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Applying Chernoff bound on normal distribution

  1. Dec 25, 2006 #1
    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.

    Last edited: Dec 25, 2006
  2. jcsd
  3. Dec 28, 2006 #2


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    Are you treating s as a constant? Can you? Isn't it a r.v., e.g. s = sn?
  4. Jan 8, 2007 #3
    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?
  5. Jan 8, 2007 #4


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    But in [itex]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}] [/itex], you have moved s out of the E[] in [itex]e^{-s \epsilon^2} [/itex] but then left it inside the E[] in [itex]E[e^{s(m_n-\mu)^2}][/itex], 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?
    Last edited: Jan 8, 2007
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