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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!

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

1 Answer 1

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There is a useful characterization of bivariate Gaussian distributions as the linear transformation of independent univariate Gaussian random variables.


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.


$$E[X \mid Y] = E[\frac{1}{2} Y + \frac{\sqrt{3}}{2} Z + 1 \mid Y] = \frac{1}{2} Y + 1.$$

$$E[X^2 \mid Y] = E[(\frac{1}{2} Y + \frac{\sqrt{3}}{2} Z + 1)^2 \mid Y] = \frac{1}{4}Y^2 + \frac{3}{4} E[Z^2] + 1 + Y = \frac{1}{4} Y^2 + Y + \frac{5}{2}.$$

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