Prove Variance of Gaussian by Differentiating Integral of N(x) =1 by $\sigma^2$ and rearranging

Question

The normalized Gaussian distribution is defined as:

$$N(x|\mu,\sigma^2)=\frac{1}{(2\pi\sigma^2)^{1/2}}exp\bigg(\frac{-1}{2\sigma^2}(x-\mu)^2\bigg)$$

Prove that:

$$Var[x]=E[(X-\mu)^2]=\sigma^2$$

by differentiating both sides of the normalization condition:

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

with respect to $\sigma^2$ and rearrange the result such that:

$$E[(X-\mu)^2] = \sigma^2$$

I've tried this proof at least 10 times...and the variance always comes out as fraction of a square...

Answer 1

For simplicity let's suppress the $\mu$ and $\sigma$ in $N(x\mid \mu,\sigma)$. The fastest way to differentiate $N$ is to use logarithmic differentiation: $$ \log N = -\frac{(x-\mu)^2}{2\sigma^2} -\log(\sqrt{2\pi}\cdot\sigma).\tag1 $$ Now differentiate both sides with respect to $\sigma$: $$ \frac1N\frac{\partial N}{\partial\sigma}=\frac{(x-\mu)^2}{\sigma^3}-\frac1\sigma $$ so that $$\frac{\partial N}{\partial\sigma}=N(x)\left[\frac{(x-\mu)^2-\sigma^2}{\sigma^3}\right].\tag2$$ You should be able to take it from here, given that $\int N(x)(x-\mu)^2\,dx=E(X-\mu)^2$ and $\int N(x)\,dx=1$.

(You can also differentiate (1) wrt $\sigma^2$ by defining $t:=\sigma^2$ and replacing all occurrences of $\sigma$ by $t^{1/2}$, then computing $\frac\partial{\partial t}$. The same result should obtain.)

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EDIT: Once you've established (2), here are the details:

$$0\stackrel{(a)}=\frac d{d\sigma}\int N(\sigma, x)\,dx \stackrel{(b)}=\int \frac{\partial N}{\partial \sigma}\,dx\stackrel{(c)}=\frac1{\sigma^3}\left[\int N(x)(x-\mu)^2dx-\int \sigma^2 N(x)dx\right] $$ Step (a) follows because $\int N(x)dx=1$. Step (b) is the Leibniz integral rule. In step (3) we substitute (2).

Answer 2

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

Now Differentiate both sides by $\sigma$:

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

$$\int_{-\infty}^{\infty} \frac{d}{d\sigma}N(x|\mu,\sigma^2)~dx=0~~~~~(1)$$

Now Working just on the differentiation:

$$\frac{d}{d\sigma}N(x|\mu,\sigma^2)=\Bigg[\bigg(\frac{\sigma^{-1}}{(2\pi)^{1/2}}\bigg)\exp\bigg(\frac{-1}{2}\sigma^{-2}(x-\mu)^2 \bigg)\Bigg]'$$

using differentiation product rule:

$$\frac{d}{d\sigma}N(x|\mu,\sigma^2)=\Bigg[\bigg(\frac{\sigma^{-1}}{(2\pi)^{1/2}}\bigg)\Bigg]'\exp\bigg(\frac{-1}{2}\sigma^{-2}(x-\mu)^2 \bigg) + \bigg(\frac{\sigma^{-1}}{(2\pi)^{1/2}}\bigg)\Bigg[\exp\bigg(\frac{-1}{2}\sigma^{-2}(x-\mu)^2 \bigg)\Bigg]'$$

Then, take the result of this differentiation, place it back into the integral (1) and substitute in these already known true equations:

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

$$E[(x-\mu)^2]=\int_{-\infty}^{\infty} (x-\mu)^{2}~ \bigg(\frac{\sigma^{-1}}{(2\pi)^{1/2}}\bigg) exp\bigg(-\frac{1}{2}\sigma^{-2} (x-\mu)^2\bigg)~dx = \int_{-\infty}^{\infty} (x-\mu)^{2}~N(x|\mu, \sigma)~dx$$

The result after simplification should yield:

$$E[(x-\mu)^2] = \sigma^2$$