# Expected value for jointly Gaussian RV

I'm studying for a midterm and came across this problem that I'm having trouble finishing.

Let X, Y be Jointly Gaussian random variables with mean E[X] = 1, E[Y] = 0, and covariance matrix $$\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}$$

A) Find $$E(X|Y)$$

B) Find $$E((X+Y)^2|Y)$$

For part A, I used a linear estimator since they are a Gaussian RV and got the following:

$$E(X|Y) = E(X) + cov(X,Y)cov^{-1}(Y)(Y-E(Y)) = 1+Y/2$$

For part B, I expanded the $$(X+Y)^2$$ term and use the linearity property of the expected value and arrived at:

$$E((X+Y)^2|Y) = E(X^2|Y) + 2YE(X|Y) + Y^2$$, and then plugged in the result from A on the second term.

The problem is, I'm not being able to figure out how to solve the first term with $$X^2$$. According to the answer sheet, the result should be:

$$var(X) − \frac{cov(X, Y)^2}{var(Y)} + (1 + Y/2)^2 + Y^2 + 2Y (1 + Y/2)$$

But I'm not getting $$E(X^2|Y) = var(X) − \frac{cov(X, Y)^2}{var(Y)} + (1 + Y/2)^2$$ no matter what I do. I've trying to also apply a linear estimator but without success.

Any help would be greatly appreciated!

• $E[X^2\mid Y]=Var[X\mid Y]+(E[X\mid Y])^2$. Nov 3, 2019 at 21:32

Let $$Y \sim N(0, 2)$$ and $$Z \sim N(0, 2)$$ be independent. Let $$X := \frac{1}{2} Y + \frac{\sqrt{3}}{2} Z + 1.$$
Then $$(X,Y)$$ is jointly Gaussian (since any linear combination of $$X$$ and $$Y$$ is Gaussian). You can check that $$E[X]=1$$ and $$E[Y] = 0$$ and $$\text{Var}(Y) = 2$$. You can also check that $$\text{Var}(X) = \frac{1}{4} \text{Var}(Y) + \frac{3}{4} \text{Var}(Z) = 2$$ and $$\text{Cov}(X,Y) = \frac{1}{2}\text{Cov}(Y,Y) = 1.$$ Thus $$(X,Y)$$ has the same distribution as your problem. This particular construction makes conditioning on $$Y$$ very easy, as we show now.
$$E[X \mid Y] = E[\frac{1}{2} Y + \frac{\sqrt{3}}{2} Z + 1 \mid Y] = \frac{1}{2} Y + 1.$$
$$E[X^2 \mid Y] = E[(\frac{1}{2} Y + \frac{\sqrt{3}}{2} Z + 1)^2 \mid Y] = \frac{1}{4}Y^2 + \frac{3}{4} E[Z^2] + 1 + Y = \frac{1}{4} Y^2 + Y + \frac{5}{2}.$$