Conditional Expectation of Product of Normals given a Linear Combination

Question

I am tasked with solving a question for a qualifying exam, but I am a little lost about this question.

Let $\eta$ and $\xi$ be two independent standard Gaussian random variables. Find $\mathbb{E}(\xi\eta \mid \xi - 2\eta)$.

My attempt:

My main idea is to show that $\operatorname{Cov}(\xi-2\eta,\xi+2\eta)=0$, making the two independent since they are Gaussian. However, computing above, you get $\operatorname{Cov}(\xi-2\eta,\xi+2\eta)=\operatorname{Var}(\xi)-4\operatorname{Var}(\eta)\neq 0$.Alternatively, my thoughts were showing $\operatorname{Cov}(\xi-\eta,\xi+\eta)=0$, which is true. But is $\xi-\eta \perp \xi+\eta \implies \xi-2\eta \perp \xi+2\eta$? The main idea being after you show that $\xi-2\eta \perp \xi+2\eta $, write $\mathbb{E}(\eta\xi \mid \xi-2\eta)=\mathbb{E}(\frac{1}{8}[\xi+2\eta]^{2}+\frac{1}{8}[\xi-2\eta]^{2}\mid \xi-2\eta) $ and proceed.

Any help is appreciated.

Answer 1

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] -\frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariate normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 - \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} - \frac{2}{25}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Answer 2

Given a linear combination like $\xi-2\eta =Y$ (say) of independent standard normal variables $(\xi,\eta)$, you can always find another linear combination $X$ of $(\xi,\eta)$ such that $X$ and $Y$ are independent.

One obvious choice is $X=2\xi+\eta$, since of course $X$ and $Y$ are jointly normal and

$$\operatorname{Cov}(X,Y)=\operatorname{Cov}(2\xi+\eta,\xi-2\eta)=0.$$

Now observe that

$$5\xi=2X+Y$$ and

$$5\eta=X-2Y.$$

Therefore,

\begin{align} 25 E\left[\xi\eta\mid \xi-2\eta \right] & = E\left[(2X+Y)(X-2Y)\mid Y \right] \\& = 2E\left[X^2\right]-3YE\left[X\right]-2Y^2 \\&= 2(5-Y^2). \end{align}

Hence, $$E\left[\xi\eta\mid \xi-2\eta \right] = \frac25 - \frac{2}{25}(\xi-2\eta)^2 \quad,\,\text{a.s.}$$