# What's stronger: Expected Value or With Probability 1?

This question is not related to "convergence".

Let $X$ be a random variable.

I wonder which of the two following conditions is stronger.

1. $\mathbb{E}\{X\} = 0$ (the expected value of $X$ is $0$)

2. $Pr\{ X=0 \} = 1$ (i.e. $X=0$ "almost everywhere")

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(1) follows from (2), so (2) is stronger. – Thomas Apr 3 '12 at 21:51
$E(X)=0$ happens a lot. Do you mean $E(|X|)=0$? – André Nicolas Apr 3 '12 at 22:21
no, I just took an arbitrary value for $\mathbb{E}\{X\}$ – Adam Apr 3 '12 at 22:23
Let $X=1$ with probability $1/2$, and let $X=-1$ with probability $1/2$. Then $E(X)=0$, but $P(X=0)\ne 1$. – André Nicolas Apr 3 '12 at 22:47

## 1 Answer

The second condition is much stronger because it defines the whole distribution, i.e. $\mathbb{P}\{X=0\}=1$ implies that the density function of $X$ is $f(x)=\cases{1&x=0 \cr 0&otherwise}$ . This tells you everything about $X$.

On the other hand, $\mathbb{E}(X)=0$ only tells you that the density $f(x)$ of $X$ satisfies $\int_{-\infty}^\infty xf(x)dx=0$ (in the continuous case) or $\sum_{-\infty}^\infty xf(x)=0$ (in the discrete case); this could be the case for many different probability densities $f(x)$. For example, any random variable whose density is symmetric around $0$ will have $\mathbb{E}(X)=0$.

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A minor cavil with respect to the last sentence: a Cauchy random variable whose density $\frac{1}{\pi(1+x^2)}$ is symmetric about $0$ has an undefined expected value. – Dilip Sarwate Apr 4 '12 at 2:39