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Math Help - transform using cdf

  1. #1
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    transform using cdf

    hello I need help
    I spent 2 hours without getting the right solution
    =
    [1-[1-P(X1<=y1)[1-P(X2<=y1)][P(X1<=y2)P(X2<=y2)]
    but the book gives a different aswer from what I got
    I think I cannot assume that Y1 and Y2 are independant right?

    Thank you
    Last edited by qwerty321; January 3rd 2010 at 01:21 AM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is the joint distribution of two order statistics from a sample of two.
    You can find that in many books.
    I think it's in Wackerly, I used to use Hogg and Craig.
    since you only have two observations...

    f( x_{(1)},x_{(2)})={2!\over 1!1!}f(x_{(1)})f(x_{(2)})I(x_{(1)}<x_{(2)})

    There are 2! ways this can happen.
    Either Y1=X1 or Y1=X2.
    The reason it's still a valid density is that we now have a restriction on the space.
    The order statistics are dependent.
    Last edited by matheagle; January 1st 2010 at 04:19 PM.
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  3. #3
    Moo
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    Hello,

    @ qwerty321 :
    No they're not independent, so your working is unfortunately wrong.
    First consider P(Y_1\geq y_1,Y_2\leq y_2) (since it's continuous random variables, it doesn't matter whether is \geq or >)

    If the min is superior to a y1 and the max is inferior to y2, it simply means that X_1,X_2 \in [y_1,y_2]

    So P(Y_1\geq y_1,Y_2\leq y_2)=P(X_1\in[y_1,y_2],X_2\in[y_1,y_2])

    And then you can use the independence between X1 and X2 to calculate this probability.

    Then P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)

    Hence P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1)

    the missing part is P(Y_2\leq y_2)=P(\max(X_1,X_2)\leq y_2)=P(X_1\leq y_2,X_2\leq y_2)

    can you finish it ?



    You have to visualize that if the min of two values is > to something, then the two values are > to something. If the max of two values is < to something, then the two values are < to something.
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  4. #4
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    matheagle i cannot understand how you gor your formula
    Moo I do not understand :

    first why you took this case:
    "
    First consider P(Y_1\geq y_1,Y_2\leq y_2) (since it's continuous random variables, it doesn't matter whether is \geq or >)"

    second:
    then how did you pass from

    "
    P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)

    to this P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1) "

    and for the missing part i guess you mean i could use the independance thing for X1 and X2

    thank you
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  5. #5
    Moo
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    Quote Originally Posted by qwerty321 View Post
    Moo I do not understand :

    first why you took this case:
    "
    First consider P(Y_1\geq y_1,Y_2\leq y_2) (since it's continuous random variables, it doesn't matter whether is \geq or >)"
    Because it's easy to deal with min if it's > to something, and to deal with max if it's < to something (see my last sentence in the first post).

    second:
    then how did you pass from

    "
    P(Y_2\leq y_2)=P(Y_2\leq y_2,Y_1\leq y_1)+P(Y_2\leq y_2,Y_1\geq y_1)

    to this P(Y_2\leq y_2,Y_1\leq y_1)=P(Y_2\leq y_2)-P(Y_2\leq y_2,Y_1\geq y_1) "
    erm... a=b+c ===> b=a-c ?

    and for the missing part i guess you mean i could use the independance thing for X1 and X2

    thank you
    Of course ^^
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  6. #6
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    thank you it finally worked
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