Chebyshev’s inequality is and is not sharp

Question

(i) Show that Chebyshev’s inequality is sharp by showing that if $0<b\leq a$ are fixed there is an $X$ with $E(X^2)=b^2$ for which $P(|X|\geq a)=b^2/a^2$.

(ii) Show that Chebyshev’s inequality is not sharp by showing $X$ has $0<E(X^2)<\infty$ then $\lim_{a\to\infty} a^2P(|X|\geq a)/E(X^2)=0$

In (i) it looks like problem is to find a $X$ for which the equality holds. But I could not find any such $X$.

In (ii) if $\lim_{a\to\infty} P(|X|\geq a)/E(X^2)=0$ it would be one line proof. But given is $\lim_{a\to\infty} a^2P(|X|\geq a)/E(X^2)=0$. If we can take $b=a-\varepsilon$, where $\varepsilon >0$ is a fixed quantity, the equality holds too. But I don't think it would be the case. It should hold for any $b\leq a$.

Need help in both part.

Note: Chebyshev’s inequality :- Suppose $\varphi : \mathbb{R}\rightarrow \mathbb{R}$ has $\varphi \geq 0$ let $A\in \mathcal{B}$ (Borel set) and let $i_A = \inf \{ \varphi (y) :y\in A \}$.

$i_A P(X\in A) \leq E(\varphi (X), X\in A)\leq E\varphi (X)$

Clearly here $\varphi (X)=X^2$

Answer 1

$$ \text{Let } X = \begin{cases} 0 & \text{with probability } 1 - \dfrac{b^2}{a^2}, \\ \\ \pm a & \text{each with probability } \dfrac{b^2}{2a^2.} \end{cases} $$ $${}$$ $$\text{Then } \operatorname{E}(X^2) = b^2 \text{ and } \Pr(|X|>a) = \dfrac{b^2}{a^2}.$$

Answer 2

Consider $X \sim N(0,1).$ Then $E(X^2)=1,$ which satisfies the criterion $0<E(X^2)<\infty.$ We want to show that

$$\lim_{a \to \infty}a^2P(|X|\geq a)/E(X^2)=0 \tag 1$$

Since $N(0,1)$ is symmetric about $0,$ we have

$$a^2P(|X|\geq a)/E(X^2)=a^2P(|X|\geq a)=2a^2P(X\geq a) \tag 2$$

It suffices to show that

$$\lim_{a\to\infty} a^2P(X\geq a)=0 \tag 3$$

Since $a^2P(X\geq a)>0$ for all $a \in \mathbb{R},$ we have

$$\lim_{a \to \infty}a^2P(X\geq a) \geq 0 \tag 4$$

Let us denote the probability density function of $N(0,1)$ distribution by $\phi(\cdot).$ Now,

$$a^2P(X\geq a)=a^2 \int_a^\infty \phi(x) \, dx < \int_a^\infty x^2\phi(x)\,dx \tag 5$$

The right-hand side integrand $x^2\phi(x)$ is non-negative. Check that this quantity shrinks to $0$ as $a \to \infty.$ i.e. for every $\varepsilon > 0, \exists M \in \mathbb{R}$ such that $a \geq M \implies \int_a^\infty x^2\phi(x)\,dx < \varepsilon,$ Hence, by $(5)$, $a \geq M \implies a^2P(X\geq a) < \epsilon.$

Since $\epsilon>0$ is arbitrary, we conclude that

$$\lim_{a \to \infty} a^2P(X\geq a) \leq 0 \tag 6$$

Clearly, $(4)$ and $(6) \implies (3) \implies (1)$ and we are done.

Answer 3

Note that $a^21_{|x|\ge a}\rightarrow 0\text{ a.s.}$, and then use the dominated convergence theorem.

Answer 4

If $\lim_{a\rightarrow\infty} a^2 P(|X|\ge a)/EX^2 \ne 0$, then there exists $c>0$, such that $$y^2 P(|X|\ge y)/EX^2>c$$ for all $y>0$.

Notice that $$EX^2=\int_0^{\infty}2y P(|X|\ge y)dy \ge \int_0^{\infty}\frac{2c}{y}dy=\infty ,$$ which contradicts the fact that $EX^2<\infty$.