If $\operatorname{var}(X) =0$ then $p( X = E (X)) = 1.$

Question

Let $X$ be a random variable. Then $\operatorname{var}(X) =0$ implies $p( X = E (X)) = 1.$

Actually i need both directions. But i was able to show that $p( X = E (X)) = 1$ implies $\operatorname{var}(X) =0.$ So all i need is the other direction.

I think we have to use the definition of the variance and set the $\operatorname{var}(X) = 0.$ But why is $p(X = E(X)) = 1$?

Answer 1

Suppose for sake of contradiction $P(X \ne E[X]) > 0$. Then there exists some $\epsilon > 0$ such that $$P(|X - E[X]| > \epsilon) > 0.$$ But then $$\text{Var}(X) = E[(X - E[X])^2] \ge E[(X - E[X])^2 \mathbf{1}_{|X - E[X]| > \epsilon}] > \epsilon^2 P(|X - E[X]| > \epsilon) > 0,$$ a contradiction.

---

If the above is hard to understand, considering the case of a discrete random variable $X$ can be helpful for intuition. If $P(X \ne E[X]) > 0$, then there is some value $c \ne E[X]$ such that $P(X=c) > 0$. Then $$\text{Var}(X) = E[(X - E[X])^2] \ge (c - E[X])^2 P(X=c) > 0.$$ This argument does not work for continuous random variables, though.

Answer 2

Shorter:

$(X-\Bbb{E}[X])^2 \geq 0$ by definition. So it follows directly from $\Bbb{E}[(X-\Bbb{E}[X])^2] = \operatorname{Var}(X) = 0$, that $(X-\Bbb{E}[X])^2 = 0$ almost surely. We conclude $X-\Bbb{E}[X] = 0$ almost surely. That is what you wanted to show.