I'm attempting to tackle an integration problem from a beginning calculus text and running up against a question that by all appearances at least is far more difficult than others.
The problem begins by mentioning (without proof) that it just so happens to be a fact that $ \int_{-\infty}^\infty e^{-x^{2}} dx = \sqrt\pi $.
Here comes a summary of the problem:
To obtain the probability density function for a normal random variable with mean $\mu$ and standard deviation $ \sigma $ "stretch" and "shift" $y = e^{-x^{2}}$ so that the resulting bell curve satisfies the following:
- The bell curve must be symmetric about the line $x = \mu$
- The x-coordinates of the bell's inflection points must be $\mu \pm \sigma $
- The total area under the graph must be exactly 1.
Notably, the problem explicitly states that zero knowledge of probability or stats is required to be able to solve this.
I know a bit about the equation for the normal distribution. I know its area under the curve is 1. But I have no idea how to derive this from what's given.
I mean, $y = e^{-x^{2}}$ already looks symmetrical around $x = 0 = \mu.$ That...sort of addresses point #1.
As for point #2, I can set the second derivative to 0 $$ -2e^{-x^2}(-2x^2 + 1) = 0$$ so $$-2e^{-x^2} = 0 $$ or $$ -2x^2 + 1 = 0$$
It would appear $-2e^{-x^2} = 0 $ has no solution, so (maybe?) we get $$ x =\mu \pm \sigma = \sqrt{\frac{1}{2}} $$
Not really sure if this is on the right track, and even if so, what to do with it.
Any help would certainly be appreciated.
(Note: This is not homework, just personal interest, so feel free to be as detailed as you like)
I think it is easier to understand in the order of 2 -> 3 -> 1.
First, for 2, you want to scale the graph so that the inflection point $x = \pm \frac{1}{\sqrt{2}}$ to be scaled to $x = \sigma$. That means you want to scale the whole graph by $\sigma \div \frac{1}{\sqrt{2}} = \sqrt{2}\sigma$ in $x$. Recall that for any graph $y=f(x)$, we can scale it by $a$ in $x$ by taking $y=f(x/a)$. So, we have $e^{-\frac{x^2}{2\sigma^2}}$.
Then, for 3, we compute $\int_{-\infty}^\infty e^{-\frac{x^2}{2\sigma^2}}\,dx$ to obtain the value we need to use to make the area equal to 1. We can use the fact $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$ for that purpose and obtain $\sqrt{2\pi}\sigma$. Therefore, we have $\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}$.
Finally, 1 is just about translation.
For 1, as you noted $e^{-x^2}$ is symmetric around $x=0$. We want to translate it to $x=\mu$. That means you want to translate the graph of $e^{x^2}$ in the direction of $x$ coordinate for $+\mu$. Recall that for any function $y=f(x)$, you can translate the graph in the direction of $x$ for $+\mu$ by taking $y = f(x-\mu)$. Thus, we obtain the final form $\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$.