anonymous
  • anonymous
Let Yn be the largest order statistic in a sample of size n from the uniform distribution on [0, t]. Show that Yn converges in probability to t, that is, that P (|Yn - t| >=e) approaches to 0 as n approaches to infinity.
Statistics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
There is so much context to this question that you should know and the typical math major hasn't committed to memory.
anonymous
  • anonymous
but all of these answers for the exam :(
anonymous
  • anonymous
What is a largest order statistic? What about this can you tell us?

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More answers

anonymous
  • anonymous
\[ \lim_{n\to \infty}\Pr(|Y_n-t|\geq \epsilon) = 0 \]
anonymous
  • anonymous
p, (1-p)*p/n
anonymous
  • anonymous
Do I have the formula right?
anonymous
  • anonymous
So is \(Y_n\) the largest value in the sample?
anonymous
  • anonymous
Start making sense of it. Don't have others do all the work for you.
anonymous
  • anonymous
\[\lim_{n \rightarrow 0} P(|Y _{n} - \theta|\ge \epsilon) \rightarrow0\]
anonymous
  • anonymous
\(n\) is approaching \(\infty\).
anonymous
  • anonymous
yes Yn is the largest samle in a sample
anonymous
  • anonymous
oh, yeah sorry n approaches to infinity
anonymous
  • anonymous
So is \(\theta\) the largest possible sample point? While \(Y_n\) is the largest in the sample.
anonymous
  • anonymous
i did some problems about distribution, z score, and others by myself, and got a right answers, but some of them are really very hard
anonymous
  • anonymous
i couldnt understand the "converges in probability to"
anonymous
  • anonymous
\[ \Pr(X\geq \epsilon) = 1-\Pr(X<\epsilon) \]
anonymous
  • anonymous
This property may or may not help us.
anonymous
  • anonymous
i think that may
anonymous
  • anonymous
It really doesn't matter what "converges in probability to" means since they gave us the definition, but I think the idea is the theoretical value you'd expect after infinite trials.
anonymous
  • anonymous
in any case thank u very much
anonymous
  • anonymous
You have an interval: \[ [Y_n,t] \]After \(n\) trials.
anonymous
  • anonymous
It's a uniform distribution, right?
anonymous
  • anonymous
Actually I think I know how we can do this.
anonymous
  • anonymous
\[(Y _{n}; \theta) \]
anonymous
  • anonymous
If we do \(n\) trials, and have \(Y_n\), then the probability that the next trial is above \(Y_n\) is\[ \frac{\theta -Y_n}{\theta} \]
anonymous
  • anonymous
yes right
anonymous
  • anonymous
The probability we surpass it by the \(m\)th trial is \[ \left(\frac{Y_n}{\theta}\right)^m\left(\frac{\theta -Y_n}{\theta}\right) \]
anonymous
  • anonymous
The probability we surpass it by the \(m\)th trial is \[ 1-\left(\frac{Y_n}{\theta}\right)^m \]
anonymous
  • anonymous
why the m-th trial?
anonymous
  • anonymous
We might not surpass it right after? I dunno.
anonymous
  • anonymous
All I can really say is that The gap between \(Y_n\) and \(\theta\) is narrowing, but I'm not sure how we'll exactly prove it.
anonymous
  • anonymous
The gap can't get any wider.
anonymous
  • anonymous
Wait a minute! I think I really got it this time!
anonymous
  • anonymous
you gave me some keys, i think i need more theory
anonymous
  • anonymous
Should be \(n\to \infty\)
anonymous
  • anonymous
but why n approaches to 0, not to infinity?
anonymous
  • anonymous
Because I copied and pasted your mistake.
anonymous
  • anonymous
oh, sorry :)
anonymous
  • anonymous
yes, it' right, but they also can be equal
anonymous
  • anonymous
Well it works for the \(|0|=0\) case as well.
anonymous
  • anonymous
ok, thank you very much for your time
goformit100
  • goformit100
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