# show that $\mathsf E(Y|X)=\mu_2+\rho\dfrac{\sigma_2}{\sigma_1}(X-\mu_1)$

Suppose $X$ and $Y$ have a joint distribution with finite means and variances respectively given by $\mu_1,\mu_2,\sigma_1^2,\sigma_2^2$ with correlation coefficient given by $\rho$.

Further suppose that $\mathsf E(Y|X)$ is a linear function in $X$.

I have to show that $\mathsf E(Y|X)=\mu_2+\rho\dfrac{\sigma_2}{\sigma_1}(X-\mu_1)$.

I also have to then show that $\mathsf E(\mathsf {Var}(Y|X))=\sigma_2^2(1-\rho^2)$.

• I've edited this using MathJax as best I could interpret the original post. Please edit it to correct any misunderstanding; it wasn't clear. – Graham Kemp Nov 19 '14 at 7:25
• Except for the p part. it is p*(o2/o1) and then all of that * (X-u1). So the u2 is not supposed to be a part of the fraction @GrahamKemp – juniormath Nov 19 '14 at 7:37
• Any better now? – Henry Nov 19 '14 at 8:07

Let $E[Y|X] = aX +b$. Taking expectations w.r.t X, $$E_X[E[Y|X]] = aE[X] + b$$ $$E[Y] = aE[X] + b$$ Therefore, $$\mu_2 = a\mu_1 + b \quad (1)$$ Now, taking variance of $E[Y|X]$ w.r.t X, $$Var_X[E[Y|X]] = a^2Var_X[X] \quad (2)$$ Rewriting E[Y|X] we get, $$E[Y|X] - aX = b$$ Taking variance w.r.t X, we get, $$Var_X[E[Y|X]] + a^2Var[X] - 2aCov(X, E[Y|X]) = 0$$ Now, $$Cov(X, E[Y|X]) = E[XE[Y|X]] - E[X]E_X[E[Y|X]] = E[XY] - E[X]E[Y] = Cov(X, Y)$$ Therefore, $$Var_X[E[Y|X]] = 2aCov(X, Y) - a^2Var[X] \quad (3)$$ Comparing (2) and (3), we get, $$a = \frac{Cov(X, Y)}{Var[X]} = \rho\frac{\sigma_2}{\sigma_1} \quad (4)$$ From (1) and (4), we get, $$b = \mu_2 - \rho\frac{\sigma_2}{\sigma_1}\mu_1$$ Substituting back a and b, we get, $$E[Y|X] = \mu_2 + \rho\frac{\sigma_2}{\sigma_1} (X - \mu_1)$$