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Math Help - Expected value derivation question

  1. #1
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    Expected value derivation question

    Hi,

    T is a function, T = \frac{\sum X_n}{\sqrt{n}}

    where, X are i.i.d Bernoulli r.v

    The end point of the derivation is,

    Prob(T \ge z) \le e^{-az} \left(\frac{e^{\frac{a}{\sqrt{n}}}+e^{\frac{-a}{\sqrt{n}}}}{2}\right)^n

    I am given,

    Prob(T \ge z) \le e^{-az} (Expectation[e^{aT}]) , so

    Prob(T \ge z) \le e^{-az} (Expectation[e^{\frac{a.X_i}{\sqrt{n}}}])

    From a previous step, I know that Prob(X = 1) = 0.5, and Prob(X = -1) = 0.5

    So the Expected values are

    e^{\frac{a.0.5}{\sqrt{n}}} for X = 1, and e^{\frac{a.-0.5}{\sqrt{n}}} for X=-1

    So the Expectation[e^{aT}] =
    \frac{e^{\frac{a}{\sqrt{n}}} + e^{\frac{-a}{\sqrt{n}}}}{2}

    At this point it looks like the answer should be

    e^{-az}\left(\frac{e^{\frac{a}{\sqrt{n}}} + e^{\frac{-a}{\sqrt{n}}}}{2}\right)

    I can't see where the nth power around the brackets should come from.

    ie. e^{-az} \left(\frac{e^{\frac{a}{\sqrt{n}}}+e^{\frac{-a}{\sqrt{n}}}}{2}\right)^n

    Any ideas?
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by notgoodatmath View Post
    Hi,
    I am given,

    Prob(T \ge z) \le e^{-az} (Expectation[e^{aT}]) , so

    Prob(T \ge z) \le e^{-az} (Expectation[e^{\frac{a.X_i}{\sqrt{n}}}])
    You're close! Except you missed one thing - observe the quoted text: when you plugged in your expression for what T equals, you inputted \frac{X_i}{\sqrt{n}} instead of \frac{\sum X_n}{\sqrt{n}}. That would give you the result for one of such random variable, not but n independently identically distributed of such random variables! Thus, accounting for the one missing component, we should instead have:

    Prob(T \ge z) \le e^{-az} (E[e^{aT}])
    = Prob(T \ge z) \le e^{-az} (E[e^{a\frac{\sum X_n}{\sqrt{n}}}])
    = Prob(T \ge z) \le e^{-az} (E[e^{a\frac{X_1 + X_2 + ... + X_n}{\sqrt{n}}}])
    = Prob(T \ge z) \le e^{-az} (E[e^{a\frac{X_1}{\sqrt{n}}} e^{a\frac{X_2}{\sqrt{n}}} ... e^{a\frac{X_n}{\sqrt{n}}}])
    = Prob(T \ge z) \le e^{-az} (E[(e^{a\frac{X_1}{\sqrt{n}}})^n])

    And there is your power of n!
    Last edited by Last_Singularity; December 27th 2008 at 09:45 PM.
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  3. #3
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    thanks for finding my n!

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