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If calculating $\mathbb E(X^r)$ where $X \sim \mathcal N(x,y) $ where $x\neq 0$ and $y\neq 1$ (so not the standard normal distribution), and $r$ is an arbitrary constant, how would one go about doing that?

I'm having trouble finding a satisfying answer on the internet. Thanks!

callculus42
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Anthony
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  • Why isn´t it the standard normal distribution ? Isn´t $X\sim\mathcal N(0,1)$ ? – callculus42 Apr 21 '17 at 17:15
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    @callculus No, it's $N(x,y)$ for $x \neq 0, y \neq 1$. – Ian Apr 21 '17 at 17:15
  • @Ian thanks for clearing that up. My syntax here isn't the best – Anthony Apr 21 '17 at 17:16
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    @ian Ah ok, != means $\neq $ – callculus42 Apr 21 '17 at 17:17
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    @Anthony You are asking for $\frac{1}{\sqrt{2 \pi y}} \int_{-\infty}^\infty z^r e^{-(z-x)^2/2y} dz$. Change variables to $w=z-x$ to get $\int_{-\infty}^\infty (w+x)^r e^{-w^2/2y} dw$. Expand the power (using the fractional binomial theorem if need be) and then another change of variable will give you the answer as a sum of Gamma functions. They will all be $\Gamma(k+1/2)$ for an integer $k$ if $r$ itself is a nonnegative integer. (The whole thing unsurprisingly blows up if $r$ is a negative integer.) – Ian Apr 21 '17 at 17:18
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    Note that you can use scaling to reduce to the case where the variance is $1$. But if $r$ is not an integer, how are you going to handle negative values of $X$? – Robert Israel Apr 21 '17 at 18:46
  • I don't understand the notation $X \sim \mathcal N(x,y)$ – wolfies Apr 21 '17 at 19:19
  • @wolfies It means $X$ is normally distributed with mean $x$ and either variance or standard deviation $y$. Usually it's variance but it depends. – Ian Apr 21 '17 at 22:12
  • @Ian How can a random variable $X$ have as its mean the realisation of itself $X=x$? The notation is appallingly awful. – wolfies Apr 22 '17 at 06:28
  • @wolfies By not using the notation $X=x$ (or $X \leq x$ etc.) anywhere. It can work fine even if it is a bit inconvenient. – Ian Apr 22 '17 at 17:06
  • Does this answer your question? Methods for Finding Raw Moments of the Normal Distribution – StubbornAtom Jun 19 '20 at 18:17

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Combine the result shown in my answer here: Expectation of a Standard Normal Random Variable, with the fact that a general normal random variable with mean $\mu$ and variance $\sigma^2$ can be expressed as a location-scale transformation of a standard normal distribution; i.e., $$\operatorname{E}[X^r] = \operatorname{E}[(\sigma Z + \mu)^r] = \sum_{k=0}^r \binom{r}{k} \sigma^k \operatorname{E}[Z^k] \mu^{r-k}.$$

heropup
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