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Math Help - cumulative distribution

  1. #1
    tmd
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    cumulative distribution

    Hi ^^

    This question asks me to compute the cdf of Y from this pdf :
    f(y) { y/25 for 0 =< y < 5 ; (2/5) - (y/25) for 5 =< y =< 10 ; 0 for otherwise }

    After I integrate the function (because F'(x) = f(x)), I got:

    F(y) { 0 for y<0 ; y^2/50 for 0 =< y < 5 ; (2y/5) - y^2/50 for 5=<y=<10 ; 1 for y => 10}
    this cdf is weird because if i plug 10 in, like (2(10)/5) - (10)^2/50) i got 2 instead of 1 . I don't understand, would someone like to explain to me thank you very much ^^
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  2. #2
    MHF Contributor matheagle's Avatar
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    You need to go back to the definition of the cdf, which is P(Y\le y)

    If y<0, then F(y)=0.

    If 0<y<5, then you are correct, P(Y\le y)=\int_{-\infty}^y f(t)dt=\int_0^y tdt/25=y^2/50

    HOWEVER, and it's good you check your work...

    If 5<y<10, then P(Y\le y)=\int_{-\infty}^y f(t)dt=\int_0^5 tdt/25 +\int_5^y \biggl({2\over 5}-{t\over 25}\biggr) dt =.5 + whatever.

    And, for y>10, you better get 1, which I did, since this is a valid denisty.
    That was the first thing I did, to see if it integrated to one.
    Too often students write down incorrect 'densities' and I spend forever on the problem.
    Last edited by matheagle; July 16th 2009 at 09:46 PM.
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  3. #3
    Master Of Puppets
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    From what I can make out from the question...

    \int_0^5 \frac{y}{25} dy = \frac{5^2}{50} - \frac{0^2}{50}

    and

    \int_5^{10} \frac{2}{5}-\frac{y}{25} dy = [\frac{2(10)}{5}-\frac{10^2}{50}] - [\frac{2(5)}{5}-\frac{5^2}{50}]

    should sum to 1.
    Last edited by pickslides; July 15th 2009 at 09:32 PM. Reason: Typo
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  4. #4
    MHF Contributor matheagle's Avatar
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    Mr Pickles, it is a valid density, but he's misunderstanding what a CDF is.
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  5. #5
    tmd
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    i see my mistake thank you very much ^^
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