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Thread: 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 $\displaystyle X_{(n)} = \text{max} \, ( X_1, \, X_2, \, ...... X_n)$.

    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)
    Last edited by mr fantastic; Jul 9th 2008 at 05:11 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???
    $\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
    Last edited by CaptainBlack; Jul 9th 2008 at 05:36 AM.
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    Quote Originally Posted by CaptainBlack View Post
    $\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 .......
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    $\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 ?
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    Quote Originally Posted by Moo View Post
    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
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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 $\displaystyle X$'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
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    Quote Originally Posted by CaptainBlack View Post
    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 <> $\displaystyle \vee$
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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 $\displaystyle \vee$ 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
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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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