Stochastic ordering of empirical mean

Question

Consider a Bernoulli distribution with mean $\mu \in (0,1)$ taking values in the set $\{0,1\}$. Suppose we draw $t \in \mathbb{N}$ independent and identically distributed (i.i.d.) samples from this distribution, and denote them with $\{X_1, X_2, \ldots, X_t\}$. For two natural numbers $t_1 < t_2$, consider the empirical means $\hat{\mu}_1 = \frac{1}{t_1} \sum_{s=1}^{t_1} X_s$ and $\hat{\mu}_2 = \frac{1}{t_2} \sum_{s=1}^{t_2} X_s$ constructed with $t_1$ and $t_2$ i.i.d. samples respectively.

Informally, it seems to me that $\hat{\mu}_2$ is closer to the true mean $\mu$ since it contains more information. I would like to know if there exist formal results along this line. More specifically, can I say:

1. Does $|\hat{\mu}_1 - \mu|$ have first-order stochastic dominance over $|\hat{\mu}_2 - \mu|$? One way that occurs to me is to compare an upper bound on the tail of $|\hat{\mu}_2 - \mu|$ with a lower bound on the tail of $|\hat{\mu}_1 - \mu|$ to obtain sufficient conditions for the dominance result to be true. However, I don't think that result would hold for all $t_1, t_2$.
2. If instead of natural numbers $t_1$ and $t_2$, I consider two random variables $T_1$ and $T_2$ taking values in $\mathbb{N}$ such that $T_1 < T_2$ almost surely, does the result of Q1. carry over to this case.

Any pointers to literature or books would be great. Thanks.

Answer 1

The answer is no. Indeed, suppose that $X_1,X_2,\dots$ are iid Bernoulli random variables (r.v.'s) each with mean $p$, $\bar X_n:=\frac1n\,\sum_1^n X_i$, and $$d_{n,p}(t):=P(|\bar X_{n+1}-p|>t)-P(|\bar X_n-p|>t). $$ The graph $\{\big(t,d_{1,1/5}(t)\big)\colon0\le t\le4/5\}$ is shown here:

We see that the (right-continuous) function $d_{1,1/5}$ takes values of both signs. In particular, $d_{1,1/5}(1/5)=4/25>0>-4/25=d_{1,1/5}(3/10)$.

Thus, the family of r.v.'s $(|\bar X_n-p|)_{n=1}^\infty$ is not stochastically monotone.