Suppose that a random sample X1,....Xn is to be taken from a uniform distribution on the interval [0,Q] and that Q is unknown. How large a random sample must be taken in order that
Pr(|max{X1,...,Xn} - Q| < 0.1Q) > 0.95
for all Q?
(From Degroot and Schervish 7.1 Exercise 1)
How do I get the Ans: 29???

2. Originally Posted by math beginner
Suppose that a random sample X1,....Xn is to be taken from a uniform distribution on the interval [0,Q] and that Q is unknown. How large a random sample must be taken in order that
Pr(|max{X1,...,Xn} - Q| < 0.1Q) > 0.95
for all Q?
(From Degroot and Schervish 7.1 Exercise 1)
How do I get the Ans: 29???
Here are some suggestions but you'll need to work through the details:

Define the random variable $\displaystyle X_{(n)} = \text{max} \, ( X_1, \, X_2, \, ...... X_n)$.

1. Note that $\displaystyle \Pr( | X_{(n)} - Q | \leq 0.1Q) = \Pr( Q - X_{(n)} \leq 0.1Q) = \Pr( - X_{(n)} \leq - 0.9Q)$ $\displaystyle = \Pr(X_{(n)} \geq 0.9Q)$.

2. You need to find the minimum integer value of n that solves $\displaystyle \int_{0.9Q}^{Q} g(x) \, dx \geq 0.95$ where $\displaystyle g(x)$ is the pdf of $\displaystyle X_{(n)}$.

3. Standard result (from order statistics): Let $\displaystyle X_1, \, X_2, \, ...... X_n$ denote continuous random variables with cdf F(x) and pdf f(x).

Define the random variable $\displaystyle X_{(n)} = \text{max} \, ( X_1, \, X_2, \, ...... X_n)$.

Then the pdf of $\displaystyle X_{(n)}$ is given by $\displaystyle g(x) = n [F(x)]^{n-1} f(x)$.

4. In your problem, $\displaystyle f(x) = \frac{1}{Q}$ for $\displaystyle 0 \leq x \leq Q$ and zero otherwise.

Therefore the pdf of $\displaystyle X_{(n)}$ is given by $\displaystyle g(x) = \frac{n}{Q^n} \, x^{n-1}$.

5. I get n = 29 (whew! That's a relief)

3. Originally Posted by math beginner
Suppose that a random sample X1,....Xn is to be taken from a uniform distribution on the interval [0,Q] and that Q is unknown. How large a random sample must be taken in order that
Pr(|max{X1,...,Xn} - Q| < 0.1Q) > 0.95
for all Q?
(From Degroot and Schervish 7.1 Exercise 1)
How do I get the Ans: 29???
$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)$

............... $\displaystyle =1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]$

but:

$\displaystyle Pr(X_i <0.9Q)=0.9$

Hence:

$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= 1- 0.9^n$

RonL

4. Originally Posted by CaptainBlack
$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)$

............... $\displaystyle =1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]$

but:

$\displaystyle Pr(X_n <0.9Q)=0.9$

Hence:

$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= 1- 0.9^n$
*Ahem* Of course, if you want to do it the easy way .......

5. Originally Posted by CaptainBlack
$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)$

............... $\displaystyle =1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]$

but:

$\displaystyle Pr(X_i <0.9Q)=0.9$

Hence:

$\displaystyle Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= 1- 0.9^n$

RonL
Pardon my question, but why is it $\displaystyle \vee$ ? the max part means that we will take only one in consideration, whereas $\displaystyle \vee$ is not necessarily exclusive and can contain 2, isn't it ?

6. Originally Posted by Moo
Pardon my question, but why is it $\displaystyle \vee$ ? the max part means that we will take only one in consideration, whereas $\displaystyle \vee$ is not necessarily exclusive and can contain 2, isn't it ?
$\displaystyle \vee$ - logical "or"

We want the probability that the maximum of the difference between the X's and Q to be less than 0.1Q, so we want the probability that one or more of them is greater than 0.9Q.

RonL

7. Originally Posted by CaptainBlack
We want the probability that the maximum of the difference between the X's and Q to be less than 0.1Q, so we want the probability that one or more of them is greater than 0.9Q.

RonL
Is it ?
We want the difference between the maximum of the X's and Q to be less than 0.1Q ('tis the same as the bold sentence). This is how I read max{...}-Q, is it false ?
And because the maximum is unique, you can't have more than 1 X satisfying the condition... ?

8. Originally Posted by Moo
Is it ?
We want the difference between the maximum of the X's and Q to be less than 0.1Q. This is how I read max{...}-Q, is it false ?
The $\displaystyle X$'s are $\displaystyle \sim U(0,Q)$

so:

$\displaystyle \max_{i=1,..,n}(|X_i-Q|)<A \Leftrightarrow |\max_{i=1,..,n}(X_i) -Q| <A$

for $\displaystyle 0 \le A \le Q$

(all the $\displaystyle X$'s are $\displaystyle \le Q$)

RonL

9. Originally Posted by CaptainBlack
The $\displaystyle X$'s are $\displaystyle \sim U(0,Q)$

so:

$\displaystyle \max_{i=1,..,n}(|X_i-Q|)<A \Leftrightarrow |\max_{i=1,..,n}(X_i) -Q| <A$

for $\displaystyle 0 \le A \le Q$

(all the $\displaystyle X$'s are $\displaystyle \le Q$)

RonL
Makes the problem clearer ! thanks

But I don't really agree...
If we want the difference with Q to be maximum, then we should take the minimum of the X's, isn't it ???

And I still don't understand your use of $\displaystyle \vee$ because it doesn't agree with the "maximum" thing...

uniqueness of maximum <> $\displaystyle \vee$

10. Originally Posted by Moo
Makes the problem clearer ! thanks

But I don't really agree...
If we want the difference with Q to be maximum, then we should take the minimum of the X's, isn't it ???

And I still don't understand your use of $\displaystyle \vee$ because it doesn't agree with the "maximum" thing...

uniqueness of maximum <> $\displaystyle \vee$
If we want Q minus the maximum of X to be less or equal than 0.1Q, then at least one X must be greater than or equal 0.9Q.

Which means (X_1>=0.9Q) or (X_2 >=0.9Q) or .. or (X_n>=0.9Q)

RonL

11. Originally Posted by CaptainBlack
If we want Q minus the maximum of X to be less or equal than 0.1Q, then at least one X must be greater than or equal 0.9Q.

Which means (X_1>=0.9Q) or (X_2 >=0.9Q) or .. or (X_n>=0.9Q)

RonL
Ok, I see now : I was really messing up things... Thanks a lot !

12. Brilliant! I understand perfectly now. Thanks so much!