I need to show the following random variable results in equality for Chebyshev's inequality. However, I believe I am making some mistakes that I cannot find. Please help me spot my mistake.
$p \in (0,1), t \in \mathbb{R}, P(X=-t)=P(X=t)=p, P(X=0)=1-2p$
Chebyshev's inequality is said to be
$P(|X-E(X)|>t) \leq \frac{Var(X)}{t^2}, t >0$
I got
$E(X) = tp – tp +0 = 0$
$Var(X)= t^2p+(-t)^2p+0 = 2pt^2$
So $P(|X-E(X)|>t) = P(|X|>t) = 0 \ (\text{if t $\geq$ 0}) \neq 2p = \frac{Var(X)}{t^2}$
What have I done wrong?
Best Answer
There is nothing wrong with your computations. It is called Chebychev's inequality and not Chebychev's equality. Chebychev's inequality gives you just an upper bound for $P(|X|>t)$. In your case, $P(|X|>t)=0$ and Chebychev's inequality gives you $P(|X|>t)\leq 2p$, which is no contradiction since $0\leq 2p$.
By the way, you have to additionally assume that $p\leq \frac{1}{2}$, in order for $P$ to be a probability measure ($P(X=0)=1-2p$ must be nonnegative, i.e., $p\leq \frac{1}{2}$).