Conditional Variance for Bivariate Normal Random Variables is Constant

Below is a problem I just did. My question for MSE is not how to solve it - but I provide it to illustrate what exactly I am asking.

Suppose X,Y are bivariate normal random variables with $$E[X] = 40$$, $$\mathrm{Var}(X) = 76$$, $$E[Y] = 30$$, $$\mathrm{Var}(Y) = 32$$, and $$\mathrm{Var}(X | Y = 28.5) = 57.$$

Calculate $$\mathrm{Var}(Y | X = 25)$$.

Although I know very little about bivariate random variables, I was able to solve this problem because I have a formula:

$$\mathrm{Var}(Y | X = x) = \sigma_{Y}^2(1 - \rho^2).$$

I am not certain, but based on convention I assume $$\rho$$ = $$\rho_{X,Y}$$ = $$\frac{\mathrm{Cov}(X,Y)}{\sigma_X \sigma_Y}$$.

Looking at the information given and my formula, I saw I could use the second formula to solve for $$\rho$$, and then re-use the formula to find the desired value. This is when I realized - the question in no way depends on the values of the conditioning variables ($$Y = 28.5, X=25)$$. This seemed strange to me. Keep in mind, my solution is just number crunching for me, I don't have a lot of background knowledge to provide intuition.

Can someone explain to me how this is intuitive that the function $$f(x) = \mathrm{Var}(Y | X = x)$$ is a constant function?

In my head when I picture a bivariate normal distribution I see what looks like an ant-hill centered over (0,0) in $$\mathbb{R}^2$$ (yes, technically I'm picturing a standard-bivariate normal). But then if I consider the cross sections cut out by fixing values of $$X$$, it seems the ones closer to the origin have a bigger hump - thus less variance? Is each cross section for different values of $$X$$ actually just like.. a scaling of the others? Thus variance stays fixed? Was this intentional in the construction of bivariate normals?

A way of seeing this is to consider how to generate pairs of random values from a bivariate Normal distribution with $$X \sim N(\mu_X,\sigma^2_X)$$ and $$Y \sim N(\mu_Y,\sigma^2_Y)$$ and covariance $$\sigma_{X,Y}=\rho \sigma_{X}\sigma_{Y}$$ between $$X$$ and $$Y$$.

One approach is to calculate the parts of $$Y$$ that depend on $$X$$ and do not depend on $$X$$ separately, and then add them together, which you can do as this is a bivariate Normal. The conditional variance of $$Y$$ given $$X=x$$ is then just the variance of the part of $$Y$$ that does not depend on $$X$$, and naturally this is not affected by the particular value $$x$$ that $$X$$ takes.

As a working algorithm:

• Generate random values for $$X\sim N(\mu_X,\sigma^2_X)$$ using your favourite piece of software
• Then $$\rho \frac{\sigma_{Y}}{\sigma_{X}}X$$ has mean $$\rho \frac{\sigma_{Y}}{\sigma_{X}}\mu_X$$ and variance $$\rho^2 \sigma_{Y}^2$$ and the covariance between $$X$$ and $$\rho \frac{\sigma_{Y}}{\sigma_{X}}X$$ is $$\rho \sigma_{X}\sigma_{Y}$$
• Now generate random values for $$Z\sim N\left(\mu_Y-\rho \frac{\sigma_{Y}}{\sigma_{X}}\mu_X,(1- \rho^2)\sigma_{Y}^2\right)$$ independent of $$X$$ so the covariance between $$X$$ and $$Z$$ is $$0$$
• Let $$Y= Z+\rho \frac{\sigma_{Y}}{\sigma_{X}}X$$. This has the result of $$Y\sim N(\mu_X,\sigma^2_X)$$ and $$(X,Y)$$ having a bivariate Normal distribution with covariance $$\rho \sigma_{X}\sigma_{Y}$$, which is what you are aiming for.

This means $$\mathrm{Var}(Y \mid X = x) = \mathrm{Var}(Z)=(1- \rho^2)\sigma_{Y}^2$$, which does not depend on the value of $$x$$

You are correct that the conditional variance of $$Y$$ given $$X=x$$ does not depend on $$x$$. [Note however that the conditional mean does depend on $$x$$.]

Regarding your mental model: you are picturing the joint density, which takes into account the randomness in $$X$$. The reason why the cross sections for fixed $$X$$ seem "smaller" as $$X$$ is farther from its mean is partly due to the fact that it is less and less likely for $$X$$ to be farther and farther from the mean. However, the conditional distribution is not just the cross section of the joint density; note that you still have to renormalize according to the randomness in $$X$$. (For discrete variables, recall $$P(Y=y \mid X=x) = P(Y=y, X=x) / P(X=x)$$.) It turns out that after renormalizing to take account for the likelihood of seeing $$X=x$$, the conditional variances are the same value $$\sigma_Y^2 (1-\rho)^2$$ in the case of the bivariate normal distribution.