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Math Help - Cumulative Frequency

  1. #1
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    Cumulative Frequency

    A random variable X is uniformly distributed between 0 and 1. Two independent observations are made,X1 and X2. Take (X1,X2 ) as a point on the lines X1 +X2 =Y in a cartesian plane. X1 +X2 =Y is triangular.
    (a) show that , for 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= Y^2


    (b) show that , for 1≤ Y≤ 2, P( X1 +X2 ≤ Y)=1- (2-Y)^2


    I know that f(x)=1 for 0≤ x≤ 1 since X is uniformly distributed. But how do I solve (a)?
    Can anyone show me the solution for (a) only so that I could solve (b) myself?


    Thanks a lot!
    Last edited by cyt91; July 6th 2010 at 02:14 PM.
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  2. #2
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    Quote Originally Posted by cyt91 View Post
    A random variable X is uniformly distributed between 0 and 1. Two independent observations are made,X1 and X2. Take (X1,X2 ) as a point on the lines X1 +X2 =Y in a cartesian plane. X1 +X2 =Y is triangular.
    (a) show that , for 0≤ Y≤ 1, P( X1 +X2 =Y)= Y2


    (b) show that , for 1≤ Y≤ 2, P( X1 +X2 =Y)=1- (2-Y)2


    I know that f(x)=1 for 0≤ x≤ 1 since X is uniformly distributed. But how do I solve (a)?
    Can anyone show me the solution for (a) only so that I could solve (b) myself?


    Thanks a lot!
    Is Y defined to be the random variable X1 + X2? What is Y2? P( X1 +X2 =Y) makes no sense, you cannot have the probability of a random variable being equal to another random variable.

    Please post the question exactly as it is writen from wherever you got it from.


    Note: Yes, the random variable Y = X1 + X2 does follow a triangular distribution. Its pdf is equal to y if 0 \leq y \leq 1, 2 - y if 1 \leq y \leq 2 and zero otherwise.
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  3. #3
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    I've edited the question. Sorry for the typo.
    How do we know that pdf of Y=X1+X2 is equal to y if 0≤ y≤ 1, 2 - y if 1≤ y≤ 2 and zero otherwise?
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    Quote Originally Posted by cyt91 View Post
    I've edited the question. Sorry for the typo.
    How do we know that pdf of Y=X1+X2 is equal to y if 0≤ y≤ 1, 2 - y if 1≤ y≤ 2 and zero otherwise?
    1. Y denotes a random variable. So P( X1 +X2 ≤ Y)=1- (2-Y)^2 still makes no sense to me for the reason previously given.

    2. Find and read the relevant example in here: http://www.dartmouth.edu/~chance/tea...k/Chapter7.pdf
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  5. #5
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    Well...that is the original question...
    How bout this question?

    The random variable X is uniformly distributed over the interval [0,9]. Find the probability density function of Y where Y=X^0.5.

    The solution given is as follows:

    f(x)= {1/9 , 0≤ x≤9
    0, otherwise

    Let g(y) be the pdf of Y.

    The random variable Y is given by Y=X^0.5

    G(y)= P( Y ≤ y)
    = P( X^0.5 ≤ y)
    = P( X ≤ y^2)
    = \int_{0}^{y^2}~f(x)~dx
    = \int_{0}^{y^2}~1/9~dx
    =(1/9)(y^2)

    They defined P( X^0.5 ≤ y). So why not P( X1 +X2 ≤ Y)?

    Thanks for the book btw.
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  6. #6
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    Quote Originally Posted by cyt91 View Post
    Well...that is the original question...
    How bout this question?

    The random variable X is uniformly distributed over the interval [0,9]. Find the probability density function of Y where Y=X^0.5.

    The solution given is as follows:

    f(x)= {1/9 , 0≤ x≤9
    0, otherwise

    Let g(y) be the pdf of Y.

    The random variable Y is given by Y=X^0.5

    G(y)= P( Y ≤ y)
    = P( X^0.5 ≤ y)
    = P( X ≤ y^2)
    = \int_{0}^{y^2}~f(x)~dx
    = \int_{0}^{y^2}~1/9~dx
    =(1/9)(y^2)

    They defined P( X^0.5 ≤ y). So why not P( X1 +X2 ≤ Y)?

    Thanks for the book btw.
    Do you understand the difference in meaning between Y and y ....?
    0≤ Y≤ 1, P( X1 +X2 ≤ Y)= Y^2
    is NOT the same as 0≤ y≤ 1, P( X1 +X2 ≤ y)= y^2.

    Correct notation is important. Incorrect use of notation leads to confusion.
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  7. #7
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    Well, that is the original question. It is given as 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= Y^2 (capital Y).
    Anyway,what is the difference between 0≤ Y≤ 1, P( X1 +X2 ≤ Y)= Y^2 and 0≤ y≤ 1, P( X1 +X2 ≤ y)= y^2.
    I'm doing self-study and I'm pretty confused with this chapter....
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