Monotonicity of Sample Mean

Question

$X_1,X_2,\ldots$ are drawn i.i.d. from a distribution with mean $\mu$. Define $\bar{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$

Prove that $\forall t \quad E[|\bar{X}_t - \mu|] \geq E[|\bar{X}_{t+1} - \mu|]$

Answer 1

Assume without loss of generality that $\mu=0$. For every $t\geqslant1$, let $S_t=X_1+\cdots+X_t$ and, for every $1\leqslant r\leqslant t+1$, let $S^{(r)}_t=S_{t+1}-X_r$. Note that $$ t\,S_{t+1}=\sum_{r=1}^{t+1}S^{(r)}_t. $$ Since each $S^{(r)}_t$ is distributed as $S_t$, the pointwise inequality $$ t\,\left|S_{t+1}\right|=\left|\sum_{r=1}^{t+1}S^{(r)}_t\right|\leqslant\sum_{r=1}^{t+1}\left|S^{(r)}_t\right| $$ implies that $$ t\,E(\left|S_{t+1}\right|)\leqslant\sum_{r=1}^{t+1}E\left(\left|S^{(r)}_t\right|\right)=(t+1)\,E(\left|S_{t}\right|). $$ Dividing both sides by $t(t+1)$ yields the result.