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Let $(X_1, X_2,...)$ be i.i.d. random variable with finite mean $E(X)$ and let $(Y_1,Y_2,...)$ be i.i.d. with mean $E(Y)<E(X)$.

I'm wondering whether there's an $n$ such that the average $\bar{X_n}=\frac{1}{n}\sum_{i=1}^nX_i$ first-order stochastically dominates the corresponding average $\bar{Y_n}$. That is, whether there is an $n$ such that $F_{\bar{Y_n}}(u)\geq F_{\bar{X_n}}(u)$ for all $u$, with strict inequality somewhere, and where $F_{\bar{Y_n}}(u)$ is the CDF of $\bar{Y_n}$ (and similar for $\bar{X_n}$).

We know from the law of large numbers that $\bar{X_n}$ and $\bar{Y_n}$ converge almost surely to $E(X)$ and $E(Y)$ respectively. This implies that they converge in distribution too. So they converge to (constant) random variables such that one (trivially) stochastically dominates the other. I'm not sure whether I can find $n$ such that $\bar{X_n}$ stochastically dominates $\bar{Y_n}$ though. It seems somewhat plausible to me that this isn't possible if $\bar{X_n}$ converges much more slowly than $\bar{Y_n}$.

If this isn't the case, what additional assumptions do I need to make it the case? By brute force computation, it seems to be true for normal distributions but I haven't managed to prove it even in this case and I'd like a much more general statement ideally. Perhaps, finite variances would help, since then they converge to $0$ as $n\rightarrow\infty$, but since the law of large numbers doesn't require finite variance, it seems weird to me that this would.

Any hints, pointers, references would be very welcome.

DM-97
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  • When the moment generating functions exist, maybe Chernoff's inequality will help, as it gives bounds for tails. $P(X>u)\leq e^{-su}M(s)$ for all $s>0$, where $M$ is the moment generating function of $X$. In particular, the moment generating function of $\bar{X_n}$ is $M(s)=E(e^{\frac{s}{n}X})^n$. Any thoughts? – DM-97 Aug 12 '21 at 14:25
  • I suspect you might find the central limit theorem useful. – lonza leggiera Aug 12 '21 at 14:45
  • Yeah, thanks. Is the thought that for large $n$, $\bar{X_n}$ and $\bar{Y_n}$ are approximately normally distributed, with means $E(X), E(Y)$ and variances $\frac{Var(X)}{n},\frac{Var(Y)}{n}$? So then the problem reduces to normal distributions (and we still need to make sure that $n$ is large enough that the approximation is sufficiently good) – DM-97 Aug 12 '21 at 16:37
  • Yes. If $\ Var(X)= Var(Y)\ $, then $\ \mathcal{N}\Big(\frac{\sqrt{n}(u-E(X))}{\sqrt{Var(X)}}\Big)<\mathcal{N}\Big(\frac{\sqrt{n}(u-E(Y))}{\sqrt{Var(Y)}}\Big)\ $ for all $\ u\ $, so if these normal approximations to $\ F_{\overline{X}n}\ $ and $\ F{\overline{Y}_n}\ $ are good enough, then $\ \overline{Y}_n\ $ should stochastically dominate $\ \overline{X}_n\ $ in this case. Details of any verification look like they're going to be fairly messy, though. – lonza leggiera Aug 13 '21 at 01:25
  • If $\ Var(X)\ne Var(Y)\ $, then $\frac{\sqrt{n}(u-E(Y))}{\sqrt{Var(Y)}}-\frac{\sqrt{n}(u-E(X))}{\sqrt{Var(X)}}\ $ has opposite signs for very large negative and very large positive values of $\ u\ $, so I suspect the result won't hold in this case. – lonza leggiera Aug 13 '21 at 01:25

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