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.
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.
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?
Can you please explain how this is the case? I can't see it
No I can't. I am completely lost with how you did this.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?
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....
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, $\displaystyle \int_{-\infty}^{\infty}x^{\rm odd \; power}f(x)dx=0$.