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Math Help - puzzling problem! please help

  1. #1
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    puzzling problem! please help

    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???
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  2. #2
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    Quote Originally Posted by math beginner View Post
    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 X_{(n)} = \text{max} \, ( X_1, \, X_2, \, ...... X_n).

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


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


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

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

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


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

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


    5. I get n = 29 (whew! That's a relief)
    Last edited by mr fantastic; July 9th 2008 at 06:11 AM. Reason: Streamlined the reply a bit. Confirmed the answer n = 29.
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  3. #3
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    Quote Originally Posted by math beginner View Post
    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???
    Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)

    ............... <br />
=1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]<br />

    but:

     <br />
Pr(X_i <0.9Q)=0.9<br />

    Hence:

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

    RonL
    Last edited by CaptainBlack; July 9th 2008 at 06:36 AM.
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    Quote Originally Posted by CaptainBlack View Post
    Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)

    ............... <br />
=1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]<br />

    but:

     <br />
Pr(X_n <0.9Q)=0.9<br />

    Hence:

    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 .......
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Pr(|max(X_1, .. X_n)-Q| \le 0.1Q)= Pr(X_1\ge 0.9Q) \vee .. \vee Pr(X_n\ge 0.9Q)

    ............... <br />
=1-[Pr(X_1 < 0.9Q) \times .. \times Pr(X_n <0.9Q)]<br />

    but:

     <br />
Pr(X_i <0.9Q)=0.9<br />

    Hence:

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

    RonL
    Pardon my question, but why is it \vee ? the max part means that we will take only one in consideration, whereas \vee is not necessarily exclusive and can contain 2, isn't it ?
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    Quote Originally Posted by Moo View Post
    Pardon my question, but why is it \vee ? the max part means that we will take only one in consideration, whereas \vee is not necessarily exclusive and can contain 2, isn't it ?
    \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
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    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... ?
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    Quote Originally Posted by Moo View Post
    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 X's are \sim U(0,Q)

    so:

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

    for 0 \le A \le Q

    (all the X's are \le Q)

    RonL
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  9. #9
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    Quote Originally Posted by CaptainBlack View Post
    The X's are \sim U(0,Q)

    so:

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

    for 0 \le A \le Q

    (all the X's are \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 \vee because it doesn't agree with the "maximum" thing...

    uniqueness of maximum <> \vee
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    Quote Originally Posted by Moo View Post
    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 \vee because it doesn't agree with the "maximum" thing...

    uniqueness of maximum <> \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
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    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 !
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  12. #12
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    Thumbs up

    Brilliant! I understand perfectly now. Thanks so much!
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