I am currently studying the Statistical Inference, 2nd, Casella & Berger.
On page 72, the authors asserts that for the normal distribution with mean $\mu$ and variance 1, $$EX^{n+1}=\mu EX^n - \frac{d}{d\mu}EX^n$$.
I cannot deduce it myself.
I know that pdf for this normal distribution is $f(x)=\frac{1}{\sqrt{2\pi}}\exp({\frac{-(x-\mu)^2}{2}})$.
Now
$$\frac{d}{d\mu}EX^n= \frac{d}{d\mu}\int_{-\infty}^{\infty}x^n\frac{1}{\sqrt{2\pi}}\exp({\frac{-(x-\mu)^2}{2}})dx=\int_{-\infty}^{\infty}\frac{d}{d\mu}(x^n\frac{1}{\sqrt{2\pi}}\exp({\frac{-(x-\mu)^2}{2}}))dx$$
$$=\int_{-\infty}^{\infty}x^n\frac{1}{\sqrt{2\pi}}\exp({\frac{-(x-\mu)^2}{2}})(x-\mu)dx=EX^{n+1}-\mu EX^n$$.
However by interchanging terms I get only $$EX^{n+1}=\mu EX^n + \frac{d}{d\mu}EX^n$$.
Am I missing something? Or is this just a typo?
You are right! I checked on my copy of CB. It's an obvious typo, easy to verify, as you correctly proved.