From almost surely to expectation

Question

Suppose you have two sequences of positive random variables, $(X_n)$ and $(Y_n)$, such that a.s. $Y_n = o(X_n)$. Do you see any sufficient condition to obtain $\mathbb E(Y_n)=o(\mathbb E(X_n))$?

Answer 1

Let $(X_n)$ a sequence of strictly positive integrable random variables and $(Y_n)$ a sequence of nonnegative integrable random variables such that

1. $X_n $ is independent from $Y_n$ for all $n ∈ \mathbb N $,
2. $Y_n = o(X_n)$ almost surely,
3. the sequence $( Y_n/X_n )$ is dominated by an integrable random variable.

Then $ \mathbb E[Y_n] = o(\mathbb E[X_n])$. To see this, we observe first that assumption (2) can be rewritten as $ \lim \frac {Y_n}{X_n} = 0 $ almost surely. Then (3) gives that $ \lim \mathbb E[\frac {Y_n}{X_n} ] = 0$. Now by independence and Jensen's inequality, $$ \mathbb E[Y_n / X_n] = \mathrm{Cov}(Y_n,1/X_n) + \mathbb E[Y_n] \mathbb E[1/X_n] = \mathbb E[Y_n]\mathbb E[1/X_n] \geq \mathbb E[Y_n]/\mathbb E[X_n]. $$ Hence $ \mathbb E[Y_n] = o(\mathbb E[X_n])$. From the last display we also see that we can replace "independent from" in (1) by $1/X_n$ and $Y_n$ being positively correlated for all $n$.

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Without the assumption (3) of a dominating integrable RV, we can find a counterexample, e.g. $ Ω = (0,1]$ with the Borel-σ-algebra and Lebesgue measure, $$ X_n =n, \qquad Y_n = n^3 1_{(0,1/n]}, \qquad \frac{Y_n}{X_n} = n^2 1_{(0,1/n]},$$ so $Y_n = o(X_n)$, but $\mathbb E[Y_n] = n^2 ∉ o(\mathbb E[X_n]) = o(n)$.

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The interesting question that remains is what happens when we do have (3) but drop the independence resp. positive correlation requirement. It does feel like $\mathbb E[Y_n/X_n] → 0 $ is incompatible with $ \mathbb E[Y_n] / \mathbb E[X_n] $ being bounded below by some positive constant if we also have the integrable bound, but I don't have a proof or counterexample.