My question is as follows: using the standard integral $$\int_{-\infty}^{\infty}e^{-ax^2}dx=\sqrt{\frac{\pi}{a}}$$
prove directly from the definition that the variance of the normal distribution,
$$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{(x-\mu)^2}{2\sigma^2}},$$ is $\sigma^2$. It is important to note that no understanding of why the integral above is true is needed to answer the question. I am instructed that integration by parts is required, but I'm struggling…
Ok, I've tried with much help from you lot, thanks greatly, but have missed something:
$$\begin{align}Var(X) &= E((X-\mu)^2) \\
& = \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}(x-\mu)^2e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx \\ & Let\;y = \frac{(x-\mu)}{\sigma} \\ & = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}y^2e^{-\frac{1}{2}y^2}dx \\
&=\frac{\sigma}{\sqrt{2\pi}}([ue^{-\frac{1}{2}u^2}]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}e^{-\frac{1}{2}u^2}du)
\\ &=\sigma
\end{align}$$
where did I go wrong?
Best Answer
Your substitution step is lacking an extra factor of $\sigma$ because you should have written $$\begin{align*} \frac{1}{\sqrt{2\pi} \sigma} \int_{x=-\infty}^\infty (x-\mu)^2 e^{-(x-\mu)^2/(2\sigma^2)} \, dx &= \frac{1}{\sqrt{2\pi} \sigma} \int_{y=-\infty}^\infty (\sigma y)^2 e^{-y^2/2} \sigma \, dy \\ &= \frac{\sigma^2}{\sqrt{2\pi}} \int_{y=-\infty}^\infty y^2 e^{-y^2/2} \, dy. \end{align*}$$ This is because $y = (x-\mu)/\sigma$ implies $x = \sigma y + \mu$, hence $dx = \sigma \, dy$.