Thread: Problem 1.9
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Old 09-17-2016, 11:43 AM
svend svend is offline
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Join Date: Sep 2016
Posts: 2
Default Re: Problem 1.9

Here's my take on Problem 1.9, part(b), which is following the same lines as the description of MaciekLeks above.

We have:

P(u \geq \alpha ) & = P(\frac{1}{N}{}\sum_n x_n \geq \alpha)  \\
 & = P(\sum_n x_n \geq N \alpha) \\
 & = P(e^{s \sum_n x_n} \geq e^{s N \alpha})

Since e^{s t} is monotonically increasing in t.

Also, e^{s t} is non negative for all t, implying Markov inequality holds:

P(e^{s \sum_n x_n} \geq e^{s N \alpha}) & \leq \frac{E(e^{s \sum_n x_n})}{e^{s N \alpha}} \\
& = \frac{E(\prod_n e^{s x_n})}{e^{s N \alpha}} \\
& = \frac{\prod_n E(e^{s x_n})}{e^{s N \alpha}}

The last line being true since [math]x_n[\math] are independent.

From there it directly follows that

P(u \geq \alpha ) \leq U(s)^N e^{-s N \alpha}
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