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Math Help - Density function solution check

  1. #1
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    Density function solution check

    Suppose Y has density function:

    f(y) = ky(1-y) when 0<=y<=1
    and
    f(y)= 0 elsewhere

    a)Find the value of k that makes f(y) a probability density function
    b)Find P(0.4<=Y<1)
    c)Find P(Y<=0.4|Y<=0.8)
    d)Find P(Y<0.4|Y<=0.8)

    sol.

    for part a I found k to be 6 (by making the integral = 1 on the interval 0,1 and solving for k

    for part b also used the integral between 0.4 and 1 and got 0.648

    c) d) is where I have the problem. I suppose both must have the same answer.

    Since this is conditional probability can we state that:

    P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

    so I find using integral P(0.4<=Y<=0.8)=0.544
    and P(y<=0.8) = P(0<=y<=0.8)=0.896

    then we divide 0.544/0.896= 0.607
    I feel that I am doing something wrong here, but can't see what.
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  2. #2
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    Quote Originally Posted by somestudent2 View Post
    Suppose Y has density function:

    f(y) = ky(1-y) when 0<=y<=1
    and
    f(y)= 0 elsewhere

    [snip]
    c)Find P(Y<=0.4|Y<=0.8)
    d)Find P(Y<0.4|Y<=0.8)

    [snip]

    c) d) is where I have the problem. I suppose both must have the same answer.

    Since this is conditional probability can we state that:

    P(Y<=0.4|Y<=0.8) = [P(y<=0.4) AND P(y<=0.8)]/P(y<=0.8) = P(0.4<=Y<=0.8)/P(y<=0.8)

    so I find using integral P(0.4<=Y<=0.8)=0.544
    and P(y<=0.8) = P(0<=y<=0.8)=0.896

    then we divide 0.544/0.896= 0.607
    I feel that I am doing something wrong here, but can't see what.
    The answer to (c) and (d) is


    \frac{\Pr(Y \leq 0.4 \text \, \text{and}\, Y \leq 0.8)}{\Pr(Y \leq 0.8)} = \frac{\Pr(Y \leq 0.4)}{\Pr(Y \leq 0.8)}  = \frac{\int_{0}^{0.4} ky(1-y) \, dy}{\int_{0}^{0.8} ky(1-y) \, dy} = \frac{\int_{0}^{0.4} y(1-y) \, dy}{\int_{0}^{0.8} y(1-y) \, dy} = .....
    Last edited by mr fantastic; April 5th 2008 at 12:50 AM.
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