Deriving Conditional Moments of a Bivariate Normal

# Expectation and Cholesky Decomposition


# Expectation and Cholesky Decomposition












Assume that the random vector $(X,Y)$ is (bivariate) normally distributed. Show that $$ \Bbb E[X|Y=y]= \Bbb E[X]+ \frac {Cov[X,Y]}{Var[Y]}(y-\Bbb E[Y])$$

Also, $$ Var[X|Y=y]= (1-\rho^2) Var[X]$$

I know i should be converting these variables into standard normal and then using Cholesky decomposition to come up with independent standard normal, I am getting pretty close to the answer but, its not neat. I might have done something wrong, Can some one please lay out the first step to convert X&Y to standard normal?? Thanks so much

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/31523

By Cholesky decomposition, you can express the normal random variables $X$ and $Y$ in the form \begin{align*} Y &= E(Y) + \sqrt{Var(Y)}\, \xi,\\ X &= E(X) + \sqrt{Var(X)}\left(\rho \xi+\sqrt{1-\rho^2} \eta\right), \end{align*} where $\rho = \frac{Cov(X, Y)}{\sqrt{Var(X)Var(Y)}}$ is the correlation, $\xi$ and $\eta$ are two independent standard normal random variables.

Then, \begin{align*} E(X \mid Y) &= E\left(E(X) + \sqrt{Var(X)}\left(\rho \xi+\sqrt{1-\rho^2} \eta\right) \mid \xi \right)\\ &=E(X) + \rho \sqrt{Var(X)}\xi\\ &=E(X) + \frac{Cov(X, Y)}{\sqrt{Var(Y)}}\xi\\ &=E(X) + \frac{Cov(X, Y)}{Var(Y)}\big(Y-E(Y) \big). \end{align*} The computation for $Var(X\mid Y)$ is similar, specifically, \begin{align*} Var(X \mid Y) &=E\left((X-E(X\mid Y))^2\mid Y \right)\\ &=E\left( (X-E(X\mid Y))^2\mid \xi\right)\\ &=E\left(Var(X)(1-\rho^2) \eta^2 \mid \xi\right)\\ &=E\left(Var(X)(1-\rho^2) \eta^2\right)\\ &=Var(X)(1-\rho^2). \end{align*}

## Answer by user16651 (score 3)

https://quant.stackexchange.com/a/31524

Another approach

$$f_{X|Y}(x,y)=\frac{f_{X,Y}(x,y)}{f_{Y}(y)}\tag 1$$ Set $$u=\frac{x-\mu_X}{\sigma_X}$$ and $$v=\frac{y-\mu_Y}{\sigma_Y}$$ we have $$f_{X|Y}(x,y)=\frac{\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\exp\left(-\frac{u^2-2\rho uv+v^2}{2(1-\rho^2)}\right)}{\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\left(-\frac{1}{2}v^2\right)}\\\qquad\qquad\qquad\qquad=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac 12\left[\frac{u-\rho v}{\sqrt{1-\rho^2}}\right]^2\right)\tag 2$$ Indeed $$f_{X|Y}(x,y)=\frac{1}{\sqrt{2\pi(1-\rho^2)}\sigma_X}\exp\left(-\frac{1}{2}\left[\frac{x-(\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)}{\sigma_X\sqrt{1-\rho^2}}\right]^2\right)\tag 3$$ as a result $$\mathbb{E}[X|Y]=\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)\tag 4$$ and $$\text{Var}(X|Y)=\sigma_X^2(1-\rho^2)\tag 5$$