Question

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.

Answer 1

There is so much context to this question that you should know and the typical math major hasn't committed to memory.

Answer 2

but all of these answers for the exam :(

Answer 3

What is a largest order statistic?

What about this can you tell us?

Answer 4

\[
\lim_{n\to \infty}\Pr(|Y_n-t|\geq \epsilon) = 0
\]

Answer 5

p, (1-p)*p/n

Answer 6

Do I have the formula right?

Answer 7

So is \(Y_n\) the largest value in the sample?

Answer 8

Start making sense of it. Don't have others do all the work for you.

Answer 9

\[\lim_{n \rightarrow 0} P(|Y _{n} - \theta|\ge \epsilon) \rightarrow0\]

Answer 10

\(n\) is approaching \(\infty\).

Answer 11

yes Yn is the largest samle in a sample

Answer 12

oh, yeah sorry n approaches to infinity

Answer 13

So is \(\theta\) the largest possible sample point? While \(Y_n\) is the largest in the sample.

Answer 14

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

Answer 15

i couldnt understand the "converges in probability to"

Answer 16

\[
\Pr(X\geq \epsilon) = 1-\Pr(X<\epsilon)
\]

Answer 17

This property may or may not help us.

Answer 18

i think that may

Answer 19

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.

Answer 20

in any case thank u very much

Answer 21

You have an interval: \[
[Y_n,t]
\]After \(n\) trials.

Answer 22

It's a uniform distribution, right?

Answer 23

Actually I think I know how we can do this.

Answer 24

\[(Y _{n}; \theta) \]

Answer 25

If we do \(n\) trials, and have \(Y_n\), then the probability that the next trial is above \(Y_n\) is\[
\frac{\theta -Y_n}{\theta}
\]

Answer 26

yes right

Answer 27

The probability we surpass it by the \(m\)th trial is \[
\left(\frac{Y_n}{\theta}\right)^m\left(\frac{\theta -Y_n}{\theta}\right)
\]

Answer 28

The probability we surpass it by the \(m\)th trial is \[
1-\left(\frac{Y_n}{\theta}\right)^m
\]

Answer 29

why the m-th trial?

Answer 30

We might not surpass it right after? I dunno.

Answer 31

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.

Answer 32

The gap can't get any wider.

Answer 33

Wait a minute! I think I really got it this time!

Answer 34

you gave me some keys, i think i need more theory

Answer 35

Should be \(n\to \infty\)

Answer 36

but why n approaches to 0, not to infinity?

Answer 37

Because I copied and pasted your mistake.

Answer 38

oh, sorry :)

Answer 39

yes, it' right, but they also can be equal

Answer 40

Well it works for the \(|0|=0\) case as well.

Answer 41

ok, thank you very much for your time

Answer 42

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