density: there should be a square in the exponent. Also, I disagree with @mjqxxxx's statement that what you have so far is correct." The first step $$E[(Y-E(Y))^n]=\int_\mathbb{R}\left(f(x)-\int_\mathbb{R}f(x)dx\right)^n\,dx$$ is wrong: it should read $$E[(Y-E(Y))^n]=\int_\mathbb{R}\left(x-\int_\mathbb{R}xf(x)dx\right)^nf(x)\,dx=\‌​int_\mathbb{R}\left(x-\mu\right)^nf(x)\,dx$$ and the last step follows immediately upon expanding $(x-\mu)^n$ via the binomial theorem, separating into a sum of integrals, and identifying $\int_\mathbb{R}x^kf(x)\,dx=E[Y^k]$. –  Dilip Sarwate Dec 19 '11 at 1:41
1  
"Am I on the right track.....?" It depends on where you want to go! As Sasha shows in his answer, $$E[(Y-\mu)^n] = \hat{m}_{n} = \begin{cases}0, & n~\text{odd},\\\sigma^n(n-1)(n-3)\cdots 3\cdot 1,& n~\text{even},\end{cases}$$ can be evaluated in straightforward fashion. On the other hand, your approach succeeded in expressing the central $\hat{m}_n$ in terms of the standard (non-central) moments $m_k = E[Y^k]$ and so now you have the task of evaluating $n$ different integrals to find the $m_k$'s. So your approach does not seem too promising to say the least. –  Dilip Sarwate Dec 19 '11 at 3:28
    
@Dilip: Thank you for pointing that out. –  Aufwind Dec 19 '11 at 4:11
add comment

1 Answer

up vote 8 down vote accepted

The $n$-th central moment $\hat{m}_n = \mathbb{E}\left( \left(X-\mathbb{E}(X)\right)^n \right)$. Notice that for the normal distribution $\mathbb{E}(X) = \mu$, and that $Y = X-\mu$ also follows a normal distribution, with zero mean and the same variance $\sigma^2$ as $X$.

Therefore, finding the central moment of $X$ is equivalent to finding the raw moment of $Y$.

In other words, $$ \begin{eqnarray} \hat{m}_n &=& \mathbb{E}\left( \left(X-\mathbb{E}(X)\right)^n \right) = \mathbb{E}\left( \left(X-\mu\right)^n \right) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sigma} (x-\mu)^n \mathrm{e}^{-\frac{(x-\mu)^2}{2 \sigma^2}} \mathrm{d} x\\ & \stackrel{y=x-\mu}{=}& \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sigma} y^n \mathrm{e}^{-\frac{y^2}{2 \sigma^2}} \mathrm{d} y \stackrel{y = \sigma u}{=} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} \sigma} \sigma^n u^n \mathrm{e}^{-\frac{u^2}{2}} \sigma \mathrm{d} u \\ &=& \sigma^n \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} } u^n \mathrm{e}^{-\frac{u^2}{2}} \mathrm{d} u \end{eqnarray} $$ The latter integral is zero for odd $n$ as an integral an odd function over a real line. So consider $$ \begin{eqnarray} && \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi} } u^{2n} \mathrm{e}^{-\frac{u^2}{2}} \mathrm{d} u = 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi} } u^{2n} \mathrm{e}^{-\frac{u^2}{2}} \mathrm{d} u \\ && \stackrel{u=\sqrt{2 w}}{=} \frac{2}{\sqrt{2\pi}} \int_0^\infty (2 w)^n \mathrm{e}^{-w} \frac{\mathrm{d} w }{\sqrt{2 w}} = \frac{2^n}{\sqrt{\pi}} \int_0^\infty w^{n-1/2} \mathrm{e}^{-w} \mathrm{d} w = \frac{2^n}{\sqrt{\pi}} \Gamma\left(n+\frac{1}{2}\right) \end{eqnarray} $$ where $\Gamma(x)$ stands for the Euler's Gamma function. Using its properties we get $$ \hat{m}_{2n} = \sigma^{2n} (2n-1)!! \qquad\qquad \hat{m}_{2n+1} = 0 $$

share|improve this answer
    
Would you be so kind to explain this a little further, please: The latter integral is zero for odd n as an integral of even and odd functions over a real line. –  Aufwind Dec 19 '11 at 2:02
    
Thanks for explaining! –  Aufwind Dec 19 '11 at 4:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.