Probability inequality for the sum of independent normal random variables

1. Jul 24, 2007

phonic

Dear all,

I wonder wheather there exsits a probability inequality for the sum of independent normal random variables ($X_i$ are i.i.d. normal random varianble with mean $\mu$ and variance $\sigma^2$):
$P\left(\frac{1}{n}\sum_{i=1}^n X_i - \mu> \epsilon\right)\leq f(\epsilon, \sigma^2,n) \right).$

We know that Bernstein inequality is for the sum of bounded random variables:
$P\left(\frac{1}{n}\sum_{i=1}^n X_i -\mu > \epsilon\right)\leq \exp\left(-\frac{n\epsilon^2}{2\sigma^2+ 2c\epsilon/3} \right).$

I wonder whether there is some similar inequality for normal variables.

Thanks!

Phonic

2. Jul 24, 2007

EnumaElish

There is an exact equality; it follows from Σ X/n ~ N(μ, σ^2/n).

Last edited: Jul 24, 2007
3. Jul 26, 2007

phonic

Tanks for your reply. Then the problem is to bound the tail probability of this normal variable. I know one inequality is (R. D. Gordon, The Annals of Mathematical Statistics, 1941(12), pp 364-366)
$P(z \geq x) = \int_x^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}z^2} dz \leq \frac{1}{x} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\mbox{\hspace{1cm}for } x>0,$
where z is a standard normal variable.

The problem of this inequality is that the function $\frac{1}{x} e^{-\frac{1}{2}x^2}$ is nor invertible (no analytical inverse function). Do you know some other bound for tail probability of a normal variable? Thanks a lot!

4. Jul 26, 2007

EnumaElish

Haven't you changed the upper bound function? Can the new function not have σ^2 or n as arguments? If it can, then you have an exact statement of the tail probability.