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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 $$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$$ where $\mu$ is the mean of $X$ and $\sigma$ is the standard deviation. When $\mu = 0$ and $\sigma = 1$ 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 $$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} dx = 1$$ 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 $$\iint_{g^2} e^{-(x^2+y^2)} dA = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)} dydx$$ in two ways. Namely, $$\iint_{g^2} e^{-(x^2+y^2)} dA = \lim_{a\to\infty} \iint_{D_a} e^{-(x^2+y^2)} dA$$ where $D_a$ is the disk with radius $a$ centered at the origin, and $$\iint_{g^2} e^{-(x^2+y^2)} dA = \lim_{a\to\infty} \iint_{S_a} e^{-(x^2+y^2)} dA$$ where $S_a$ is the square with vertices $(\pm a, \pm a)$. Problem. (20 points) (a) Use polar coordinates to show that $$\lim_{a\to\infty} \iint_{D_a} e^{-(x^2+y^2)} dA = \pi$$ (b) Show that $$\lim_{a\to\infty} \iint_{S_a} e^{-(x^2+y^2)} dA = \left[ \int_{-\infty}^{\infty} e^{-x^2} dx \right] \left[ \int_{-\infty}^{\infty} e^{-y^2} dy \right]$$ (c) Use parts (a) and (b) to conclude that $$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$ (d) Use part (c) and substitution to conclude that $$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx = 1$$ (* This shows that the standard normal distribution is a probability density function.) (e) (Bonus: 5 points) Let $\mu$ and $\sigma$ be constants. Show that $$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} dx = 1$$

          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
$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}$$
where $\mu$ is the mean of $X$ and $\sigma$ is the standard deviation. When $\mu = 0$ and $\sigma = 1$ 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
$$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} dx = 1$$
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
$$\iint_{g^2} e^{-(x^2+y^2)} dA = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)} dydx$$
in two ways. Namely,
$$\iint_{g^2} e^{-(x^2+y^2)} dA = \lim_{a\to\infty} \iint_{D_a} e^{-(x^2+y^2)} dA$$
where $D_a$ is the disk with radius $a$ centered at the origin, and
$$\iint_{g^2} e^{-(x^2+y^2)} dA = \lim_{a\to\infty} \iint_{S_a} e^{-(x^2+y^2)} dA$$
where $S_a$ is the square with vertices $(\pm a, \pm a)$.
Problem. (20 points)
(a) Use polar coordinates to show that
$$\lim_{a\to\infty} \iint_{D_a} e^{-(x^2+y^2)} dA = \pi$$
(b) Show that
$$\lim_{a\to\infty} \iint_{S_a} e^{-(x^2+y^2)} dA = \left[ \int_{-\infty}^{\infty} e^{-x^2} dx \right] \left[ \int_{-\infty}^{\infty} e^{-y^2} dy \right]$$
(c) Use parts (a) and (b) to conclude that
$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$
(d) Use part (c) and substitution to conclude that
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx = 1$$
(* This shows that the standard normal distribution is a probability density function.)
(e) (Bonus: 5 points) Let $\mu$ and $\sigma$ be constants. Show that
$$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2} dx = 1$$

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