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.)



Is there monotonicity of measure concentration?
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 $Y\ge X$.


