# where is $\sum_k(-1)^k$ going in this formula?

$\varphi _{X}(w)=\sum ^{\infty}_{k=0}(-1)^k(wb)^{2k}$,and the mth moment of X is $E[X^m]=(-j)^m\frac{d^m \varphi _{X}(w)}{dw^m}=(-j)^m\sum_k (-1)^k \times 2k \times(2k-1)\times ... \times(2k-(m-1))\times b^{2k}w^{2k-m}$

From the formula we derive,we found the $2k=m$,and the solution said so $E[X^m]=(-j)^m \times m! \times b^m$,i know where are $(-j)^m$ and $m!$ and $b^m$ from and the reason that $w^{2k-m}$ is dissappeared,but where is the $\sum_k(-1)^k$ going?

Your formula to compute the $m^{th}$ moment is not correct. I assume that $j$ is defined such that $j^2 = -1$ (which is not the standard notation by the way), that $b$ is some positive constant, and that $\phi_X(w) = E[e^{jwX}]$ is well defined for $|w|<\frac{1}{b}$.
To answer your question, the correct formula is: $$E[X^m] = (-j)^m \frac{d^m}{dw^m} \bigg|_{w=0} \phi_X(w)$$ i.e. you have to evaluate the $m^{th}$ derivative at $w=0$ to get the $m^{th}$ moment. If you are careful with your summation terms, you will see that in general the $m^{th}$ derivative will look like $$\sum_{k = p}^{+\infty} (-1)^k \frac{(2k)!}{(2k-2p)!}b^{2k}w^{2k-2p}, \quad m=2p$$ or $$\sum_{k = p+1}^{+\infty} (-1)^k \frac{(2k)!}{(2k-2p-1)!}b^{2k}w^{2k-2p-1}, \quad m=2p+1$$ So evaluating the formula at $w=0$ makes all the terms disappear in the case $m=2p+1$ (all the terms contains a $w$ to some positive power) $$E[X^m] = 0, \quad m \; odd$$ and for $m=2p$, all the terms except the first one i.e. $m=2k$ contains terms with $w$ to some positive power. So at $w=0$, only the first term is left and you get $$E[X^m] = (-1)^{m/2} (-j)^m m! \: b^{m}, \quad m \; even$$
• a little confused,why is $E[X^m]$ =0 when m is odd? – Shine Sun May 15 '18 at 13:27
• The terms in the series all have $w^n$ for $n \geq 1$, so at $w=0$ they all vanish. Try it for $m=1$ by taking 1 derivative from the characteristic function you gave me. – monty47 May 15 '18 at 14:00
• That was an example, my general proof shows that $E[X^m]=0$ for any m odd (morally for the same reasons as the case m=1). Is there anything you do not understand in the proof that I can clarify? – monty47 May 15 '18 at 22:51
• but if m=3,the $E[X^3]=(-1)^{3/2}(-j)^33!b^3$,why is it 0? – Shine Sun May 15 '18 at 23:00