# is it by Chebychev's inequality? [closed]

Let x be random variables with mean zero and variance 1. Let n be natural number, t>0, C>0. Let also $P(x^2\geq n)\geq C/(n^t)$. Show, if $t\geq4$, then $P(x^2\geq n)\leq C/(n^2)$.

I got so fare, that with $t\geq 4$, $\frac{C}{n^t}\leq \frac{C}{n^2}\leq P(x^2\geq n)\leq \frac{Ex^4}{n^2}$. But how to find bound on $Ex^4$?

• Is this a homework? – MCH Dec 4 '11 at 18:20
• No, Its in the paper I am reading now, and I cannot get it. Thank you – user7488 Dec 4 '11 at 18:31
• I think this is a consequence of the variance being 1 and the mean being 0. Ignore all the other hypotheses; they're just serving to confuse you. – Peter Shor Dec 4 '11 at 18:55
• This question might be more suitable for Mathematics. – Kaveh Dec 4 '11 at 23:52
• @PeterShor sometimes the closing doesn't happen quickly enough :) – Suresh Venkat Dec 5 '11 at 18:03

## 1 Answer

Suppose to the contrary that $$P(x^2 \geq n_i) \geq C/n_i^2$$ for some infinite sequence $n_1$, $n_2$, $\ldots$, $n_i$, $\ldots$, with $P(x^2 > n_i) > 2P(x^2 > n_{i+1})$. Then we have $$Var(x) = \int_{x=0}^\infty x^2 d \mu.$$ But now, let's set $n_0 = 0$ and break this integral up into $$Var(x) = \sum_{j=0}^\infty \int_{n_j}^{n_{j+1}}x^2 d \mu.$$ We have$$\int_{n_j}^{n_{j+1}} x^2 d\mu \geq n_j^2 \int_{n_j}^{n_{j+1}} d\mu = n_j^2(P(x\geq n_j)-P(x \geq n_{j+1})) \geq C/2.$$ This shows that the integral diverges, contradicting the variance being 1.