# Borel-Cantelli lemmas

• August 14th 2009, 03:47 AM
funnyinga
Characteristic Function of a symmetric RV
A random variable $X$ is called symmetric if $X$ and $-X$ have the same distribution. Show that the characteristic function of a symmetric random variable takes values only in the reals.
• August 14th 2009, 07:05 AM
matheagle
If the distribution is symmmetric then all the odd moments are zero...

$0=E(X)=E(X^3)=...$

Then use $e^{\theta}=1+\theta +{(\theta)^2\over 2!}+{(\theta)^3\over 3!}+....$

So $E(e^{itX})=1+itE(X) +{E(itX)^2\over 2!}+{E(itX)^3\over 3!}+....$

$=1-{t^2E(X^2)\over 2!}+...$

which no longer has those eye's in it, see my point?
• August 14th 2009, 11:08 PM
funnyinga
Quote:

Originally Posted by matheagle
If the distribution is symmmetric then all the odd moments are zero...

$0=E(X)=E(X^3)=...$

Can you please explain how this is the case? I can't see it (Worried)

Quote:

Then use $e^{\theta}=1+\theta +{(\theta)^2\over 2!}+{(\theta)^3\over 3!}+....$

So $E(e^{itX})=1+itE(X) +{E(itX)^2\over 2!}+{E(itX)^3\over 3!}+....$

$=1-{t^2E(X^2)\over 2!}+...$

which no longer has those eye's in it, see my point?
No I can't. (Worried) I am completely lost with how you did this.

The characteristic function is $\mathbb{E} e^{itX} = \int_{- \infty}^{\infty} e^{itx} d F(x)$, right? I can't see what you have done here....
• August 14th 2009, 11:24 PM
matheagle
Calc 2, I obtained the Taylor Series for $e^{\theta}$

I then let $\theta =itX$

$i^2=-1$, $i^4=1$ and so on.

If the distribution function is symmetric then the odd moments are 0, so those terms drop out of the series.
In the continuous case f(-x)=f(x) means that f(x) is an even function.
An even times an odd function is odd.

So, $\int_{-\infty}^{\infty}x^{\rm odd \; power}f(x)dx=0$.