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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")

  • 2
    Let $X=1$ with probability $1/2$, and let $X=-1$ with probability $1/2$. Then $E(X)=0$, but $P(X=0)\ne 1$.2012-04-03

1 Answers 1

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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 $\mathbb{E}(X)=0$.

  • 1
    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.2012-04-04