Convergence of expectation of inverse sum of random variables

Question

Suppose $X_1,..,X_n,...$ are positive random variables, such that $(X_1+...+X_n)/n\to\mu$ almost surely as $n\to\infty$.

I intuitively believe that $\mathbb{E}\Big[\frac{Y}{(X_1+...+X_n)/n}\Big]\to\mathbb{E}[Y]/\mu$ for any random variable $Y$, as the integrand converges to $Y/\mu$ almost surely.

Is this true? What convergence theorem can be used to prove this sort of proposition?

Answer 1

One way to prove this is to use Slutsky's theorem, which states that $\frac{Y}{(X_1+...+X_n)/n}$ converges in distribution to $Y/\mu$ provided that $\mu\neq 0$. This implies the property you stated.

I suspect that this is false since almost sure convergence of $\left\{Z_n\right\}$ to a constant $\mu$ does not imply the convergence of $\mathbb{E}\left[Z_n\right]$ to $\mu$. Take for example $Z_n$ such that $$P(Z_n=n^2)=1/n^2$$ $$P(Z_n=0)=1-1/n^2$$ Then $\left\{Z_n\right\}$ converges almost surely to $0$ since $\sum_nP(|Z_n-0|>\varepsilon)<\infty$ for all $\varepsilon>0$, but $\mathbb{E}\left[Z_n\right]=1$ does not converge to $0$.