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 $$y\in(0,1) \text{ such that } \int_0^y G_n(s)\,ds = \int_0^y G_{n+1}(s)\, ds?$$

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

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