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I have two random variables $X$ and $Y$ which follows Normal distribution , whose pdf's are given by

$f(x)= \frac{1}{2 \sqrt{2 \pi} \sigma}[e^{\frac{-(x-1)^2}{2 \sigma^2}}+e^{\frac{-(x+1)^2}{2 \sigma^2}}]$

$f(y)= \frac{1}{ \sqrt{2 \pi} \sqrt{1+ \sigma^2}}[e^{\frac{-x^2}{2(1+\sigma^2)}}]$

Both Random Variables $X$ and $Y$ has Mean Zero and Variance $1+\sigma^2$.

I am interested to evaluate the Higher moments for both $X$ and $Y$.

Is the higher moments for both $X$ and $Y$ are equal. What is the difference between these two Random variables.

Thank you.

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Your first variable is not normal; it is a mixture of normals. –  whuber Mar 11 '13 at 18:47
    
That's homework? –  ziggystar Mar 11 '13 at 20:37
    
@ziggystar the homework tag now maps to self-study, with similar intent –  Glen_b Mar 11 '13 at 23:24
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1 Answer

up vote 3 down vote accepted

By reading off the arguments of the exponentials, it is evident that $X$ is a mixture of Normals centered at $\pm 1$, whereas $Y$ is a Normal with the same mean and variance as $X$. Here are their density functions when $\sigma=1/2$:

PDFs

$X$ is blue with two peaks; $Y$ is red with one peak. (When $\sigma$ is much larger than $1/2$, $X$ will have only a single "merged" peak, but it will be flatter at the top than $Y$'s peak.)

From this picture it is apparent that

  • All odd moments will be zero, because both variables are symmetric about the origin.

  • The red curve ($Y$) has fatter tails than the blue ($X$), implying its higher even moments will be greater.


The easiest way to do the calculations is with the characteristic or moment generating functions. The MGF of a Normal$(\mu, \tau)$ distribution is

$$\exp{(t \mu +\frac{t^2 \tau ^2}{2})};$$

when expanded as a MacLaurin series in $t$, the coefficient of $t^n$ is $1/n!$ times the $n$th moment. Plugging in $\mu=0$ and $\tau = \sqrt{1+\sigma^2}$ gives the MGF of $Y$ as

$$1+\left(\frac{1}{2}+\frac{\sigma ^2}{2}\right) t^2+\left(\frac{1}{8}+\frac{\sigma ^2}{4}+\frac{\sigma ^4}{8}\right) t^4+\left(\frac{1}{48}+\frac{\sigma ^2}{16}+\frac{\sigma ^4}{16}+\frac{\sigma ^6}{48}\right) t^6+\left(\frac{1}{384}+\frac{\sigma ^2}{96}+\frac{\sigma ^4}{64}+\frac{\sigma ^6}{96}+\frac{\sigma ^8}{384}\right) t^8+\ldots$$

whereas that of $X$ is the average of MGFs of its components, which works out to

$$1+\left(\frac{1}{2}+\frac{\sigma ^2}{2}\right) t^2+\left(\frac{1}{24}+\frac{\sigma ^2}{4}+\frac{\sigma ^4}{8}\right) t^4+\left(\frac{1}{720}+\frac{\sigma ^2}{48}+\frac{\sigma ^4}{16}+\frac{\sigma ^6}{48}\right) t^6+\left(\frac{1}{40320}+\frac{\sigma ^2}{1440}+\frac{\sigma ^4}{192}+\frac{\sigma ^6}{96}+\frac{\sigma ^8}{384}\right) t^8+\ldots.$$

The agreement with the preceding MGF in the first two terms is evident. The coefficients of $t^4$ give the fourth moments: comparing, we see that of $Y$ exceeds that of $X$ by $4!(\frac{1}{8}-\frac{1}{24})\gt 0$, as expected. Similar comparisons bear out (and quantify) our impression that $Y$ has the greater higher moments.

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That hand-drawn look on the axes and tick labels seems 'friendly'. Is that one the xkcd-style? –  Glen_b Mar 11 '13 at 23:28
    
@Glen_b Yes. I am learning to distinguish graphics that need to be clear and precise from those that benefit from some "friendly" simplification. This, evidently, is one of the latter. –  whuber Mar 11 '13 at 23:34
    
Thank you so much for explanation –  Gold Mar 12 '13 at 9:25
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