Is there enough information for computing $E(y_1y_2|x_1,x_2)$?

Question

Suppose we know

• $(x_1,y_1)$ and $(x_2,y_2$) are i.i.d.
• $E(y_1|x_1)=\alpha+\beta x_1$ and $E(y_2|x_2)=\alpha+\beta x_2$

The context, as you can tell, is a simple linear regression. Is the above info sufficient yet to compute (i.e. write in terms of $x_1,x_2,\alpha,\beta$): $$ E(y_1y_2|x_1,x_2)? $$ I've seen some lecture notes, which may contain either transcription or instructor's errors, that claim YES but I can't see why.

Answer 1

Let $Y=(Y_1,Y_2)'$ and $X=(X_1,X_2)'$. If all r.v.s are continuous, then \begin{gather} f_{Y\mid X}(y\mid x)=\frac{f_{Y,X}(y,x)}{f_X(x)}=\frac{f_{Y_1,X_1}(y_2,x_2)f_{Y_2,X_2}(y_2,x_2)}{f_{X_1}(x_1)f_{X_2}(x_2)}\\=f_{Y_1\mid X_1}(y_1\mid x_1)f_{Y_2\mid X_2}(y_2\mid x_2). \end{gather} Thus, $$ \mathsf{E}[Y_1Y_2\mid X_1=x_1,X_2=x_2]=\mathsf{E}[Y_1\mid X_1=x_1]\mathsf{E}[Y_2\mid X_2=x_2]. $$

Answer 2

Going only by definitions: The conditional expectation is $$ E(y_1 y_2|x_1, x_2) = \int \! y_1 y_2 \, f_{Y_1,Y_2|X_1,X_2}(y_1, y_2|x_1, x_2) \, dy_1 \, dy_2 \;, $$ where the conditional p.d.f. is $$ f_{Y_1,Y_2|X_1,X_2}(y_1, y_2|x_1, x_2) = \frac{f_{X_1,Y_1,X_2,Y_2}(x_1, y_1, x_2, y_2)}{f_{X_1,X_2}(x_1,x_2)} \;. $$ Using the independence of $(x_1, y_1)$ and $(x_2, y_2)$: $$ f_{Y_1,Y_2|X_1,X_2}(y_1, y_2|x_1, x_2) = \frac{f_{X_1,Y_1}(x_1, y_1) \, f_{X_2,Y_2}(x_2, y_2)}{f_{X_1}(x_1) \, f_{X_2}(x_2)} = f_{Y_1|X_1}(y_1|x_1) \, f_{Y_2|X_2}(y_2|x_2) \;. $$ Inserting into the integral: $$ E(y_1 y_2|x_1, x_2) = \int \! y_1 \, f_{Y_1|X_1}(y_1|x_1) \, dy_1 \int \! y_2 \, f_{Y_2|X_2}(y_2|x_2) \, dy_2 = E(y_1|x_1) \, E(y_2|x_2) $$ This is exactly the same result you could get more quickly by realising that $y_1$ and $y_2$ are independent, and thus have zero covariance!