If $E[X]=0$ then $E[|X|]$ is finite

Question

I read a textbook and it says: $E[X]=0$ implies $E[|X|]<\infty$ but I cannot understand the logic. Could anyone tell me why so? Many thanks!

Answer 1

By definition: $E[X]$ is only defined (EDIT: as a real number) for $X$ such that $E[|X|] < \infty$.

Answer 2

Let $X^+=\max\{X,0\}$ and $X^-=\max\{-X,0\}$. Then $E(X)$ is defined as $E(X^+)-E(X^-)$ when at least one of $E(X^+)$ and $E(X^+)$ is finite. The statement $E(X)=0$ (or $E(X)=$ any finite real number) then implicitly implies that both $E(X^+)$ and $E(X^+)$ are finite. We infer $E(|X|)=E(X^+)+E(X^-)$ is also finite.

Answer 3

\begin{align} EX&=\int_{-\infty}^{\infty}xf(x)dx\\ &=\int_{-\infty}^{0}xf(x)dx+\int_{0}^{\infty}xf(x)dx\\ &=A+B \end{align} $A+B=0$ implies $A=-B$. Note also that both $A$ and $B$ are finite. \begin{align} E|X|&=\int_{-\infty}^{\infty}|x|f(x)dx\\ &=-\int_{-\infty}^{0}xf(x)dx+\int_{0}^{\infty}xf(x)dx\\ &=-A+B\\ &=2B \end{align}

Answer 4

Well, consider a tempting example where not requiring $E[|X|]$ to be finite causes problems. Consider a variable with probability density function of $\frac{1}2\frac{1}{(|x|+1)^2}$. If we integrate $$\lim_{L\rightarrow\infty}\int_{-L}^{L}x\cdot \frac{1}2\frac{1}{(|x|+1)^2}\,dx$$ then we get $0$ as the expectation of $x$. However, if we take the limit in a different manner, say as $$\lim_{L\rightarrow\infty}\int_{-2L}^{L}x\cdot \frac{1}2\frac{1}{(|x|+1)^2}\,dx$$ then we get $-\log(2)$ as the expectation - which is bad because we should've gotten the same answer!

Essentially, if we wish to make any sense of the statement $E[X]$ it is necessary to take a certain infinite integral in a well-defined sense - and this can only happen if at least one of $E[\max(x,0)]$ and $E[\min(x,0)]$ are finite. If one is infinite then $E[X]$ would also be, so if $E[X]=0$ is finite, so are both of those integrals meaning $E[|X|]$ is finite.

Answer 5

Suppose on the contrary that $0 = E[X]$, but $\infty = E[|X|]$. Then, we have

$$\infty = E[|X|] = E[X^+ + X^-] = E[X^+] + E[X^-]$$

$$\to E[X^+] = \infty \ \text{or} \ E[X^-] = \infty$$

Then

$$E[X] = E[X^+ - X^-] = E[X^+] - E[X^-]$$

1. $$= \infty \ \text{if} \ E[X^+] = \infty \ \text{but} \ E[X^-] < \infty$$

2. $$= -\infty \ \text{if} \ E[X^-] = \infty \ \text{but} \ E[X^+] < \infty$$

3. $$\text{is indeterminate otherwise}$$

All cases contradict $E[X] = 0$, or any other real number for that matter. This is why in the first place using '$E[X]$' often requires that $E[|X|] = \infty$. Exceptions include when the random variable takes values in $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$