As mentioned in a previous post, I've been trying to work through ALL of the problems in Jacod and Protter's Probability Essentials. The following problem has been giving me issues:

Let $Z \sim N(0,1)$. Show that $E[X^{2n+1}]=0$ and $E[X^{2n}]=\frac{(2n)!}{2^n n!}$

The issue I'm having is proceeding form my Taylor Expansion for $\varphi_{_Z}(u)$. Here's my work:

Since we're dealing with $N(0,1)$, we have $$\varphi_{_Z}(u)=e^{-\frac{1}{2}u^2}$$ Recalling that the Taylor expansion of $e^x$ wrapped around 0, we have $$\varphi_{_Z}(u)=e^{-\frac{1}{2}u^2}=\sum_{k\geq0}(-\frac{u^2}{2})^k\cdot\frac{1}{k!}=\sum_{k\geq 0}(-1)^k\frac{u^{2k}}{2^kk!}$$ $$\implies \varphi_{_Z}(u)=\sum_{k\geq 0}(-1)^k\frac{u^{2k}}{2^kk!}$$ Next, recall by a theorem in the book that $$\frac{d^m}{du^m}\varphi_{_Z}(0)=i^mE[Z^m]$$ where $i\in \mathbb{C}$. It follows that $$\frac{d^{2k}}{du^{2k}}\varphi_{_Z}(0)=\sum_{k\geq0}(-1)^k\frac{(2k)!}{2^k k!}=i^{2k}E[Z^{2k}]$$ and $$\frac{d^{2k+1}}{du^{2k+1}}\varphi_{_Z}(0)=\sum_{k\geq0}(-1)^k(0)=0$$ Since $(-1)^k=i^{2k}$, it follows that we have $$\sum_{k\geq0}i^{2k}\frac{(2k)!}{2^k k!}=i^{2k}E[Z^{2k}]$$ $$\implies E[X^{2k}]=\frac{(2k)!}{2^k k!}$$

Is it justified to get rid of the sum? I'm basically at the answer, but keeping the sum is killing me. Any suggestions?


You made a mistake while differentiating. Notice \begin{align} \varphi_{_Z}(u)&=\sum_{k\geq 0}(-1)^k\frac{u^{2k}}{2^kk!}\\ &=\frac{u^{0}}{2^00!}-\frac{u^{2}}{2^11!}+\dotsb+(-1)^{n}\frac{u^{2n}}{2^nn!}+(-1)^{n+1}\frac{u^{2(n+1)}}{2^{n+1}(n+1)!}+\dotsb, \end{align} So \begin{align}\frac{d^{2n}}{du^{2n}}\varphi_{_Z}(0)&=0+0+\dotsb+(-1)^{n}\frac{(2n)!0^0}{2^nn!}+(-1)^{n+1}\frac{(2(n+1))!0^{2}}{2^{n+1}(n+1)!}+\dotsb\\ &=(-1)^{n}\frac{(2n)!}{2^nn!}. \end{align} That is, $$(-1)^{n}\frac{(2n)!}{2^nn!}=i^{2n}E[Z^{2n}],$$ such that $$E[Z^{2n}]=\frac{(2n)!}{2^n n!}$$ follows immediately.

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