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Suppose $X$ and $Y$ are nonnegative random variables such that $\mathrm{Pr}(X\geq t)\leq\mathrm{Pr}(Y\geq t)$ for all $t\geq0$. Now take $X_1,\ldots,X_n$ to be independent with the same distribution as $X$, and similarly for $Y_1,\ldots,Y_n$ with $Y$. I would like to know if the following is necessarily true: $$ \mathrm{Pr}\bigg(\frac{1}{n}\sum_{i=1}^nX_i\geq t\bigg)\leq\mathrm{Pr}\bigg(\frac{1}{n}\sum_{i=1}^nY_i\geq t\bigg) \qquad \forall t\geq0. $$ By assumption, it's true for $n=1$. For large $n$, the sample averages will approach $\mathbb{E}[X]$ and $\mathbb{E}[Y]$, respectively, which satisfy $$ \mathbb{E}[X] =\int_0^\infty\mathrm{Pr}(X\geq t)~dt \leq\int_0^\infty\mathrm{Pr}(Y\geq t)~dt =\mathbb{E}[Y]. $$ So intuitively, the inequality of interest should also hold for large $n$. Can anything funny happen for moderately sized $n$?

(As the title suggests, I'm interested in this as a tool analyze measure concentration.)

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Slightly tongue-in-cheek: with distributions like Cauchy or Levy the sample averages will not approach the means. –  Waldemar Oct 8 '13 at 11:51
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up vote 8 down vote accepted

Yes, it is true for any $n$. The easiest way to see it is by using the fact that your condition means precisely that $X$ and $Y$ can be realized on the same probability space $\Omega$ in such a way that $Y\ge X$.

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So $X$ and $Y$ don't need to be nonnegative? –  Dustin G. Mixon Oct 8 '13 at 4:20
    
No, they don't. By the way, if your $X$ and $Y$ are compactly supported, one can always make them positive just by adding an appropriate constant. –  R W Oct 8 '13 at 4:34
    
Can something be said when all (X_i) and (Y_i) are assumed to be independent? –  Joris Bierkens Oct 8 '13 at 6:44
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@Joris Bierkens: No. Let $\overline X= (X_i)$ and $\overline Y = (Y_i)$ be the corresponding sequence valued random variables. The question involves just the distributions of $\overline X$ and $\overline Y$, but not their joint distribution. –  R W Oct 8 '13 at 9:05
    
@R.W. - If we realize $X$ and $Y$ in the same probability space, can't we draw the $X_i$'s and $Y_i$'s with $2n$ independent outcomes of the probability space? I think the difficulty would not be independence between the $X_i$'s and $Y_i$'s, but rather a more exotic joint distribution. –  Dustin G. Mixon Oct 8 '13 at 12:13
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