Application of Double Integrals to Probability Theory

The normal distribution plays a very important role in probability theory. The probability density function for a normally distributed random variable $x$ is defined by

$f(x)=\frac{1}{\sqrt[\phantom{2}]{2}}{e}^{-\frac{(x-{)}^2}{2}{}^2}$

where $$ is the mean of $x$ and $$ is the standard deviation. When $=0$ and $=1$ this is called the standard normal distribution.

In this Mini$-$Project we will use double integrals to show that $f$ in fact defines a probability density function. That is$,$ we will show that

${\displaystyle \underset{-}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}{e}^{-\frac{(x-{)}^2}{2}{}^2}dx=1$

Setup

We start by working with the function $g(x,y)=e^{-(x^2+y^2)}.$ We will make use of the fact that we can define the improper integral

${}_{R^2}{e}^{-(x^2+y^2)}dA={\displaystyle \underset{-}{\overset{}{}}}{\displaystyle \underset{-}{\overset{}{}}}{e}^{-(x^2+y^2)}dydx$

in two ways. Namely,

${}_{R^2}{e}^{-(x^2+y^2)}dA=\underset{a}{lim}{}_{D_a}{e}^{-(x^2+y^2)}dA$

where $D_a$ is the disk with radius a centered at the origin, and

${}_{R^2}{e}^{-(x^2+y^2)}dA=\underset{a}{lim}{}_{S_a}{e}^{-(x^2+y^2)}dA$

where $S_a$ is the square with vertices $(+-a,+-a).$

$2$

Problem. $(20$ points$)$

$($a$)$ Use polar coordinates to show that

$\underset{a}{lim}{}_{D_a}{e}^{-(x^2+y^2)}dA=$

$($b$)$ Show that

$\underset{a}{lim}{}_{S_a}{e}^{-(x^2+y^2)}dA=[{\displaystyle \underset{-}{\overset{}{}}}{e}^{-x^2}dx]*[{\displaystyle \underset{-}{\overset{}{}}}{e}^{-y^2}dy]$

$($c$)$ Use parts $($a$)$ and $($b$)$ to conclude that

${\displaystyle \underset{-}{\overset{}{}}}{e}^{-x^2}dx=\sqrt[\phantom{2}]{}$

$($d$)$ Use part $($c$)$ and substitution to conclude that

${\displaystyle \underset{-}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}{e}^{-\frac{x^2}{2}}dx=1$

$(*$ This shows that the standard normal distribution is a probability density function.$)$

$($e$)($Bonus: $5$ points$)$ Let $$ and $$ be constants. Show that

${\displaystyle \underset{-}{\overset{}{}}}\frac{1}{\sqrt[\phantom{2}]{2}}{e}^{-\frac{(x-{)}^2}{2}{}^2}dx=1$