I am struggling to derive $E[X|Y=y]=\mu_X+\sigma_X\rho(\frac{y-\mu_y}{\sigma_Y})$ when X and Y follow bivariate normal distribution.
I have read this, but I don't get how to get the following steps:
$$
\frac{\int x \exp\left(-\frac{1}{2(1-\rho^2)}\left(\frac{x-\mu_X}{\sigma_X}-\rho\frac{y-\mu_Y}{\sigma_Y}\right)^2 \right)dx }{\int \exp\left(-\frac{1}{2(1-\rho^2)}\left(\frac{x-\mu_X}{\sigma_X}-\rho\frac{y-\mu_Y}{\sigma_Y}\right)^2 \right) dx} = \frac{\int (x+\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}) \exp(-\frac{1}{2(1-\rho^2)}(\frac{x}{\sigma_X})^2 )dx }{\int exp(-\frac{1}{2(1-\rho^2)}(\frac{x}{\sigma_X})^2 ) dx}
$$
$$
E[X|Y=y] = \int x \frac{f(x, y)}{f_Y(y)}dx
$$
Substitute the formula in and get:
\begin{align}
\frac{f(x, y)}{f_Y(y)} &= \frac{1}{\sigma_X\sqrt{2\pi}\sqrt{1-\rho^2}}exp\left( -\frac{1}{2(1-\rho^2)} (\frac{x- (\mu_X+\frac{\sigma_y}{\sigma_Y}\rho(y-\mu_Y))}{\sigma_X})^2
\right)
\end{align}
Best Answer
Approach 1
With two independent standard normals $\varepsilon_X,\varepsilon_Y\,,$ the correlated normals $X,Y$ can be written as \begin{align} X&=\mu_X+\sigma_X\Big(\rho\varepsilon_Y+\sqrt{1-\rho^2}\varepsilon_X\Big)\,,\\ Y&=\mu_Y+\sigma_Y\varepsilon_Y\,. \end{align} Therefore, $$ X=\mu_X+\sigma_X\Big(\rho\frac{Y-\mu_Y}{\sigma_Y}+\sqrt{1-\rho^2}\varepsilon_X\Big)\,. $$ This implies $$ \boxed{\quad\mathbb E\big[X\big|Y=y\big]=\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}\,.\quad} $$
Approach 2
The covariance matrix of $X$ and $Y$ is $$ \Sigma=\begin{pmatrix}\sigma_X^2 &\sigma_X\sigma_Y\rho\\\sigma_X\sigma_Y\rho&\sigma_Y^2\end{pmatrix} $$ which has inverse $$ \Sigma^{-1}=\frac{1}{\sigma_X^2\sigma_Y^2(1-\rho^2)}\begin{pmatrix}\sigma_Y^2 &-\sigma_X\sigma_Y\rho\\-\sigma_X\sigma_Y\rho&\sigma_X^2\end{pmatrix}\,. $$ The joint density of $X,Y$ is then \begin{align} f_{X,Y}(x,y)&=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}}\\ &\quad\times\exp\Big(-\frac{\sigma_Y^2(x-\mu_X)^2-2\sigma_X\sigma_Y\rho(x-\mu_X)(y-\mu_Y)+\sigma_X^2(y-\mu_Y)^2}{2\sigma_X^2\sigma_Y^2(1-\rho^2)}\Big)\,. \end{align} From (looking at the denominator of the term in the exponential) \begin{align} &\sigma_Y^2(x-\mu_X)^2-2\sigma_X\sigma_Y\rho(x-\mu_X)(y-\mu_Y)+\sigma_X^2(y-\mu_Y)^2\\[2mm] &=\sigma_Y^2\Big(x-\mu_X-\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}\Big)^2+\sigma_X^2(1-\rho^2)(y-\mu_Y)^2 \end{align} and the unconditional distribution of $Y$, $$ f_Y(y)=\frac{1}{\sqrt{2\pi}\sigma_Y}\exp\Big(-\frac{(y-\mu_Y)^2}{2\sigma_Y^2}\Big), $$ it follows that \begin{align} \frac{f_{X,Y}(x,y)}{f_Y(y)}&=\frac{1}{\sqrt{2\pi}\sigma_X\sqrt{1-\rho^2}}\exp\Big(-\frac{(x-\mu_X-\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y})^2}{2\sigma_X^2(1-\rho^2)}\Big)\,. \end{align} This means that, conditional on $Y$, the random variable $X$ is normal with
standard deviation $\sigma_X\sqrt{1-\rho^2}\,,$ and
mean $\mu_X+\sigma_X\rho\frac{y-\mu_Y}{\sigma_Y}\,,$ as expected.