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In Bishops book, "Statistical Pattern Recognition", there is one exercise, which states to derive the second order moment of the Gaussian Distribution:

$E[x^2] = \int_{-\infty}^{\infty} N(x|\mu, \sigma^2 )x^2dx = \mu^2 + \sigma^2 $

As a hint, the book states to take the derivative of

$\int_{-\infty}^{\infty} N(x|\mu, \sigma^2 )dx = 1 $

with respect to $\sigma^2$.

This is what I have tried and failed miserably and where I need some advice in the right direction. My current attempt:

$ \int_{-\infty}^{\infty} \exp(-\frac{1}{2\sigma^2} (x-\mu)^2)dx = \sqrt{2\pi\sigma^2} $

$ \Longrightarrow \int_{-\infty}^{\infty} \frac{d}{d\sigma^2} \exp(-\frac{1}{2\sigma^2} (x-\mu)^2)dx = \frac{d}{d\sigma^2} \sqrt{2\pi\sigma^2} $

I then tried to substitute $z = \sigma^2$, before taking the derivative.

$ \Longrightarrow \int_{-\infty}^{\infty} \frac{d}{dz} \exp(-\frac{1}{2z} (x-\mu)^2)dx = \frac{d}{dz} \sqrt{2\pi z} $

$ \Longrightarrow \int_{-\infty}^{\infty} \frac{1}{2z^2}(x-\mu)^2 \exp(-\frac{1}{2z} (x-\mu)^2)dx = (2\pi z)^{-0.5} $

I am sure, that up to this point couple of mistakes already must have happend, which is why I will not write down my subsequent transformations.

kklaw
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    Because this is purely an exercise in differentiation, it's not really on topic here on CV. But you will succeed if you take the hint literally: differentiate the original integral, not some equation based on it. Start with $\mu=0$ to avoid unnecessary complications and be more aggressive: use the variable $1/\sigma^2$ (the "natural parameter") rather than $\sigma^2.$ – whuber Sep 01 '22 at 20:28
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    I think you've almost got it (bar a few small mistakes), just bring the rhs back to the lhs - ie so that you end up calculating the second moment of gaussian on lhs – seanv507 Sep 01 '22 at 20:29

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