# 21.2 - Joint P.D.F. of X and Y

21.2 - Joint P.D.F. of X and Y

We previously assumed that:

1. $$Y$$ follows a normal distribution,
2. $$E(Y|x)$$, the conditional mean of $$Y$$ given $$x$$ is linear in $$x$$, and
3. $$\text{Var}(Y|x)$$, the conditional variance of $$Y$$ given $$x$$ is constant.

Based on these three stated assumptions, we found the conditional distribution of $$Y$$ given $$X=x$$. Now, we'll add a fourth assumption, namely that:

4. $$X$$ follows a normal distribution.

Based on the four stated assumptions, we will now define the joint probability density function of $$X$$ and $$Y$$.

Definition. Assume $$X$$ is normal, so that the p.d.f. of $$X$$ is:

$$f_X(x)=\dfrac{1}{\sigma_X \sqrt{2\pi}} \text{exp}\left[-\dfrac{(x-\mu_X)^2}{2\sigma^2_X}\right]$$

for $$-\infty<x<\infty$$. And, assume that the conditional distribution of $$Y$$ given $$X=x$$ is normal with conditional mean:

$$E(Y|x)=\mu_Y+\rho \dfrac{\sigma_Y}{\sigma_X}(x-\mu_X)$$

and conditional variance:

$$\sigma^2_{Y|X}= \sigma^2_Y(1-\rho^2)$$

That is, the conditional distribution of $$Y$$ given $$X=x$$ is:

\begin{align} h(y|x) &= \dfrac{1}{\sigma_{Y|X} \sqrt{2\pi}} \text{exp}\left[-\dfrac{(Y-\mu_{Y|X})^2}{2\sigma^2_{Y|X}}\right]\\ &= \dfrac{1}{\sigma_Y \sqrt{1-\rho^2} \sqrt{2\pi}}\text{exp}\left[-\dfrac{[y-\mu_Y-\rho \dfrac{\sigma_Y}{\sigma_X}(X-\mu_X)]^2}{2\sigma^2_Y(1-\rho^2)}\right],\quad -\infty<x<\infty\\ \end{align}

Therefore, the joint probability density function of $$X$$ and $$Y$$ is:

$$f(x,y)=f_X(x) \cdot h(y|x)=\dfrac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \text{exp}\left[-\dfrac{q(x,y)}{2}\right]$$

where:

$$q(x,y)=\left(\dfrac{1}{1-\rho^2}\right) \left[\left(\dfrac{X-\mu_X}{\sigma_X}\right)^2-2\rho \left(\dfrac{X-\mu_X}{\sigma_X}\right) \left(\dfrac{Y-\mu_Y}{\sigma_Y}\right)+\left(\dfrac{Y-\mu_Y}{\sigma_Y}\right)^2\right]$$

This joint p.d.f. is called the bivariate normal distribution.

Our textbook has a nice three-dimensional graph of a bivariate normal distribution. You might want to take a look at it to get a feel for the shape of the distribution. Now, let's turn our attention to an important property of the correlation coefficient if $$X$$ and $$Y$$ have a bivariate normal distribution.

Theorem

If $$X$$ and $$Y$$ have a bivariate normal distribution with correlation coefficient $$\rho_{XY}$$, then $$X$$ and $$Y$$ are independent if and only if $$\rho_{XY}=0$$. That "if and only if" means:

1. If $$X$$ and $$Y$$ are independent, then $$\rho_{XY}=0$$
2. If $$\rho_{XY}=0$$, then $$X$$ and $$Y$$ are independent

Recall that the first item is always true. We proved it back in the lesson that addresses the correlation coefficient. We also looked at a counterexample i that lesson that illustrated that item (2) was not necessarily true! Well, now we've just learned a situation in which it is true, that is, when $$X$$ and $$Y$$ have a bivariate normal distribution. Let's see why item (2) must be true in that case.

### Proof

Since we previously proved item (1), our focus here will be in proving item (2). In order to prove that $$X$$ and $$Y$$ are independent when $$X$$ and $$Y$$ have the bivariate normal distribution and with zero correlation, we need to show that the bivariate normal density function:

$$f(x,y)=f_X(x)\cdot h(y|x)=\dfrac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \text{exp}\left[-\dfrac{q(x,y)}{2}\right]$$

factors into the normal p.d.f of $$X$$ and the normal p.d.f. of $$Y$$. Well, when $$\rho_{XY}=0$$:

$$q(x,y)=\left(\dfrac{1}{1-0^2}\right) \left[\left(\dfrac{X-\mu_X}{\sigma_X}\right)^2+0+\left(\dfrac{Y-\mu_Y}{\sigma_Y}\right)^2 \right]$$

which simplifies to:

$$q(x,y)=\left(\dfrac{X-\mu_X}{\sigma_X}\right)^2+\left(\dfrac{Y-\mu_Y}{\sigma_Y}\right)^2$$

Substituting this simplified $$q(x,y)$$ into the joint p.d.f. of $$X$$ and $$Y$$, and simplifying, we see that $$f(x,y)$$ does indeed factor into the product of $$f(x)$$ and $$f(y)$$:

\begin{align} f(x,y) &= \dfrac{1}{2\pi \sigma_X \sigma_Y \sqrt{1-\rho^2}} \text{exp}\left[-\dfrac{1}{2}\left(\dfrac{X-\mu_X}{\sigma_X}\right)^2--\dfrac{1}{2}\left(\dfrac{Y-\mu_Y}{\sigma_Y}\right)^2\right]\\ &= \dfrac{1}{\sigma_X \sqrt{2\pi} \sigma_Y \sqrt{2\pi}}\text{exp}\left[-\dfrac{(x-\mu_X)^2}{2\sigma_X^2}\right] \text{exp}\left[-\dfrac{(y-\mu_Y)^2}{2\sigma_Y^2}\right]\\ &= \dfrac{1}{\sigma_X \sqrt{2\pi}}\text{exp}\left[-\dfrac{(x-\mu_X)^2}{2\sigma_X^2}\right]\cdot \dfrac{1}{\sigma_Y \sqrt{2\pi}}\text{exp}\left[-\dfrac{(y-\mu_Y)^2}{2\sigma_Y^2}\right]\\ &=f_X(x)\cdot f_Y(y)\\ \end{align}

Because we have shown that:

$$f(x,y)=f_X(x)\cdot f_Y(y)$$

we can conclude, by the definition of independence, that $$X$$ and $$Y$$ are independent. Our proof is complete.

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