If $E(X)= μ$ and $E((X- μ)^4) =b$, then for $t>0$, $P\{|X- μ|\ge t\}\le b/t^4$

Question

Let $X$ be a random variable for which $E(X)= μ$ and $E((X- μ)^4) =b$. Prove that for every $t>0$, $$P\{|X- μ|\ge t\}\le b/t^4.$$

I know I need to use Chebyshev inequality but if I do I get $P\{|X- μ|\ge t\}\le rt(b)/t^2$. So if I can show that $rt(b)/t^2 \le 1$ then I am done but surely this isn't always true?

Answer 1

It's not clear how you arrived at your answer. The "simplest" form of Chebyshev-Markov for a non-negative random variable $Y$ is:

$$P(Y\geq t) \leq \frac{E[Y]}{t}.$$

Since $\{Y\geq t\}=\{Y^4\geq t^4\}$, for non-negative $t$,

$$P(Y^4\geq t^4)\leq \frac{E[Y^4]}{t^4},$$

where for you $Y:=|X-t|$.