Law of Iterated Expectation applied to a ratio?

Question

Consider the random variables $\epsilon_1, X_1, X:=(X_1,\dots,X_n)$ with $X_1,...,X_n$ i.i.d. Is it true that $$\mathbb E\left(\frac{\epsilon_1}{X_1}\right)=\mathbb E\left(\frac{\mathbb E(\epsilon_1\mid X)}{X_1} \right)$$ Sketch of proof?

Answer 1

If $X$ and $X_1$ were not related as they are, this would not generally be true. Take for example independent random variables $X$ and $Y$ and let $Z = 1/Y.$ Then $$ \mathbb{E}\left[ \frac{Y}{Z} \right] = \mathbb{E}[Y^2] $$ and $$ \mathbb{E}\left[ \frac{\mathbb{E}[Y \mid X]}{Z} \right] = \mathbb{E}\left[ \frac{\mathbb{E}[Y]}{Z} \right] = \mathbb{E}[Y]\mathbb{E}\left[\frac{1}{Z}\right] = \mathbb{E}[Y]^2 $$

However, what is true is that for any random variables $A$ and $B$, and measurable function $f$, $$ \mathbb{E}[f(A) B] =\mathbb{E}\big[\, f(A)\, \mathbb{E}[B \mid A]\big] $$ To prove this last statement, use the law of iterated expectation and that $$\mathbb{E}[f(A) B \mid A] = f(A) \mathbb{E}[B \mid A]$$

In your case, the latter point seems more appropriate: Set $A = X = (X_1, ..., X_n)$, set $f(X_1, ..., X_n) = 1/X_1$, and set $B = \epsilon_1$.

Answer 2

Assuming $\mathbb E|\epsilon_1/X_1|<+\infty$, the tower property (or law of total expectation) implies that \begin{align}\mathbb E\left(\frac{\epsilon_1}{X_1}\right)&=\mathbb E\left(\mathbb E\left(\frac{\epsilon_1}{X_1}\mid X\right)\right)=^{X_1 \text{ is known given }X}\\[0.2cm]&=\mathbb E\left( \frac{1}{X_1} \mathbb E\left(\epsilon_1\mid X\right)\right)\end{align} so, it seems correct to me.