# Probability for a random variable to exceed its expectation [closed]

Let $$X$$ be a real random variable, defined on some probability space $$(\Omega,\mathcal{T},\mathbb{P})$$, with finite expectation $$\mathbb{E}(X)$$.

I don't make any other assumption.

I am convinced that $$\mathbb{P}\left(X\geqslant\mathbb{E}(X)\right)>0$$ (that is : non zero), but I don't see how to prove it rigorously.

Thanks in advance for any hint / help ...

• Also there is no better absolute bound. Consider a random variable which takes value $n$ with probability $1/n$ and $0$ otherwise, for large $n$. Commented Jul 22 at 13:33
• @MichaelLugo: Similarly, you can't set a (useful) upper bound, because the delta distribution always returns exactly its expected value. Commented Jul 23 at 16:18

If $$\mathbb P\big\{X\geq \mathbb E[X]\big\}=0$$, then $$\mathbb P\big\{X<\mathbb E[X]\big\}=1$$. Therefore, $$X<\mathbb E[X]\quad \text{a.s.},$$ and thus, $$\mathbb E[X]<\mathbb E\big[\mathbb E[X]\big]=\mathbb E[X],$$ which is a contradiction.

Abbreviate $$\mu:=\Bbb E[X]$$.

Suppose $$\Bbb P\left(X\ge \mu\right)=0$$. Then $$\mu-X$$ is a non-negative random variable with mean $$0$$. By Markov's inequality $$\Bbb P(\mu-X> 1/n)\le{\Bbb E[\mu-X]\over1/n}={\mu-\mu\over 1/n}=0$$ for each positive integer $$n$$. It follows that $$\mu-X=0$$ a.s.; that is, $$\Bbb P(X=\mu) =1$$.

In short, apart from the degenerate case in which $$X$$ is a.s. constant, both $$\Bbb P(X>\mu)$$ (and by the same token $$\Bbb P(X<\mu)$$) must be strictly positive.

Directly from the definition:

Suppose for a contradiction that $$\mathbb{P}[X\geq \mathbb{E}[X]]=0$$. Since the probability of $$X$$ exceeding its expectation is zero this is equal to

$$\mathbb{E}[X] =\int_\Omega X dp = \int_{\{X:X<\mathbb{E}[X]\}} X dP$$

moreover we know the probability of the event $$\{X<\mathbb{E}[X]\}$$ is 1 since it is the complement of the event we are interested in, hence we can bound this integral in the following way:

$$\mathbb{E}[X]=\int_{\{X:X<\mathbb{E}[X]\}} X dP< \mathbb{E}[X]\int_{\{X:X<\mathbb{E}[X]\}} dP= \mathbb{E}[X]$$

This boils down to this question about strict monotonocity