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I was tying to prove the following: Given $(X,Y)$ a centered gaussian vector in $\mathbb{R}^2$ with the following covariance matrix $$ \Sigma = \begin{bmatrix} \sigma^2_x & \sigma_{x,y} \\ \sigma_{x,y} & \sigma^2_y \end{bmatrix} $$ such that $\det(\Sigma)>0$ so it admits a density. Find the expression of the density of $E[X\mid Y]$.

I am supposed to find that the conditional law of $X|Y=y$ is $N(\frac{\sigma^2_y y}{\sigma_{x,y}},\frac{\det(\Sigma)}{\sigma^2_y} )$ but I can't get it. It must be something dumb, can you please check my math ?

PS: please excuse the "typo" $\mathbb{E}[X\mid Y=y]$

calculus

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  • $\begingroup$ The theory says that $E(X\mid Y)=aY+b$ and that $(a,b)$ is entirely determined by the pair of identities $$E(X)=E(E(X\mid Y))=aE(Y)+b\quad E(XY)=E(E(X\mid Y)Y)=aE(Y^2)+bE(Y)$$ Thus, $E(X\mid Y)$ is normal with mean $aE(Y)+b$ and variance $a^2\sigma^2_Y$. Can you finish? $\endgroup$ – Did Jun 28 '15 at 19:54
  • $\begingroup$ Sure I can, but that does not prove that $E[X|Y]$ is a gaussian. Anyway, I know this trick but I wanted to see where my mistake is in this calclus $\endgroup$ – statquant Jun 28 '15 at 20:03
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    $\begingroup$ Sorry, I do not check handwritten proofs presented in images. $\endgroup$ – Did Jun 28 '15 at 20:08
  • $\begingroup$ I can understand that :) $\endgroup$ – statquant Jun 28 '15 at 20:09
  • $\begingroup$ You have $\sigma_x$ and $\sigma_y$ where you need $\sigma_X$ and $\sigma_Y$. The distinction between capital and lower case is there for a reason. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 28 '15 at 20:17
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The joint density is $$ \text{constant}\cdot \exp\left( \frac{-1} 2 (\mathbf x-\mu)'\Sigma^{-1} (\mathbf x-\mu) \right) $$ where $\mathbf x=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\mu =\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}$ is the expected value. We have $$ \Sigma^{-1} = \frac 1 {\det\Sigma} \begin{bmatrix} \sigma_Y^2 & -\rho\sigma_{X,Y} \\ -\rho\sigma_{X,Y} & \sigma_X^2 \end{bmatrix} $$ where $\sigma_{X,Y}=\rho\sigma_X\sigma_Y$ so $\rho$ is the correlation. (I used capital $X$ and capital $Y$ in the subscripts for a reason.) Multiplying matrices, we get this form of the joint density:

\begin{align} \text{constant}\cdot \exp\left( -\frac{1}{2(1-\rho^2)} \left[ \frac{(x-\mu_X)^2}{\sigma_X^2} + \frac{(y-\mu_Y)^2}{\sigma_Y^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} \right] \right) \end{align}

View the function in square brackest as a function of $x$ alone: $$ \frac{(x-\mu_X)^2}{\sigma_X^2} - \frac{2\rho(x-\mu_X)(y-\mu_Y)}{\sigma_X \sigma_Y} + \text{constant} $$ where "constant" means not depending on $x$.

Now complete the square, getting $$ \frac{(x-\text{something})^2}{\text{something}}. $$ Note well:

  • The "something" in the numerator will depend on $y$ and will be the conditional expected value.
  • From the "something" in the denominator you can figure out the conditional variance. It will not depend on $y$. It should come to $(1-\rho^2)\sigma_X^2$, i.e. it is the part of the variance of $X$ that is not "explained" by the fact that $Y$ varies.
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