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Math Help - Markov inequality

  1. #1
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    Markov inequality

    If X is a Poisson random variable with mean \lambda show that for i < \lambda,

    P(X \leq i) \leq e^{-\lambda}(\frac{e\lambda}{i})^i

    I've tried all the inequalities, but I keep getting stuck. Any push in the right direction would be greatly appreciated.
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    Quote Originally Posted by Anonymous1 View Post
    If X is a Poisson random variable with mean \lambda show that for i < \lambda,

    P(X \leq i) \leq e^{-\lambda}(\frac{e\lambda}{i})^i

    I've tried all the inequalities, but I keep getting stuck. Any push in the right direction would be greatly appreciated.
    Since P(X \leq i) = \sum_{j=0}^i \frac{1}{j!} j^i e^{-\lambda},
    the desired inequality is equivalent to
    \sum_{j=0}^i \frac{1}{j!} j^i \leq (\frac{e \lambda}{i})^i.

    To that end, note that
    (\frac{e \lambda}{i})^i = (\frac{\lambda}{i})^i e^i
    =(\frac{\lambda}{i})^i \sum_{j=0}^{\infty}\frac{i^j}{j!}
    > (\frac{\lambda}{i})^i \sum_{j=0}^i \frac{i^j}{j!} ...(truncating)
    =\sum_{j=0}^i \frac{\lambda^i}{j! i^{i-j}}
    \geq \sum_{j=0}^i \frac{\lambda^i}{j! \lambda^{i-j}} ... since i \leq \lambda and i-j \geq 0
    =\sum_{j=0}^i \frac{\lambda^j}{j!}
    and we are done.
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