$\mathbb{P(|\mathbf{X}|\leq 2) = 1}$ if $\mathbf{X}$ has bounded moments?

Question

Suppose that a random variable $\mathbf{X}$ has bounded moments: $\mathbb{E}(\mathbf{X}^k)\le k^{2}2^k$. I would like to show that $\mathbb{P(|\mathbf{X}|\leq 2) = 1}$. I am considering using Markov's inequality but I am not sure on how to proceed since the inequality signs of the problem and Markov's inequality are different. Any direction or suggestions would be appreciated.

Answer 1

Hint: $E\sum \frac {X^{2k}} {(2+c)^{2k}} <\infty$ for any $c>0$. Hence $\sum \frac {X^{2k}} {(2+c)^{2k}}$ converges a.s. This implies $ \frac {X^{2k}} {(2+c)^{2k}} \to 0$ a.s, Hence also in probability. In particular, $P(|X^{2k} | >(2+c)^{2k}) \to 0$ as $ k \to \infty$. Can you finish?

Answer 2

As shown in Limit of $L^p$ norm, we have $E[|X|^k]^{1/k}$ converging to the $L^\infty$ norm of $X$, which is the essential supremum of $|X|$. But we have $E[|X|^k]^{1/k} \le k^{2/k} \cdot 2$ which by some easy calculus converges to $2$.

Answer 3

As OP considered using Markov's inequality here is one way to do it. For some small $\epsilon>0$

$$\begin{split}\mathbb P(|X|>2+\epsilon)&=\mathbb P(X^{2k}>(2+\epsilon)^{2k})\le\frac{\mathbb E(X^{2k})}{(2+\epsilon)^{2k}}\le\frac{(2k)^22^{2k}}{(2+\epsilon)^{2k}}\end{split}$$

Then multiply by the far sides of the above inequality by negative one, and add one to both sides.

$$\begin{split}\mathbb P(|X|\le2+\epsilon)=1-\mathbb P(|X|>2+\epsilon)&\ge1-\frac{(2k)^22^{2k}}{(2+\epsilon)^{2k}}\end{split}$$

Taking the limit as $k\rightarrow\infty$ gives $\mathbb P(|X|\le2+\epsilon)\ge1$. Since $\epsilon$ is positive I have some doubts whether the statement holds.