Thread: Problem 1.9
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Old 11-09-2018, 06:02 PM
Ulyssesyang Ulyssesyang is offline
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Default Re: Problem 1.9

Quote:
Originally Posted by svend View Post
Here's my take on Problem 1.9, part(b), which is following the same lines as the description of MaciekLeks above.

We have:

\begin{split}
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})
\end{split}

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

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

\begin{split}
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}}
\end{split}

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}
I just think you should note that there are two expectations, one is based on e^su_n, while other is based on u_n. Of course, you can refer the law of unconscious statisticians to prove that both are same.
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