I want to show $$E(Z^{2k})=\frac{(2k)!}{2^kk!} \text{ for } Z\sim N(0,1).$$

I showed $$E(Z^{2k})=(2k-1)\cdot E(Z^{2k-2})$$ and guessed by trying (since $E(Z^2)=1$): $$E(Z^{2k})=(2k-1)!!$$ How do I prove this by induction?

$k=1$ is true: $1=E(Z^2)=(2-1)!!=1$. Now $k\mapsto k+1$:


Thanks for any help!


1 Answer 1


You are nearly done. $$(2k)!=(2k)!!(2k-1)!!=2^kk!(2k-1)!!$$

  • $\begingroup$ Thank you but first I want to show $$E(Z^{2k})=(2k-1)!!$$ by induction. Can you help me? $\endgroup$
    – Uhmm
    Apr 10, 2023 at 17:07
  • $\begingroup$ This follows from your line $E(Z^{2k})=(2k-1)\cdot E(Z^{2k-2})$ $\endgroup$
    – DanDan面
    Apr 10, 2023 at 17:17

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