I am trying to prove or disprove the following:
let X be a random variable such that $E[X^4]=1$ then $P(|X|\geq 2) \le \frac{1}{16}$?
It's obviously related to Markov's inequality. Well, my attempt was to alter the original proof:
$$ E[X^n]=\int_{0}^{\infty} x^n*f(x)dx \geq \int_{c}^{\infty}x^nf(x)dx\geq\int_{c}^{\infty}c^nf(x)dx=c^nP(X\geq c) $$
then
$$ P(X\geq c) \leq \frac{E[X^n]}{c^n}$$
Given the following "proof" (I'm just not sure about it's validity), it's easy to see that, given that $ E[X^4] = 1$:
$$ P(X\geq 2) \leq \frac{1}{2^4}=\frac{1}{16}$$
Is that a valid proof?
Without using the Markov inequality (or actually proving it in the process), using the expectation of indicator function on an event to denote the probability of the event, we get: \begin{align*} \mathbb P[|X|\geq 2]&=\mathbb E[\mathbf 1(|X|\geq 2)]\\ &=\mathbb E[\mathbf 1(|X|^4\geq 16)]\\ &=\mathbb E\left[\mathbf 1\left(\frac{|X|^4}{16}\geq 1\right)\right]\\ &\leq \mathbb E\left[ \frac{|X|^4}{16}\right]\\ &=\frac{\mathbb E[X^4]}{16}\\ &=\frac{1}{16} \end{align*}
Since $\frac{|X|^4}{16}$ is a positive random variable.
To see why $\mathbf 1(a\geq 1)\leq a$ for $a\geq 0$, it is sufficient to observe that if $a<1$, then $\mathbf 1(a\geq 1)=0\leq a$ and for $a\geq 1$, $\mathbf 1(a\geq 1)=1\leq a$