1. ## Proving Chebyshev's Inequality

Prove Chebyshev's inequality, which states that if $\displaystyle X$ is a random variable with expectation $\displaystyle \mu$ and finite variance $\displaystyle \sigma^2$, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2}$, for any $\displaystyle \varepsilon > 0$

2. Originally Posted by Jimmy_W
Prove Chebyshev's inequality, which states that if $\displaystyle X$ is a random variable with expectation $\displaystyle \mu$ and finite variance $\displaystyle \sigma^2$, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2}$, for any $\displaystyle \varepsilon > 0$
Have you Googled for a proof ....?

3. Hi Mr Fantastic,
I have just realised I left out part of my question , which provides me with a road to go down, but I am really lost so I found that none of the proofs on google really helped me out.

The question should have been:

Prove Chebyshev's inequality, which states that if $\displaystyle X$ is a random variable with expectation $\displaystyle \mu$ and finite variance $\displaystyle \sigma^2$, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2}$, for any $\displaystyle \varepsilon > 0$
by first showing that $\displaystyle \mathbb{E}[X - \mu)^2] \geq \mathbb{E}[(X- \mu)^2 I_{\{(X- \mu)^2 \geq \varepsilon^2\}}].$

Its this last part that I carelessly left out that has me stumped.

4. I am really stumped on this one. But the little work that I have been able to do on it is:

$\displaystyle (X - \mu)^2 \geq \varepsilon^2 \geq \mu - \varepsilon \geq X$
and $\displaystyle \mu + \varepsilon \leq X$

and now I am lost

5. This is what I have put together.
Could somebody please look over it for me?

$\displaystyle (X - \mu)^2 \geq \varepsilon^2 \geq \mu - \varepsilon \geq X$
and $\displaystyle \mu + \varepsilon \leq X$

So... $\displaystyle (- \infty, \mu - \varepsilon) \cup (\mu + e, \infty)$

Now
$\displaystyle \mathbb{E} [(X - \mu)^2] = \int_{-\infty}^{\infty} (X - \mu)^2 \ f(x) \ dx$

$\displaystyle \geq \ \int (X - \mu)^2 \ f(x) \ dx$

$\displaystyle \geq \int \varepsilon^2 \ f(x) \ dx$ (because $\displaystyle ((X - \mu)^2 \geq \varepsilon^2) \in (-\infty, \infty)$

$\displaystyle = \varepsilon^2 \ \int f(x) dx$

$\displaystyle = \varepsilon^2 \ \mathbb{P}(|X - \mu| \geq \varepsilon$

therefore

$\displaystyle \sigma^2 \geq \varepsilon^2 \ \mathbb{P}((|X - \mu| \geq \varepsilon)$

and

$\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2}$