If $X\sim\mathcal N(0,1)$ Why is $\Phi_X^{(j)}(0)=0$ for $j$ odd ?
($\Phi_X^{(j)}(0):j^{th}$ derivative of the characteristic function of the r.v. $X$)
We computed $\displaystyle\Phi_X(t)=e^{-\frac{t^2}{2}}$
then the power series of $\displaystyle e^{-\frac{t^2}{2}}=\sum\limits_{j=0}^{\infty}(-1)^j\frac{\frac{t^{2j}}{2^j}}{j!}$
and the Taylor series of $\displaystyle\Phi_X(t)=\sum\limits_{j=0}^{\infty}\Phi_X^{(j)}(0)\frac{t^j}{j!}$
Then we concluded that $\Phi_X^{(2j+1)}(0)=0$ (the odd orders derivatives)
but why is this true ? Is this a sufficient justification ?
Thanks.
Note that $\Phi_X^{(j)}(0) = E[X^j]$.
That $\Phi_X^{(2j+1)}(0) = 0$ is not really related to the mgfs/cfs/power series.
That $\Phi_X^{(2j+1)}(0) = 0$ is because $E[X^{2j+1}] = 0$.
The odd moments of standard normal are all zero.