# Are sample means ordered by strict second-order stochastic dominance throughout the support?

Consider random variables $$X_1,X_2,\dots$$.

Each $$X_i$$ is independent and identically distributed on $$[0,1]$$ with a cumulative distribution $$F$$ that has a positive density $$f(x)>0$$ throughout the support $$x\in[0,1]$$.

My question is about the order of the distributions of the sample means $$Y_n=\frac{X_1+\cdots+X_n}{n}$$ in terms of strict stochastic second-order stochastic dominance. Denote the distribution of $$Y_n$$ by $$G_n$$.

It is known that $$Y_n$$ strictly second-order stochastically dominates $$Y_{n+1}$$, that is, $$\int_0^y G_n(s)\,ds \geq \int_0^y G_{n+1}(s)\, ds \text{ for all } y\in[0,1]$$ with a strict inequality for some $$y\in(0,1)$$.

My question is whether it is ever possible to have a $$y\in(0,1)$$ where the inequality holds with equality? In other words, given that $$F$$ has positive density throughout the support, can there be $$y\in(0,1) \text{ such that } \int_0^y G_n(s)\,ds = \int_0^y G_{n+1}(s)\, ds?$$

• I have the intuition that if $X \sim \mathcal U_{[0,1]}$ then for all $n$, $\mathbb P(Y_n > 0.5) = 0.5$ – winperikle Apr 7 at 13:30
• Yes this is clearly true because for the uniform distribution, all $Y_n$ are symmetrically distributed around the mean of .5. – ChooCheeDuck Apr 7 at 14:42