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

[tex]m_n = \frac{1}{n} \sum_{t=1}^n X_t[/tex]

[tex]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}] [/tex]

The question is how to calculate this expectaion

[tex] E[e^{s(m_n-\mu)^2}] [/tex]

Can anybody give some hints? Thanks a lot!

Since [tex]m_n \sim N(\mu, \frac{\sigma^2}{n} ) [/tex],

[tex] E[(m_n-\mu)^2}] = \frac{\sigma^2}{n} [/tex]. But

[tex] E[e^{s(m_n-\mu)^2}] [/tex] seems not easy.

Phonic

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

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

[tex]m_n = \frac{1}{n} \sum_{t=1}^n X_t[/tex]

[tex]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}] [/tex]

The question is how to calculate this expectaion

[tex] E[e^{s(m_n-\mu)^2}] [/tex]

Can anybody give some hints? Thanks a lot!

Since [tex]m_n \sim N(\mu, \frac{\sigma^2}{n} ) [/tex],

[tex] E[(m_n-\mu)^2}] = \frac{\sigma^2}{n} [/tex]. But

[tex] E[e^{s(m_n-\mu)^2}] [/tex] seems not easy.

Phonic

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