Probability of a non negative distribution

Question

Let $X$ be a non-negative random variable with $\text{Var}(X)<\frac{1}{2}$. Show that then $P\big(-1+E(X)\le X\le 2E(X)\big)\ge \frac{1}{2}$ where $P$ is the probability measure and Var is the variance and $E(X)$ is the expectation of $X$.

My approach : I was thinking to compute $P(-1+E(X)\le X)$ and $P(X\ge 2E(X))$ so that $P\big(-1+E(X)\le X\le 2E(X)\big)=P(-1+E(X)\le X)-P(X > 2E(X))$ and $P(-1+E(X)\le X)=1-P(-1+E(X)> X)=1-P(X-E(X)<-1)$ and $P(X-E(X)<-1)\le \frac{Var(X)}{(1)^2}\le \frac{1}{2}$(By Chebyshev Inequality and given condition) hence $P(X\ge -1+E(X))\ge \frac{1}{2}$ and by similar work since $X$ is non negative hence assuming $E(X)>0$ we get that $P(X>2E(X))\le \frac{1}{2}$ hence $P\big(-1+E(X)\le X\le 2E(X)\big)\ge 0$ but how will i get it $\ge \frac{1}{2}$.

Any type of help will be appreciated. Thanks in advance.

Answer 1

Some comments: In the link, the Chebysev's inequality is $$P(|X-\mu|\geq k) \leq \frac{\sigma^2}{k^2}$$ or $$P(|X-\mu|< k) \geq 1 -\frac{\sigma^2}{k^2} $$

If $EX>1$, one has $$P(-1<X - EX \leq EX) = P(-1\leq X-EX \leq 1) + P(1\leq X-EX \leq EX)\geq 1-\frac{VarX}{1}\geq \frac{1}{2}.$$

If $EX<1$, one has $$P(-1<X - EX \leq EX) = P(-1\leq X-EX \leq -EX) + P(-EX\leq X-EX \leq EX)\geq 1-\frac{VarX}{(EX)^2}\geq \frac{1}{2}.$$

Answer 2

Let $\mu=\Bbb E(X)$ and $\sigma^2=\mathrm{Var}(X)$.

From Chebyshev's Inequality we have $$ \Bbb P(|X-k|\leq t)\geq 1-\frac{E[(X-k)^2]}{t^2} $$ that is $$ \Bbb P(-t+k<X<t+k)\geq 1-\frac{E[(X-k)^2]}{t^2} $$ By setting $a=-t+k$ and $b=t+k$ we have the general Chebyshev's Inequality $$ \Bbb P(a\leq X \leq b)=\Bbb P\left(\left|X-\frac{a+b}{2}\right|\leq \frac{b-a}{2}\right)\geq 1- \frac{\sigma^2+\left(\mu-\frac{a+b}{2}\right)^2}{\left(\frac{b-a}{2}\right)^2} $$ Now set $a=\mu-1$ and $b=2\mu$ $$ \Bbb P(\mu-1\leq X\leq2\mu)\geq 1- \frac{\sigma^2+\left(\mu-\frac{3\mu-1}{2}\right)^2}{\left(\frac{\mu+1}{2}\right)^2}\geq 1- \frac{1/2+\left(\frac{1-\mu}{2}\right)^2}{\left(\frac{1+\mu}{2}\right)^2} $$ Now the fraction term in the last inequality is $\leq 1/2$ if $\mu>1$ (otherwise, if $\mu\leq 1$, then we just truncate the lower bound at zero, and it is easy to arrive to the final inequality).

Then $$\Bbb P(\mu-1\leq X\leq2\mu)\geq 1/2$$