Can I use the Chebychev inequality to prove this statement?

Question

Let $X$ be a random variable with $\Bbb{E}(X^2)<\infty$ and $a>0$ I need to show that $$\Bbb{P}(X>a)\leq \frac{\Bbb{E}(X^2)}{a^2}$$

My idea was the following. Let me first remark that $$\{X>a\}\subset \{|X-\Bbb{E}(X)|\geq a\}$$Indeed if $X>a>0$ then $|X|>a$. Now consider $a\leq|X-\Bbb{E}(X)|\leq |X|+|\Bbb{E}(X)|$. This is equivalent to say that $a-|X|<0\leq|\Bbb{E}(X)|$ which is always true. Therefore we have $$\Bbb{P}(X>a)\leq \Bbb{P}(|X-\Bbb{E}(X)|\geq a)\stackrel{\text{Chebychev}}{\leq}\frac{Var(X)}{a^2}=\frac{\Bbb{E}(X^2)-\Bbb{E}(X)^2}{a^2}\leq\frac{\Bbb{E}(X^2)}{a^2}$$

Is this proof correct so? I know there is maybe a simpler one, but this was the first thing coming to my mind so it would be nice if you first could take a look at this prove and then if it's correct discuss about alternatives

Thanks a lot

Answer 1

To prove the result just use Markovs inequality as the other answer shows. In your proof you use the inequality you are trying to prove! Note you use Chebyshev inequality :

$$ P(|Y-\mu|\geq \sigma k)\leq \frac{1}{k^2}.$$

But if we define the centered random variable $X=Y-\mu$, then the above says (note $\text{Var}(Y)=\sigma$)

$$ P(|X| \geq \sigma k) \leq \frac{1}{k^2},$$

choose $k=\frac{a}{\sigma}$ gives

$$ P(|X| \geq a) \leq \frac{\sigma^2}{a^2}. $$

You want to prove this last inequality, but as you can see it is equivalent to Chebyshevs inequality (so you cant you use Chebyshev to prove it).

Answer 2

Perhaps a shorter path is this: $$\mathbb{P}(X>a)\leq \mathbb{P}(|X|>a)=\mathbb{P}(X^2>a^2)\leq \frac{E(X^2)}{a^2}$$ The last inequality is trivial, for any nonnegative random variable $Y$ and $b>0$ $$E(Y)=\int_{Y>b}Y\,d\mathbb{P}+\int_{Y\leq b}Y\,d\mathbb{P}\geq b\,\mathbb{P}(Y>b)$$ As pointed out by Clement C. this trivial inequality is known as Markov's inequality.