If $X$ is a random variable with $E[X]= \mu, \text {satisfying} P(X \le 0)=0,$ show that $P(X > 2 \mu) \le \frac {1}{2}$

Question

If $X$ is a random variable with $E[X]= \mu, \text {satisfying } P(X \le 0)=0,$ show that $P(X > 2 \mu) \le \dfrac {1}{2}$

Attempt: $P(X > {2\mu}) = \int_{2\mu}^{\infty} f(x) dx \le \int_{0}^{2\mu} f(x) dx +\int_{2\mu}^{\infty} f(x) dx$

Some other results which I could deduce:

$\mu = \int_0^{\infty}x f(x) dx=\int_0^{1} x f(x) dx + \int_1^{\infty} x f(x) dx \ge \int_0^{1} x f(x) dx + \int_1^{\infty} f(x) dx$

Thr rightmost term is $= \int_0^{2\mu} [x f(x) +f(x) ]dx + \int_{2\mu}^{\infty} [x f(x) +f(x) ]dx$

Could someone please give me a clue on how to move ahead. Thanks a lot for your help

Answer 1

We have \begin{align} 2\mu I_{X > 2\mu} \le X \end{align}

Now, take expectation on the both sides and you will get the result.

Note that this is called Markov's inequality.

Answer 2

Based on $P\left(X\leq0\right)=0$ it can be proved that: $$\mathbb{E}\max\left(0,X\right)=\mathbb{E}X=\mu>0$$

Assume that: $$P\left(X>2\mu\right)>\frac{1}{2}$$

Then on base of the following evident inequality: $$\max\left(0,X\right)\geq2\mu\mathbf{1}_{X>2\mu}$$ we find that: $$\mu=\mathbb{E}\max\left(0,X\right)\geq\mathbb{E}2\mu\mathbf{1}_{X>2\mu}=2\mu P\left(X>2\mu\right)>\mu$$ A contradiction is found so our assumption must be wrong.

Conclusion:$$P(X>2\mu)\leq\frac12$$