For positive random variables $(X_1, Y_1)$ and $(X_2, Y_2)$, suppose that $(X_1, Y_1)$ and $(X_2, Y_2)$ have the same distribution and (the two pairs) are independent. Also suppose that $E[Y_1|X_1] = \theta X_1$. Let $Z=\frac{Y_1 + Y_2}{X_1+X_2}$. Find $E[Z]$.
Solution attempt: Using Law of Iterated Expectations (LIE), we have that $E[Y_1]=\theta E[X_1]$.
We can also write $Z=\frac{Y_1 + Y_2}{X_1+X_2}$ as $\frac{Y_1}{X_1 + X_2} + \frac{Y_2}{X_1 + X_2}$.
So, $E[Z]=E[\frac{Y_1}{X_1 + X_2} + \frac{Y_2}{X_1 + X_2}] = E[\frac{Y_1}{X_1 + X_2}] + E[\frac{Y_2}{X_1 + X_2}]$. Now, I tried to use LIE again to get:
$E[Z] = E_{X_1+X2}E[\frac{Y_1}{X_1 + X_2} | X_1+X_2]+... = E_{X_1+X2}[\frac{1}{X_1+X_2}E[Y_1|X_1+X_2]] +...$
Now I have no idea what to do. Can I still treat the $\frac{1}{X_1+X_2}$ as a constant and take it out? I don't think so. How do I proceed from here? Any help appreciated!
Use LIE three times, then simply because of independence of the two coordinates w.r.t. each other.
$\begin{align} \mathsf E[Z] & = \mathsf E_{X_1}\left[\mathsf E_{X_2\mid X_1}\left[\mathsf E_{Y_1\mid X_1,X_2}\left[\mathsf E_{Y_2\mid X_1,X_2,Y_1}\left[\frac{Y_1+Y_2}{X_1+X_2}\middle| X_1,X_2,Y_1\right]\middle| X_1,X_2\right]\middle| X_1\right]\right] \\[2ex] & = \mathsf E_{X_1}\left[\mathsf E_{X_2\mid X_1}\left[\mathsf E_{Y_1\mid X_1,X_2}\left[\frac{Y_1+\mathsf E_{Y_2\mid X_2}[Y_2\mid X_2]}{X_1+X_2}\middle| X_1,X_2\right]\middle| X_1\right]\right] \\ & \vdots \\[1ex] & = \mathsf E_{X_1}\left[\mathsf E_{X_2}\left[\frac{\mathsf E_{Y_1\mid X_1}\left[Y_1\mid X_1\right]+\mathsf E_{Y_2\mid X_2}\left[\mathsf Y_2\mid X_2\right]}{X_1+X_2}\right]\right] \\ & \vdots \end{align}$
Can you take it from here?