# Proof that $E(|X_1 - X_2|)$ is bound by twice the mean

Let $X_1$, $X_2$ be iid random variables.

How do I show that for non-negative variables $E(|X_1 - X_2|)$ is bound from above by twice the expected value of $X_1$ (or $X_2$)?

• Clearly this is not true when the common expectation is not positive! – whuber Jan 5 '18 at 14:41
• yes thats why I wrote "nonnegative variables" – Manuel R Jan 5 '18 at 15:32
• Simply apply the triangle inequality. – Zhanxiong Jan 5 '18 at 15:54
• $|X_1-X_2| \leq |X_1| + |X_2|$. You actually don't need the independence assumption to reach your conclusion. – Zhanxiong Jan 5 '18 at 15:58
• @Manuel S, I added a self-study tag to the post. – Lucas Roberts Jan 5 '18 at 16:39

as pointed out in the comments by @zhanxiong, the triangle inequality is sufficient here, take:

$|X_1 -X_2| \leq |X_1| +|X_2|$ and take expectations to get

$\mathbb{E}(|X_1 -X_2|) \leq \mathbb{E}(|X_1|) +\mathbb{E}(|X_2|)$. However, you cannot equate the two marginal expectations to be the same value without assuming they have the same marginal distributions. Assuming this you can conclude $\mathbb{E}(|X_1 -X_2|) \leq 2\mathbb{E}(|X_1|)$ and the non-negative assumption allows you to say that $\mathbb{E}(|X_1|) = \mathbb{E}(X_1)$. So that if $\mu=\mathbb{E}(X_1)$ then you have: $\mathbb{E}(|X_1 -X_2|) \leq 2\mu$.

As mentioned by @zhanxiong in the comments this doesn't use independence but does use that the marginals have the same distributions.

As @whuber points out in a comment this is not possible if the variables could be negative. Here is a simple example showing what happens if they are negative expectations; let $X_i \sim Rademacher(p=1/3)$ where $\Pr(X_i=1)=1/3$ and $\Pr(X_i=-1)=2/3$. The expectation of each is $\mathbb{E}(X_i) =-1/3$. Then with the i.i.d. assumption we can calculate the four probabilities:

1. $\Pr(X_1=1, X_2=1) = 1/9$ and $|1-1| = 0$
2. $\Pr(X_1=1, X_2=-1) = 2/9$ and $|1-(-1)| =2$
3. $\Pr(X_1=-1, X_2=1) = 2/9$ and $|-1-1| =2$
4. $\Pr(X_1=-1, X_2=-1) = 4/9$ and $|-1-(-1)|=0$

Multiplying the probabilities of each event (l.h.s.) with the values on the r.h.s. and summing all four numbers gives you $\mathbb{E}(|X_1 -X_2|)=8/9$ and twice the expectation is: $2\mathbb{E}(X_i) = -2/3$. Clearly, $8/9 \nleq -2/3$ so the claim needs additional conditions/assumptions (non-negativity) to be true.

• I'm curious, why did you chose that particular Rademacher for an example? Wouldn't it be simpler to just consider $X_i$ such that $P(X_i = -1)=1$? Also, doesn't $E(X_1) = E(X_2)$ follow from the iid assumption of the question? – psarka Jan 5 '18 at 16:22
• @psarka, no particular reason, your example of $P(X_i=-1)=1$ would work too but I prefer non-deterministic random variables-they are more fun. For your second question, yes $\mathbb{E}(X_1)=\mathbb{E}(X_2)$ follows from the i.i.d. assumption.However, the point made by Zhanxiong in the comments is that the i.i.d. assumption is a stronger assumption than necessary to prove the claim. I was trying to show/argue how to show it without i.i.d. assumption. – Lucas Roberts Jan 5 '18 at 16:38