<p>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 $P(x^2 &gt; n_i) &gt; 2P(x^2 &gt; n_{i+1})$. 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. </p>

CC BY-SA 3.0
2011-12-04T19:07:54.033Z
2011-12-04T19:56:15.653Z
2011-12-04T19:56:15.653Z
Peter Shor 
<p>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 $P(x^2\geq n)\leq C/(n^2)$.</p>

<p>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 $Ex^4$?</p>

CC BY-SA 3.0
2011-12-04T18:20:08.603Z
Is this a homework?
Mahdi Cheraghchi
CC BY-SA 3.0
2011-12-04T18:55:11.503Z
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 
CC BY-SA 3.0
2011-12-04T23:52:56.517Z
This question might be more suitable for [math.se].
Kaveh
CC BY-SA 3.0
2011-12-05T12:07:51.803Z
For the right choice of $C$, it is indeed Chebychev's inequality, but I don't see how you can use Chebychev to show $P(x^2 \geq n) \leq C/n^2$ asymptotically for arbitrary $C$.
Peter Shor 
CC BY-SA 3.0
2011-12-05T12:09:14.390Z
@Kaveh: it probably would have been more suitable for math. I'm surprised it didn't get closed for being too elementary, but I suppose questions that are too elementary in the field of mathematics are not too elementary in the field of computer science, and vice versa.
Peter Shor 
CC BY-SA 3.0
2011-12-05T18:03:33.910Z
@PeterShor sometimes the closing doesn't happen quickly enough :)
Suresh Venkat
CC BY-SA 3.0
2011-12-05T18:41:29.480Z
@David: You do not need $t\geq 4$ in the proof. See my comment above---it's a completely unnecessary hypothesis. All you need is variance 1 and mean 0. I'm also assuming that what you need is that $\forall C \,\exists n_0$ such that $P(x^2 \geq n) \leq C/n^2$ for $n \geq n_0$. If I've misinterpreted the question, you should ask the correct question on [math SE](http://math.stackexchange.com/), where it would be more suitable.
Peter Shor 
<pr.probability>
is it by Chebychev's inequality?
David
# is it by Chebychev's inequality?

(Asked by: username_0 on 2011-12-04)

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 $Ex^4$?

**Tags:** pr.probability
> **Comments:**
> **username_1**: Is this a homework?  
> **username_2**: 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.  
> **username_3**: This question might be more suitable for [math.se].  
> **username_2**: For the right choice of $C$, it is indeed Chebychev's inequality, but I don't see how you can use Chebychev to show $P(x^2 \geq n) \leq C/n^2$ asymptotically for arbitrary $C$.  
> **username_2**: @Kaveh: it probably would have been more suitable for math. I'm surprised it didn't get closed for being too elementary, but I suppose questions that are too elementary in the field of mathematics are not too elementary in the field of computer science, and vice versa.  
> **username_4**: @PeterShor sometimes the closing doesn't happen quickly enough :)  
> **username_2**: @David: You do not need $t\geq 4$ in the proof. See my comment above---it's a completely unnecessary hypothesis. All you need is variance 1 and mean 0. I'm also assuming that what you need is that $\forall C \,\exists n_0$ such that $P(x^2 \geq n) \leq C/n^2$ for $n \geq n_0$. If I've misinterpreted the question, you should ask the correct question on [math SE](http://math.stackexchange.com/), where it would be more suitable.  

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## 1 answers

### Answer by username_2 on 2011-12-04. Score: 8

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.


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