1. Characteristic Function of a symmetric RV

A random variable $\displaystyle X$ is called symmetric if $\displaystyle X$ and $\displaystyle -X$ have the same distribution. Show that the characteristic function of a symmetric random variable takes values only in the reals.

2. If the distribution is symmmetric then all the odd moments are zero...

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

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

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

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

which no longer has those eye's in it, see my point?

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

$\displaystyle 0=E(X)=E(X^3)=...$
Can you please explain how this is the case? I can't see it

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

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

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

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

The characteristic function is $\displaystyle \mathbb{E} e^{itX} = \int_{- \infty}^{\infty} e^{itx} d F(x)$, right? I can't see what you have done here....

4. Calc 2, I obtained the Taylor Series for $\displaystyle e^{\theta}$

I then let $\displaystyle \theta =itX$

$\displaystyle i^2=-1$, $\displaystyle i^4=1$ 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, $\displaystyle \int_{-\infty}^{\infty}x^{\rm odd \; power}f(x)dx=0$.