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Say we have $X \sim N(0,1)$. I was wondering how we can use Stein's Lemma to derive the moments of the r.v. $X$ by calculating the first few moments.

How would you summarize it in the form $EX^n$ if the moments we calculated end up being quite different in form? Or are they?

So what I've tried to do so far is that I used the formula:

$E\bigl(g(X)(X-\mu)\bigr)=\sigma^2 E\bigl(g'(X)\bigr)$

And from there, I tried to see if I can make any substitutions but I don't think you can. Furthermore, I really don't see how you can even calculate moments from this thing. From class, I think you have to use Taylor expansion but I really don't see how you can get to that here either. I really need some help on this.

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Let $X \sim N(\mu,\sigma^2)$.

  1. Take $g := 1$. Then Stein's lemma gives $$\mathbb{E}(1 \cdot (X-\mu)) = 0, \tag{1}$$ i.e. $\mathbb{E}(X) = \mu$.
  2. For $g(x):= (x-\mu)^{n-1}$, $n \geq 2$, we have, by Stein's lemma, $$\mathbb{E}((X-\mu)^n) = \sigma^2 (n-1) \mathbb{E}((X-\mu)^{n-2}).$$ If $n$ is odd, i.e. $n=2k+1$, then we get by iteration $$\begin{align*} \mathbb{E}((X-\mu)^n) &= (\sigma^2 (n-1)) (\sigma^2 (n-3)) \cdots (\sigma^2 \mathbb{E}(X-\mu)) \\ &\stackrel{(1)}{=} 0. \end{align*}$$ On the other hand, if $n=2k$ is even, then $$\begin{align*} \mathbb{E}((X-\mu)^n) &= (\sigma^2 (n-1)) (\sigma^2 (n-3)) \cdots (\sigma^2 \cdot 1) \\ &= (\sigma^2)^{\frac{n}{2}} (n-1) (n-3) (n-5) \cdots 1. \end{align*}$$
  3. If we are not interested in the centered moments $\mathbb{E}((X-\mu)^n)$, but in $\mathbb{E}(X^n)$, then we can use the identity $$X^n = ((X-\mu)+\mu)^n = \sum_{k=0}^n {n \choose k} (X-\mu)^k \mu^{n-k}$$ to express $\mathbb{E}(X^n)$ in terms of $\mu$ and $\sigma^2$ using the results from step 2.
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