Applying Chernoff bound on normal distribution

[phonic]

Dear all,

I am trying to find out a good bound on the deveation of a normal distributed variable from its mean.

The noramly distributed variables X_t \sim N(\mu, \sigma^2), t= 1,2,...,n are iid. Applying the Chebyshev inequality on the mean of these n iid variables:

m_n = \frac{1}{n} \sum_{t=1}^n X_t

P(m_n - \mu \geq \epsilon ) = \frac{1}{2} P(e^{s(m_n - \mu)^2} \geq e^{s\epsilon^2} ) \leq \frac{1}{2}e^{-s \epsilon^2} E[e^{s(m_n-\mu)^2}]

The question is how to calculate this expectaion
E[e^{s(m_n-\mu)^2}]

Can anybody give some hints? Thanks a lot!

Since m_n \sim N(\mu, \frac{\sigma^2}{n} ),
E[(m_n-\mu)^2}] = \frac{\sigma^2}{n}. But
E[e^{s(m_n-\mu)^2}] seems not easy.

Phonic

 

[EnumaElish]

Are you treating s as a constant? Can you? Isn't it a r.v., e.g. s = s_n?

 

[phonic]

s can be considered as a constant number. Since the Markov inequality holds for any s.

Is there some bounds on the tail probability of a normal distribution?

 

[EnumaElish]

But in P(m_n - \mu \geq \epsilon ) = \frac{1}{2} P(e^{s(m_n - \mu)^2} \geq e^{s\epsilon^2} ) \leq \frac{1}{2}e^{-s \epsilon^2} E[e^{s(m_n-\mu)^2}], you have moved s out of the E[] in e^{-s \epsilon^2} but then left it inside the E[] in E[e^{s(m_n-\mu)^2}], is that legit? More to the point, can you also take it out of the latter, and if you can, would that make the job easier?