Symmetric Moment Generating Function

Hi guys,

I was wondering if there is a more formal proof to show that if a random variable with symmetric pdf around zero, i.e. f(-x)=f(x) then that RV also has symmetric MGF, M(-t)=M(t).

What I did was to show that all the moments are equal for both cases.

For example:

\(\displaystyle \displaystyle M(-t)= \int_{-\infty}^{\infty} e^{-tx} f(x)\ dx \Rightarrow M\prime(0) =\int_{-\infty}^{\infty} xf(x)\ dx = 0\) because of symmetry around 0, thus the first moment is the equal for both cases.

In the same way one can find that all odd moments are zero because of symmetry and all even moments are equal.

But is there a more formal way to show that \(\displaystyle \displaystyle E(e^{tx})=E(e^{-tx})\) ?

This is not an answer to your question, but a hint about LaTeX. When you want an operator to be more than a single character, like e^(tx), enclose the multiple characters in curly brackets :
e^{tx},\;\;\;\;\;\sqrt{a^2+b^2} -->

\(\displaystyle \displaystyle e^{tx},\;\;\;\;\;\sqrt{a^2+b^2}\)

DrPhil said:

This is not an answer to your question, but a hint about LaTeX. When you want an operator to be more than a single character, like e^(tx), enclose the multiple characters in curly brackets :
e^{tx},\;\;\;\;\;\sqrt{a^2+b^2} -->

\(\displaystyle \displaystyle e^{tx},\;\;\;\;\;\sqrt{a^2+b^2}\)

Click to expand...

Much appreciated, thank you.