# Is there monotonicity of measure concentration?

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.)

• Slightly tongue-in-cheek: with distributions like Cauchy or Levy the sample averages will not approach the means. Oct 8, 2013 at 11:51

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 $Y\ge X$.
• So $X$ and $Y$ don't need to be nonnegative? Oct 8, 2013 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.
• @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. - 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. Oct 8, 2013 at 12:13