Prove (a) (and thus, by symmetry, (b), also) of Theorem...

Prove (a) (and thus, by symmetry, (b), also) of Theorem 3.2.5. Hint: Recall that $$ f_X(x)=\int_{-\infty}^{\infty} f(x, y) d y $$ where $f(x, y)$ is given by $$ \begin{aligned} f(x, y) & =\frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp \left\{-\frac{1}{2\left(1-\rho^2\right)}\left[\left(\frac{x-\mu_X}{\sigma_X}\right)^2\right.\right. \\ & \left.\left.-\frac{2 \rho\left(x-\mu_X\right)\left(y-\mu_Y\right)}{\sigma_X \sigma_Y}+\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2\right]\right\} \end{aligned} $$ To simplify the integration of (3.212) let $u=\left(x-\mu_X\right) / \sigma_X$ and $v=$ $\left(y-\mu_Y\right) / \sigma_Y$. Then, since $d y=\sigma_Y d v,(3.212)$ reduces to $$ f_X(x)=\frac{\int_{-\infty}^{\infty} \exp \left[-\left(u^2-2 \rho u v+v^2\right) / 2\left(1-\rho^2\right)\right] d v}{2 \pi \sigma_X \sqrt{1-\rho^2}} . $$ Adding and subtracting $\rho^2 u^2$ gives $$ \begin{aligned} u^2-2 \rho u v+v^2 & =v^2-2 \rho u v+\rho^2 u^2-\rho^2 u^2+u^2 \\ & =(v-\rho u)^2+u^2\left(1-\rho^2\right) \end{aligned} $$ and thus, (3.214) becomes $$ f_X(x)=\frac{e^{-\frac{u^2}{2}}}{2 \pi \sigma_X \sqrt{1-\rho^2}} \int_{-\infty}^{\infty} \exp \left\{\frac{-(v-\sigma u)^2}{2\left(1-\rho^2\right)}\right\} d v . $$ Now let $z=(v-\rho u) / \sqrt{1-\rho^2}$, and using the fact that $$ \int_{-\infty}^{\infty} \exp \left(\frac{-z^2}{2}\right) d z=\sqrt{2 \pi}, $$ show that $$ f_X(x)=\frac{\exp \left[-\frac{1}{2}\left(\frac{x-\mu_X}{\sigma_X}\right)^2\right]}{\sqrt{2 \pi} \sigma_X} . $$

Prove (a) (and thus, by symmetry, (b), also) of Theorem 3.2.5. Hint: Recall that
$$
f_X(x)=\int_{-\infty}^{\infty} f(x, y) d y
$$
where $f(x, y)$ is given by
$$
\begin{aligned}
f(x, y) & =\frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \exp \left\{-\frac{1}{2\left(1-\rho^2\right)}\left[\left(\frac{x-\mu_X}{\sigma_X}\right)^2\right.\right. \\
& \left.\left.-\frac{2 \rho\left(x-\mu_X\right)\left(y-\mu_Y\right)}{\sigma_X \sigma_Y}+\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2\right]\right\}
\end{aligned}
$$
To simplify the integration of (3.212) let $u=\left(x-\mu_X\right) / \sigma_X$ and $v=$ $\left(y-\mu_Y\right) / \sigma_Y$. Then, since $d y=\sigma_Y d v,(3.212)$ reduces to
$$
f_X(x)=\frac{\int_{-\infty}^{\infty} \exp \left[-\left(u^2-2 \rho u v+v^2\right) / 2\left(1-\rho^2\right)\right] d v}{2 \pi \sigma_X \sqrt{1-\rho^2}} .
$$
Adding and subtracting $\rho^2 u^2$ gives
$$
\begin{aligned}
u^2-2 \rho u v+v^2 & =v^2-2 \rho u v+\rho^2 u^2-\rho^2 u^2+u^2 \\
& =(v-\rho u)^2+u^2\left(1-\rho^2\right)
\end{aligned}
$$
and thus, (3.214) becomes
$$
f_X(x)=\frac{e^{-\frac{u^2}{2}}}{2 \pi \sigma_X \sqrt{1-\rho^2}} \int_{-\infty}^{\infty} \exp \left\{\frac{-(v-\sigma u)^2}{2\left(1-\rho^2\right)}\right\} d v .
$$
Now let $z=(v-\rho u) / \sqrt{1-\rho^2}$, and using the fact that
$$
\int_{-\infty}^{\infty} \exp \left(\frac{-z^2}{2}\right) d z=\sqrt{2 \pi},
$$
show that
$$
f_X(x)=\frac{\exp \left[-\frac{1}{2}\left(\frac{x-\mu_X}{\sigma_X}\right)^2\right]}{\sqrt{2 \pi} \sigma_X} .
$$

Key Concepts

Bivariate Normal Distribution Properties

The bivariate normal distribution is characterized by its joint density function, which contains a quadratic form in the exponent involving correlation. One key property is that its marginal distributions are also normal, a fact that is demonstrated by integrating out one variable from the joint density.

Completing the Square

Completing the square is an algebraic method used to rewrite a quadratic expression in a form that is easier to integrate, especially when it appears in an exponential. This technique isolates terms involving a single variable and allows the integral to be split into factors, one of which turns into a standard normal integral.

Gaussian Integrals

Gaussian integrals refer to integrals of the form ?exp(?z²/2) dz over the entire real line, which are known to evaluate to ?(2?). Recognizing and using this result is essential for normalizing the marginal density derived from a bivariate normal distribution.

Marginal Density Function

This concept involves obtaining the probability distribution of a subset of variables by integrating the joint distribution over the nuisance variables. In this context, it explains how the marginal density of one normally distributed variable can be computed from the bivariate normal joint distribution by integrating out the other variable.

Change of Variables in Integration

Changing variables is a technique used to simplify an integral by substituting variables with scaled or shifted versions. In the proof, the substitution of standardized variables facilitates handling the quadratic form in the exponent and makes the integral resemble the standard Gaussian integral.

Transcript

00:01 For this problem, i have started by copying what we've been given for this problem.

00:05 What we are looking at is an improper integral over all real numbers, and we are taking that double integral of e to the negative x squared plus y squared da.

00:16 Now, we have a couple ways of looking at this, and these are all equivalent.

00:20 First, we can think of this as being over all real numbers.

00:23 Second, we can think of it as x and y both going from negative infinity to positive infinity.

00:30 The base just expands without bounds in any direction.

00:34 Another way to view that, and they're equivalent, just a different way of thinking about it, is to imagine that that base is a circle.

00:41 And it's a circle with a radius a, and we're going to let a just expand and expand and expand out to infinity.

00:46 So it's an ever -increasing circle.

00:49 What we want to show for this problem to start with is that, no matter how we write it here, the volume is going to equal pi.

00:57 So we're going to take one of these, in particular we're going to take this third case.

01:00 And we're going to show that it evaluate, when we evaluate it, we'll get pi for a result.

01:07 Now, since we are looking at the base as an ever -expanding circle, we are going to go to polar coordinates.

01:13 Now, as a reminder, when we go to polar coordinates, there are three basic things that we do a lot for substitutions.

01:20 First, x squared plus y squared equals r squared.

01:23 The second one is x equals r times cosine of theta.

01:27 And the third is y equals r times sine of theta.

01:32 Now out of these three, we're going to look at the first one.

01:35 Because if you look, we have that x squared plus y squared already in our integral up in the exponent.

01:41 So when we go to substitute this in, we're going to have a double integral.

01:44 We'll come back and put the limits in in in just a moment.

01:47 We have e to the negative r squared.

01:51 We have a d r.

01:53 Sorry, i would do theta first.

01:54 D theta, dr.

01:55 And i have to put in an extra factor of r.

01:57 You always have to do that any time you are using polar coordinates when you're switching over to those.

02:02 You have to put that extra r in.

02:04 Okay, now what are my limits? well, first, theta.

02:08 Our base is a circle expanding without bounds.

02:11 It's a full circle, so theta is going from zero to two pi.

02:15 And my radius is expanding without bounds, so it's going to go from zero to infinity.

02:21 Okay, let's begin by integrating that innermost integral.

02:26 So the outer one is going to stay the same.

02:28 For the inner one, i am integrating with respect to theta, and i don't really have a theta there.

02:33 So this is just going to become theta r, e to the negative r squared.

02:38 And i'm evaluating this.

02:40 I'll write it this way from theta equals 0 to 2 pi, d r.

02:48 Okay, first, theta is 2 pi.

02:50 This gives me 2 pi r, e to the negative r squared.

02:56 And if theta is zero, this just becomes zero.

02:59 So that's my new integral.

03:03 Now, this next piece, you might be able to do this in your head.

03:07 I'm going to write it out just in case anybody who's watching this video, if you're a little unsure or hesitant to do this, i'm going to write this piece out.

03:14 But we're going to do a u substitution.

03:16 I would really like to have e to the u because taking the integral of e to the u is very simple.

03:22 So in order to do that, i'm going to let you equal my exponent, negative r squared.

03:29 That means that d -u would be negative 2r -dr.

03:34 Well, i already have the 2 and the r and the dr.

03:36 I just need a negative.

03:38 So i can multiply by two negatives.

03:40 I'll put one on the outside.

03:42 And that way, i haven't changed the value of my integral.

03:46 So when i plug everything in, i have a negative on the outside.

03:50 I have r going from zero to infinity.

03:55 I have a pie.