Show that $E(Y\mid X=x)$ is a linear function in $x$

Question

Let $Y$ and $X$ be bivariate normal distributed with expectationvector $\mu=(\mu_Y,\mu_X)^T$ and covariance matrix $\Sigma=\begin{pmatrix}\sigma_Y^2 & p_{XY}\\p_{XY} & \sigma_X^2\end{pmatrix}$. Show that the conditional expectation $E(Y\mid X=x)$ is a linear function in $x$.

Hello!

To my knowledge it is $$ E(Y\mid X=x)=\int_{\mathbb{R}}y\cdot f_{Y\mid X}(y\mid x)\, dy. $$ So first I tried to determine $f_{Y\mid X}(y\mid x)$ by $$ f_{Y\mid X}(y\mid x)=\frac{f_{Y,X}(y,x)}{f_X(x)}. $$

To my calculation this is $$ f_{Y\mid X}(y\mid x)=\frac{\sigma_X}{\sqrt{2\pi}\sqrt{\sigma_Y^2\sigma_X^2-p_{XY}^2}}\exp\left(-\frac{1}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)}\cdot(\sigma_X^2(y-\mu_Y)^2-2p_{XY}(x-\mu_X)(y-\mu_Y)+\sigma_Y^2(x-\mu_X)^2)+\frac{1}{2}\frac{(x-\mu_X)^2}{\sigma_X^2}\right) $$

Is that right?

In case it is: How can I know determine

$$ \int y\cdot f_{Y\mid X}(y\mid x)\, dy, $$ i.e. how can I determine $$ \frac{\sigma_X}{\sqrt{2\pi}\sqrt{\sigma_Y^2\sigma_X^2-p_{XY}^2}}\int_{\mathbb{R}}y\cdot\exp\left(-\frac{1}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)}\cdot(\sigma_X^2(y-\mu_Y)^2-2p_{XY}(x-\mu_X)(y-\mu_Y)+\sigma_Y^2(x-\mu_X)^2)+\frac{1}{2}\frac{(x-\mu_X)^2}{\sigma_X^2}\right)\, dy? $$

Edit:

If I set $$ c:=-\frac{\sigma_X^2}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)}, d:=\frac{2p_{XY}(x-\mu_X)}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)}, q:=\frac{\sigma_Y^2(x-\mu_X)^2}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)} $$ and $$ w:=\frac{(x-\mu_X)^2}{2\sigma_X^2} $$ then I have to calculate the following: $$ \frac{\sigma_X\exp(-q+w)}{\sqrt{2\pi}\sqrt{\sigma_Y^2\sigma_X^2-p_{XY}^2}}\int y\cdot\exp(c(y-\mu_Y)^2+d(y-\mu_Y))\, dy $$

Answer 1

If you insist on working with densities, then note that the joint density of $(X,Y)$ is given by $$ f_{X,Y}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\Big(-\frac{1}{2(1-\rho^2)}r(x,y)\Big), $$ with $$ r(x,y)=\frac{(x-\mu_X)^2}{\sigma_X^2}-2\rho\frac{(x-\mu_X)}{\sigma_X}\frac{(y-\mu_Y)}{\sigma_Y}+\frac{(y-\mu_Y)^2}{\sigma_Y^2}. $$ Here $\rho$ is the correlation between $X$ and $Y$, i.e. $p_{XY}=\rho \sigma_X\sigma_Y$. Calculations (do these) now show that $r(x,y)$ can be written as $$ r(x,y)=-\frac{1}{2\sigma_X^2}(x-\mu_X)^2-\frac{1}{2\sigma_Y^2(1-\rho^2)}\Big(y-\mu_Y-\frac{\rho\sigma_Y}{\sigma_X}(x-\mu_X)\Big)^2 $$ and hence $f_{X,Y}$ factors into the product of $$ \frac{1}{\sqrt{2\pi}\sigma_X}\exp\Big(-\frac{1}{\sigma_X^2}(x-\mu_X)^2\Big)\tag{1} $$ and $$ \frac{1}{\sqrt{2\pi}\sigma_Y(1-\rho^2)}\exp\left(-\frac{1}{2\sigma_Y^2(1-\rho^2)}\Big(y-\mu_Y-\frac{\rho\sigma_Y}{\sigma_X}(x-\mu_X)\Big)^2\right)\tag{2} $$ showing that the conditional density of $Y$ given $X=x$ is given by $(2)$, i.e. $Y\mid X=x$ is normally distributed with mean $\mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X)$ and variance $\sigma_Y^2(1-\rho^2)$.

Answer 2

You might want to work directly on random variables: there exists some parameters $(a,b)$ and a standard normal random variable $Z$ independent of $X$ such that $Y-\mu_Y=a(X-\mu_X)+bZ$ (can you show this?), thus $E(Z\mid X)=E(Z)=0$ hence $E(Y\mid X)=\mu_Y+a(X-\mu_X)$. To compute $a$, note that $\mathrm{Cov}(Y,X)=a\cdot\mathrm{var}(X)+b\cdot\mathrm{Cov}(Z,X)$ and $\mathrm{Cov}(Z,X)=0$, hence $\rho_{XY}=a\cdot\sigma^2_X$.

Answer 3

If you want to calculate directly, try massaging your expression for $f_{Y\mid X}(y\mid X=x)$ into the form $$f_{Y\mid X}(y\mid X=x) = \frac{\exp\left(-\frac{1}{2}\left(\frac{y-a}{b}\right)^2\right)}{b\sqrt{2\pi}}$$ where $a$ and $b$ are functions of $x$ and the other parameters but do not depend on $y$. Then see if you can determine the expected value without doing the integration that you propose to carry out.