I am looking to prove the following:
Given iid random variable's $X = X_1, X_2, \dots$, and mean $E[X] = \bar{X}$ ,show that:
$$E[|X|] \geq E[|X – \bar{X}|] \tag{1}\label{1}$$
This intuitively makes sense-clearly centering a random variable will bring it closer to $0$.
If there was no absolute value, applying linearity of expectation would show us:
$$E[X – \bar{X}] = E[X] – E[\bar{X}] = \bar{X} – \bar{X} = 0$$
And then noting that $E[|X|] \geq 0$, we can show that $E[|X|] \geq E[|X – \bar{X}|]$.
Is this all that is needed? In other words, is there a more formal way of proving $(1)$? I have considered using the fact that the absolute value function is convex, but am not sure if that is necessary to prove this.
Context
Updating this to include some additional context. I am asking this as part of a larger proof of the Weak Law of Law Numbers with infinite variance (part of Discrete Stochastic Processes Theory for Applications by Gallager). In it he states:
Theorem 1.7.4 (WLLN): For each integer $n \geq 1$, let $S_n = X_1 + \dots + X_n$, where $X_1, X_2, \dots$ are IID rvs satisfying $E[|X|] < \infty$. Then for any $\epsilon > 0$,
$$lim_{n \rightarrow \infty} P\Big( \Big| \frac{S_n}{n} – E[X]\Big| > \epsilon \Big) = 0$$
Proof: We use a truncation argument. Let $b$ be a positive number (which we later take to be increasing with $n$), and for each variable $X_i$, define a new rv $\hat{X}_i$ by:
$$
\hat{X}_i = \begin{cases}
X_i & for \; \; E[X] – b \leq X_i \leq E[X] + b \\
E[X] + b & for \; \; X_i > E[X] + b \\
E[X] – b & for \; \; X_i < E[X] + b
\end{cases}
$$
The truncated variables are IID. Show that each $\hat{X}_i$ has a finite mean $E[\hat{X}]$ and a finite variance $\sigma_{\hat{X}}^2$. Show that the variance is upper bounded by the second moment around the original mean $\bar{X}$, i.e. show that $\sigma_{\hat{X}}^2 \leq bE[|\hat{X} – \bar{X}|] \leq 2bE[|X|]\tag{2}\label{2}$
It is this last step that prompted me to ask this question. I am not able to show that:
$$\sigma_{\hat{X}}^2 \leq bE[|\hat{X} – \bar{X}|] \leq 2bE[|X|]$$
I can prove the left side of the inequality, $\sigma_{\hat{X}}^2 \leq bE[|\hat{X} – \bar{X}|]$. Specifically, as $E[\hat{X}] \rightarrow E[X]$ the variance $\sigma_{\hat{X}}^2$ is maximized. Hence:
$$\sigma_{\hat{X}}^2 = Var(\hat{X}) = E[(\hat{X} – E[\hat{X}])^2]$$
Where this is maximized when $E[\hat{X}] = E[X]$:
$$\sigma_{\hat{X}}^2 \leq E[(\hat{X} – E[X])^2]$$
And we can show that $|\hat{X} – E[X]| \leq b$. Using this inequality we finally show:
$$\sigma_{\hat{X}}^2 \leq E[b^2] = bE[|\hat{X} – E[X]|]$$
So, that leaves me trying to prove the right hand side of the inequality in (2):
$$bE[|\hat{X} – \bar{X}|] \leq 2bE[|X|]$$
Which is why I asked the question of how to prove (1). My reasoning was that if I could prove (1), I could then prove the right hand side of the inequality in (2).
Best Answer
This is false. Here is a counter-example:
$$X = \left\{\begin{array}{ll} 1 & \mbox{ with prob $1/4$} \\ 0 & \mbox{ with prob $3/4$} \end{array}\right.$$
Then
On the other hand, if $Y$ is any random variable with finite variance then $$ E[Y^2] \geq E[(Y-E[Y])^2]$$
Further, for any random variable $Y$ with finite mean we have $$E[|Y-E[Y]|] \leq 2E[|Y|]$$ Proof: $$|Y-E[Y]| \leq |Y| + |E[Y]| \leq |Y| + E[|Y|]$$ where the final inequality holds by Jensen's inequality with the convex function $|y|$. Taking expectations of both sides gives the result. $\Box$
[The example random variable $X$ given at the start of this answer indeed satisfies this inequality because $3/8\leq 2(1/4)$.]