A random variable $ X $ satisfies $ | X | \le 1 $ a.s. Show that $ \operatorname {Var} ( X ) \le 1 $, and that the maximum is attained.
I am apply Chebyshev's inequality.
If $ X $ is any random variable, then for any $ b > 0 $ we have $$ \Pr ( | X − \mathbb E [ X ] | \ge b ) \le \frac { \operatorname {Var} ( X ) } { b ^ 2 } \text , $$ But I am not getting anything please help me how to solve this.
Thank you.
Clearly, if $|X| \leq 1$ a.s., then we almost surely have $X^2 \leq 1$.
Therefore, $\mathbb{E}[X^2] \leq 1$, i.e. the second (and therefore also the first) moment of $X$ exists.
So, $0 \leq \mathbf{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 \leq 1 - 0 = 1$, where we used $(\mathbb{E}[X])^2 \geq 0$ and $\mathbb{E}[X^2] \leq 1$.
Note that $\mathbf{Var}(X)$ is non-negative by definition.
You also attain the bound if $\mathbb{P}(X = 1) = \mathbb{P}(X = -1) = \frac{1}{2} $.
Clearly we would then have $|X| \leq 1$ a.s. and also $\mathbb{E}[X] = 0$.
So, $\mathbf{Var}(X) = \mathbb{E}[X^2] = \mathbb{E}[1] = 1$.