Books Probability and Statistical Inference

The Bivariate Normal Distribution

Many applications are concerned with the conditional distribution of one of the random variables, say $𝑌$ , given the the value of the other random variable $𝑋=𝑥$ . Assume, we can make the following assumptions:

$𝑌|𝑋=𝑥$ follows a normal distribution.
The conditional mean ${\mu }_{𝑌|𝑥}$ is a linear function of $𝑥$ .
The conditional variance ${\sigma }_{𝑌|𝑥}^{2}$ is a constant. That is, it is independent of $𝑥$ . (Recall that ${\sigma }_{𝑌|𝑥}^{2}$ can be regarded as a function of $𝑥$ .)

Assumption (b) implies that

{\mu }_{𝑌|𝑥}={\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋}).

Now, we will study the implication of assumption (c). We have

\begin{aligned}{\sigma }_{𝑌|𝑥}^{2}=\mathbb{E}[{(𝑌-{\mu }_{𝑌|𝑥})}^{2}|𝑋=𝑥] & ={\int }_{-\infty }^{\infty }{(𝑦-{\mu }_{𝑌|𝑥})}^{2}h(𝑦|𝑥)\, \mathrm{d}𝑦 \\ & =\underbrace{{\int }_{-\infty }^{\infty }{(𝑦-{\mu }_{𝑌|𝑥})}^{2}h(𝑦|𝑥)\, \mathrm{d}𝑦}_{\text{independent of}\, 𝑥\, \text{by assumption}}\underbrace{{\int }_{-\infty }^{\infty }{𝑓}_{𝑋}(𝑥)\, \mathrm{d}𝑥}_{=1} \\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{(𝑦-{\mu }_{𝑌|𝑥})}^{2}\underbrace{h(𝑦|𝑥){𝑓}_{𝑋}(𝑥)}_{=𝑓(𝑥,𝑦)}\, \mathrm{d}𝑦\, \mathrm{d}𝑥 \\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{(𝑦-{\mu }_{𝑌|𝑥})}^{2}𝑓(𝑥,𝑦)\, \mathrm{d}𝑦\, \mathrm{d}𝑥 \\ & \text{Regard as a joint expectation} \\ & =\mathbb{E}[{(𝑌-{\mu }_{𝑌|𝑋})}^{2}] \\ & \text{Note here we write}\, 𝑋\, \text{instead of}\, 𝑥\, \text{since}\, 𝑋\, \text{is also random} \\ & =\mathbb{E}\left[{\left(𝑌-{\mu }_{𝑌}-\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑋-{\mu }_{𝑋})\right)}^{2}\right] \\ & =\mathbb{E}[{(𝑌-{\mu }_{𝑌})}^{2}] \\ & -2\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}\mathbb{E}[(𝑋-{\mu }_{𝑋})(𝑌-{\mu }_{𝑌})]+{\left(\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}\right)}^{2}\mathbb{E}[{(𝑋-{\mu }_{𝑋})}^{2}] \\ & ={\sigma }_{𝑌}^{2}(1-{\rho }^{2})\end{aligned}

In summary, we have concluded that the formula for the conditional variance is

{\sigma }_{𝑌|𝑥}^{2}={\sigma }_{𝑌}^{2}(1-{\rho }^{2}).

Therefore,

𝑌|𝑋=𝑥∼𝒩\left({\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋}),{\sigma }_{𝑌}^{2}(1-{\rho }^{2})\right).

The conditional PDF $h(𝑦|𝑥)$ of $𝑌|𝑥$ is given by

h(𝑦|𝑥)=\frac{1}{\sqrt{2\pi {\sigma }_{𝑌}^{2}(1-{\rho }^{2})}}\exp \left[-\frac{{\left[𝑦-{\mu }_{𝑌}-\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋})\right]}^{2}}{2{\sigma }_{𝑌}^{2}(1-{\rho }^{2})}\right]

(4.8)

Note that so far we only assume that $𝑋$ has mean ${\mu }_{𝑋}$ and variance ${\sigma }_{𝑋}^{2}$ ; nothing is said about the distribution of $𝑋$ . If we can further assume that $𝑋$ follows a normal distribution, i.e., $𝑋∼𝒩({\mu }_{𝑋},{\sigma }_{𝑋}^{2})$ , then the PDF of $𝑋$ is

{𝑓}_{𝑋}(𝑥)=\frac{1}{\sqrt{2\pi {\sigma }_{𝑋}^{2}}}\exp \left[-\frac{{(𝑥-{\mu }_{𝑋})}^{2}}{2{\sigma }_{𝑋}^{2}}\right].

Combined with (4.8), we can show that the joint PDF of $𝑋$ and $𝑌$ is given by

𝑓(𝑥,𝑦)=\frac{1}{2\pi {\sigma }_{𝑋}{\sigma }_{𝑌}\sqrt{1-{\rho }^{2}}}\exp \left[-\frac{𝑞(𝑥,𝑦)}{2}\right]

where

𝑞(𝑥,𝑦)=\frac{1}{1-{\rho }^{2}}\left[{\left(\frac{𝑥-{\mu }_{𝑋}}{{\sigma }_{𝑋}}\right)}^{2}-2\rho \left(\frac{𝑥-{\mu }_{𝑋}}{{\sigma }_{𝑋}}\right)\left(\frac{𝑦-{\mu }_{𝑌}}{{\sigma }_{𝑌}}\right)+{\left(\frac{𝑦-{\mu }_{𝑌}}{{\sigma }_{𝑌}}\right)}^{2}\right].

(4.9)

Remark

We introduce $𝑞(𝑥,𝑦)$ to ease the typing of the expression of $𝑓(𝑥,𝑦)$ for it is too long otherwise.

Note that in (4.9), we deliberately organize in terms of $(𝑥-{\mu }_{𝑋})/{\sigma }_{𝑋}$ and $(𝑦-{\mu }_{𝑌})/{\sigma }_{𝑌}$ to normalize (or standardize) the data of $𝑥$ and $𝑦$ . Recall that these terms are related to the standard scores of $𝑋$ and $𝑌$ , respectively.

We formally define the bivariate normal distribution as follows.

Definition 4.4

If the joint PDF of $𝑋$ and $𝑌$ is of the form (4.10), then the distribution of $(𝑋,𝑌)$ is called the bivariate normal distribution with parameters ${\mu }_{𝑋}$ , ${\sigma }_{𝑋}^{2}$ , ${\mu }_{𝑌}$ , ${\sigma }_{𝑌}^{2}$ , and $\rho$ where ${\sigma }_{𝑋}>0$ , ${\sigma }_{𝑌}>0$ , and $|\rho |<1$ .

𝑓(𝑥,𝑦)=\frac{1}{2\pi {\sigma }_{𝑋}{\sigma }_{𝑌}\sqrt{1-{\rho }^{2}}}\exp \left[-\frac{𝑞(𝑥,𝑦)}{2}\right]

(4.10)

where

𝑞(𝑥,𝑦)=\frac{1}{1-{\rho }^{2}}\left[{\left(\frac{𝑥-{\mu }_{𝑋}}{{\sigma }_{𝑋}}\right)}^{2}-2\rho \left(\frac{𝑥-{\mu }_{𝑋}}{{\sigma }_{𝑋}}\right)\left(\frac{𝑦-{\mu }_{𝑌}}{{\sigma }_{𝑌}}\right)+{\left(\frac{𝑦-{\mu }_{𝑌}}{{\sigma }_{𝑌}}\right)}^{2}\right].

Remark

Recall that we arrived at the bivariate normal PDF by first imposing structure on the conditional distribution of $𝑌$ given $𝑋=𝑥$ , then assuming that $𝑋$ is normal, and only then obtaining the joint density. That order matches how the distribution is often motivated: it is natural and keeps intuition in view. By contrast, many texts begin with the full joint formula and then derive properties; that arrangement can be harder to read and can obscure the motivating structure and the secondary insights that arise when one builds the model in stages.

For a logically tight treatment, however, we now take Definition 4.4 as the definition of the bivariate normal distribution. Until this point, we had no official definition; accordingly, any property used informally in the heuristic derivation must not be carried over without proof. Once Definition 4.4 is in place, we must reestablish those properties starting from the density alone. In particular, we may not treat ${\mu }_{𝑋}$ , ${\sigma }_{𝑋}^{2}$ , ${\mu }_{𝑌}$ , ${\sigma }_{𝑌}^{2}$ , and $\rho$ merely as suggestive symbols for the means and variances of $𝑋$ and $𝑌$ and the correlation between them; we must derive those interpretations from Definition 4.4.

Reestablishing these facts from the definition may look arduous; in practice the work is largely routine, because the constructive argument that produced the joint density already suggests what to integrate and how the pieces fit together.

Proposition 4.1

If $𝑋$ and $𝑌$ have the bivariate normal distribution with parameters ${\mu }_{𝑋}$ , ${\sigma }_{𝑋}^{2}$ , ${\mu }_{𝑌}$ , ${\sigma }_{𝑌}^{2}$ , and $\rho$ , then we have the following properties:

$𝑋∼𝒩({\mu }_{𝑋},{\sigma }_{𝑋}^{2})$ and $𝑌∼𝒩({\mu }_{𝑌},{\sigma }_{𝑌}^{2})$ .
$𝑋|𝑌=𝑦∼𝒩\left({\mu }_{𝑋}+\rho \frac{{\sigma }_{𝑋}}{{\sigma }_{𝑌}}(𝑦-{\mu }_{𝑌}),{\sigma }_{𝑋}^{2}(1-{\rho }^{2})\right)$ and $𝑌|𝑋=𝑥∼𝒩\left({\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋}),{\sigma }_{𝑌}^{2}(1-{\rho }^{2})\right)$ .
The correlation coefficient of $𝑋$ and $𝑌$ is $\rho$ .

Proof

(a) We only prove $𝑋∼𝒩({\mu }_{𝑋},{\sigma }_{𝑋}^{2})$ since the other case is analogous. We can factorize $𝑓(𝑥,𝑦)$ as follows:

\begin{aligned}𝑓(𝑥,𝑦) & =\underbrace{\frac{1}{\sqrt{2\pi {\sigma }_{𝑋}^{2}}}\exp \left[-\frac{{(𝑥-{\mu }_{𝑋})}^{2}}{2{\sigma }_{𝑋}^{2}}\right]}_{={\tilde{𝑓}}_{𝑋}(𝑥)} \\ & \cdot \underbrace{\frac{1}{\sqrt{2\pi {\sigma }_{𝑌}^{2}(1-{\rho }^{2})}}\exp \left[-\frac{{\left[𝑦-{\mu }_{𝑌}-\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋})\right]}^{2}}{2{\sigma }_{𝑌}^{2}(1-{\rho }^{2})}\right]}_{=\tilde{h}(𝑥,𝑦)} \\ & ={\tilde{𝑓}}_{𝑋}(𝑥)\tilde{h}(𝑥,𝑦).\end{aligned}

(4.11)

The goal is to show that

{\int }_{-\infty }^{\infty }\tilde{h}(𝑥,𝑦)\, \mathrm{d}𝑦=1.

And this is evident when we note that $\tilde{h}(𝑥,𝑦)$ is exactly the PDF of some random variable following $𝒩\left({\mu }_{𝑋}+\rho \frac{{\sigma }_{𝑋}}{{\sigma }_{𝑌}}(𝑦-{\mu }_{𝑌}),{\sigma }_{𝑋}^{2}(1-{\rho }^{2})\right)$ . Therefore,

{\int }_{-\infty }^{\infty }𝑓(𝑥,𝑦)\, \mathrm{d}𝑦={\int }_{-\infty }^{\infty }{\tilde{𝑓}}_{𝑋}(𝑥)\tilde{h}(𝑥,𝑦)\, \mathrm{d}𝑦={\tilde{𝑓}}_{𝑋}(𝑥){\int }_{-\infty }^{\infty }\tilde{h}(𝑥,𝑦)\, \mathrm{d}𝑦={\tilde{𝑓}}_{𝑋}(𝑥).

This shows that ${\tilde{𝑓}}_{𝑋}(𝑥)$ is indeed the PDF of $𝑋$ ; and by its form, we conclude that $𝑋∼𝒩({\mu }_{𝑋},{\sigma }_{𝑋}^{2})$ .

(b) We only prove $𝑌|𝑋=𝑥∼𝒩\left({\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑌}),{\sigma }_{𝑌}^{2}(1-{\rho }^{2})\right)$ . Now that we know $𝑋∼𝒩({\mu }_{𝑋},{\sigma }_{𝑋}^{2})$ , we can use (4.11) to show that

h(𝑦|𝑥)=\tilde{h}(𝑥,𝑦)=\frac{1}{\sqrt{2\pi {\sigma }_{𝑌}^{2}(1-{\rho }^{2})}}\exp \left[-\frac{{\left[𝑦-{\mu }_{𝑌}-\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑋})\right]}^{2}}{2{\sigma }_{𝑌}^{2}(1-{\rho }^{2})}\right].

This implies that $𝑌|𝑋=𝑥∼𝒩\left({\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑌}),{\sigma }_{𝑌}^{2}(1-{\rho }^{2})\right)$ .

{\mu }_{𝑌|𝑥}={\mu }_{𝑌}+\rho \frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}(𝑥-{\mu }_{𝑌}),

which is a linear function of $𝑥$ . Therefore, the coefficient of $𝑥$ must be $\tilde{\rho }\frac{{\sigma }_{𝑌}}{{\sigma }_{𝑋}}$ ; and hence $\rho =\tilde{\rho }$ .

If we fix $𝑧>0$ and consider the level set $𝑓(𝑥,𝑦)=𝑧$ , then (4.10) rearranges to a quadratic equation in $𝑥$ and $𝑦$ :

𝑞(𝑥,𝑦)+2\ln \left(𝑧\cdot 2\pi {\sigma }_{𝑋}{\sigma }_{𝑌}\sqrt{1-{\rho }^{2}}\right)=0.

We show that this curve is an ellipse. Equivalently, every plane parallel to the $𝑥𝑦$ -plane intersects the graph of the bivariate normal density in an ellipse; thus the contours of the density are ellipses.

Written in the general form

𝐴{𝑥}^{2}+𝐵𝑥𝑦+𝐶{𝑦}^{2}+𝐷𝑥+𝐸𝑦+𝐹=0,

where the coefficients are

\begin{aligned}𝐴 & =\frac{1}{{\sigma }_{𝑋}^{2}(1-{\rho }^{2})} \\ 𝐵 & =-\frac{2\rho }{{\sigma }_{𝑋}{\sigma }_{𝑌}(1-{\rho }^{2})} \\ 𝐶 & =\frac{1}{{\sigma }_{𝑌}^{2}(1-{\rho }^{2})} \\ 𝐷 & =\frac{2}{{\sigma }_{𝑋}{\sigma }_{𝑌}(1-{\rho }^{2})}\left(-\frac{{\sigma }_{𝑌}{\mu }_{𝑥}}{{\sigma }_{𝑋}}+\rho {\mu }_{𝑦}\right) \\ 𝐸 & =\frac{2}{{\sigma }_{𝑋}{\sigma }_{𝑌}(1-{\rho }^{2})}\left(-\frac{{\sigma }_{𝑋}{\mu }_{𝑦}}{{\sigma }_{𝑌}}+\rho {\mu }_{𝑥}\right) \\ 𝐹 & =\frac{1}{1-{\rho }^{2}}\left(\frac{{\mu }_{𝑋}^{2}}{{\sigma }_{𝑋}^{2}}-\frac{2\rho {\mu }_{𝑋}{\mu }_{𝑌}}{{\sigma }_{𝑋}{\sigma }_{𝑌}}+\frac{{\mu }_{𝑌}^{2}}{{\sigma }_{𝑌}^{2}}\right)+2\ln \left(𝑧\cdot 2\pi {\sigma }_{𝑋}{\sigma }_{𝑌}\sqrt{1-{\rho }^{2}}\right)\end{aligned}

The discriminant is

\Delta ={𝐵}^{2}-4𝐴𝐶=\frac{4({\rho }^{2}-1)}{{\sigma }_{𝑋}^{2}{\sigma }_{𝑌}^{2}{(1-{\rho }^{2})}^{2}}<0.

Since $\Delta <0$ , the conic is an ellipse.

The formula for the center $({𝑐}_{𝑥},{𝑐}_{𝑦})$ of the ellipse is

\begin{aligned}{𝑐}_{𝑥} & =\frac{𝐵𝐸-2𝐶𝐷}{4𝐴𝐶-{𝐵}^{2}} \\ {𝑐}_{𝑦} & =\frac{𝐵𝐷-2𝐴𝐸}{4𝐴𝐶-{𝐵}^{2}}.\end{aligned}

After a lengthy but routine calculation, we obtain

({𝑐}_{𝑥},{𝑐}_{𝑦})=({\mu }_{𝑋},{\mu }_{𝑌}).

That is, the centers of all the ellipses are at $({\mu }_{𝑋},{\mu }_{𝑌})$ , independent of $𝑧$ .

Example 4.5

Let $(𝑋,𝑌)$ have the bivariate normal distribution with parameters ${\mu }_{𝑋}=10$ , ${\sigma }_{𝑋}^{2}=9$ , ${\mu }_{𝑌}=12$ , ${\sigma }_{𝑌}^{2}=16$ , and $\rho =0.6$ . Figure 4.4 shows the joint PDF as a surface over the $𝑥𝑦$ -plane on the left. On the right, contours of that density are drawn in the plane; superimposed is the conditional mean ${\mu }_{𝑌|𝑥}$ of $𝑌$ given $𝑋=𝑥$ , i.e., the line $𝑦={\mu }_{𝑌}+\rho {\sigma }_{𝑌}/{\sigma }_{𝑋}(𝑥-{\mu }_{𝑋})$ . It passes through the point $({\mu }_{𝑋},{\mu }_{𝑌})$ , which is the center of all the ellipses.

Figure 4.4 (a) Joint PDF of $𝑋$ and $𝑌$ . (b) Contours in the $𝑥𝑦$ -plane with the conditional mean ${\mu }_{𝑌|𝑥}$ .

Recall that if two random variables are independent, then their correlation coefficient is $0$ . But the converse is not true. However, for the bivariate normal distribution, the independence is equivalent to the zero correlation.

Theorem 4.1

Let

𝑋

and

𝑌

have the bivariate normal distribution with correlation coefficient

\rho

. Then,

𝑋

and

𝑌

are independent if and only if

\rho =0

Proof

We only need to show the reverse direction. Assume that $\rho =0$ . It follows from (4.11) that

𝑓(𝑥,𝑦)=\underbrace{\frac{1}{\sqrt{2\pi {\sigma }_{𝑋}^{2}}}\exp \left[-\frac{{(𝑥-{\mu }_{𝑋})}^{2}}{2{\sigma }_{𝑋}^{2}}\right]}_{={𝑓}_{𝑋}(𝑥)}\cdot \underbrace{\frac{1}{\sqrt{2\pi {\sigma }_{𝑌}^{2}}}\exp \left[-\frac{{(𝑦-{\mu }_{𝑌})}^{2}}{2{\sigma }_{𝑌}^{2}}\right]}_{={𝑓}_{𝑌}(𝑦)}.

This shows that the joint PDF can be factored as the product of the marginal PDFs. Therefore, $𝑋$ and $𝑌$ are independent.

Exercises

Exercise 4.8

Let

𝑓(𝑥,𝑦)=\frac{1}{2\pi }{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}[1+𝑥𝑦{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-2)/2}],\quad 𝑥,𝑦\in \mathbb{R}.

Show that

$𝑓(𝑥,𝑦)$ is a joint PDF.
The marginal PDFs ${𝑓}_{𝑋}(𝑥)$ and ${𝑓}_{𝑌}(𝑦)$ are each normal.

Note

This example tells us that $𝑋$ and $𝑌$ may each be normal; and yet their joint distribution may not be a bivariate normal distribution. Figure 4.5 shows the joint PDF of $𝑋$ and $𝑌$ in this example.

Figure 4.5 Joint PDF $𝑓(𝑥,𝑦)$ of $𝑋$ and $𝑌$ . $𝑋$ and $𝑌$ are each normal, but $𝑓(𝑥,𝑦)$ is not a bivariate normal PDF.

Proof

(a) The statement that $𝑓(𝑥,𝑦)\geq 0$ for all $𝑥,𝑦\in \mathbb{R}$ is not that obvious. The key is to show that

|𝑥𝑦|{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-2)/2}\leq 1.

We have

|𝑥𝑦|{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-2)/2}\leq |𝑥𝑦|{𝑒}^{-(2|𝑥𝑦|-2)/2}=|𝑥𝑦|{𝑒}^{1-|𝑥𝑦|}

where the inequality follows from the fact that ${𝑥}^{2}+{𝑦}^{2}\geq 2|𝑥𝑦|$ . Let $𝑡=|𝑥𝑦|$ and define $𝑔(𝑡)=𝑡{𝑒}^{1-𝑡}$ . We now estimate $𝑔(𝑡)$ . Taking the derivative of $𝑔(𝑡)$ ,

{𝑔}^{\prime }(𝑡)=(1-𝑡){𝑒}^{1-𝑡},\quad 𝑡>0.

Note that

{𝑔}^{\prime }(𝑡)\begin{cases}>0,\quad 𝑡<1 \\ =0,\quad 𝑡=1 \\ <0,\quad 𝑡>1\end{cases}

Therefore, $𝑔$ attains its maximum at $𝑡=1$ and the maximum value is $𝑔(1)=1$ . In summary, we shown that

|𝑥𝑦|{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-2)/2}\leq |𝑥𝑦|{𝑒}^{1-|𝑥𝑦|}=𝑔(𝑡)\leq 1

where the equality holds if and only if $|𝑥|=|𝑦|=1$ . This implies that indeed $𝑓(𝑥,𝑦)\geq 0$ for all $𝑥,𝑦\in \mathbb{R}$ . Moreover, the only two points where $𝑓$ vanishes are $𝑓(1,-1)=0$ and $𝑓(-1,1)=0$ .

We also need to show that $𝑓(𝑥,𝑦)$ integrates to $1$ . Write

𝑓(𝑥,𝑦)=\frac{1}{2\pi }[{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}+𝑥𝑦{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-1)}].

We divide the integral into two parts:

\begin{aligned} & {\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }𝑓(𝑥,𝑦)\, \mathrm{d}𝑥\, \mathrm{d}𝑦 \\ = & \frac{1}{2\pi }\left(\underbrace{{\int }_{-\infty }^{\infty }{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}\, \mathrm{d}𝑥\, \mathrm{d}𝑦}_{={𝐼}_{1}}+\underbrace{{\int }_{-\infty }^{\infty }𝑥𝑦{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-1)}\, \mathrm{d}𝑥\, \mathrm{d}𝑦}_{={𝐼}_{2}}\right).\end{aligned}

(4.12)

By using the polar coordinate transformation, we have

{\int }_{-\infty }^{\infty }{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}\, \mathrm{d}𝑥\, \mathrm{d}𝑦=2\pi .

And

{\int }_{-\infty }^{\infty }{𝑒}^{-{𝑥}^{2}/2}\, \mathrm{d}𝑥=\sqrt{2\pi }.

(4.13)

(Recall that we have done above calculations when we were studying the normal distributions.) Therefore, ${𝐼}_{1}=2\pi$ . We now evaluate ${𝐼}_{2}$ . (The goal is to show that ${𝐼}_{2}=0$ .)

Again, using the polar coordinate transformation, we have

\begin{aligned}{𝐼}_{2}={\int }_{-\infty }^{\infty }𝑥𝑦{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-1)}\, \mathrm{d}𝑥\, \mathrm{d}𝑦 & ={\int }_{-\infty }^{\infty }{\int }_{0}^{2\pi }\sin \theta \cos \theta {𝑟}^{3}{𝑒}^{-({𝑟}^{2}-1)}\, \mathrm{d}\theta \, \mathrm{d}𝑟 \\ & ={\int }_{-\infty }^{\infty }{𝑟}^{3}{𝑒}^{-({𝑟}^{2}-1)}\underbrace{{\int }_{0}^{2\pi }\sin \theta \cos \theta \, \mathrm{d}\theta }_{=0}\, \mathrm{d}𝑟 \\ & =0.\end{aligned}

Plugging the values of ${𝐼}_{1}$ and ${𝐼}_{2}$ into (4.12) yields

{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }𝑓(𝑥,𝑦)\, \mathrm{d}𝑥\, \mathrm{d}𝑦=1.

(b) We only compute the marginal ${𝑓}_{𝑋}(𝑥)$ . We divide the integral into two parts:

{𝑓}_{𝑋}(𝑥)={\int }_{-\infty }^{\infty }𝑓(𝑥,𝑦)\, \mathrm{d}𝑦=\frac{1}{2\pi }\left(\underbrace{{\int }_{-\infty }^{\infty }{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}\, \mathrm{d}𝑦}_{={𝐼}_{1}}+\underbrace{{\int }_{-\infty }^{\infty }𝑥𝑦{𝑒}^{-({𝑥}^{2}+{𝑦}^{2}-1)}\, \mathrm{d}𝑦}_{={𝐼}_{2}}\right).

By (4.13), we obtain

{𝐼}_{1}={\int }_{-\infty }^{\infty }{𝑒}^{-({𝑥}^{2}+{𝑦}^{2})/2}\, \mathrm{d}𝑦={𝑒}^{-{𝑥}^{2}/2}{\int }_{-\infty }^{\infty }{𝑒}^{-{𝑦}^{2}/2}\, \mathrm{d}𝑦=\sqrt{2\pi }{𝑒}^{-{𝑥}^{2}/2}.

As for ${𝐼}_{2}$ , we may immediately conclude that ${𝐼}_{2}=0$ since the integrand is an odd function of $𝑦$ . Therefore,

{𝑓}_{𝑋}(𝑥)=\frac{1}{2\pi }\left(\sqrt{2\pi }{𝑒}^{-{𝑥}^{2}/2}+0\right)=\frac{1}{\sqrt{2\pi }}{𝑒}^{-{𝑥}^{2}/2}.

This shows that $𝑋∼𝒩(0,1)$ , i.e., $𝑋$ follows a standard normal distribution. Similarly, we can show that $𝑌∼𝒩(0,1)$ .