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Math Help - Functions of a random variable question

  1. #1
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    Functions of a random variable question

    I have this question i cant nut out

    f(x)=3/32(4-x^2) when -2<x<2
    hence F(x)=0 when -2>x
    (3/8)x-(x^3)/32+1/2 when -2<x<2
    1 when 2<x


    let Y=(X^2)+1, find the CDF of Y

    Heres my (probably wrong method or something) working out:
    Fy(y)=P(Y<y)=P((x^2+1)<y)=P(X<sqrt(Y-1))=Fx(sqrt(y-1))

    therefore Fy(y)=3/8(sqrt(y-1))^(1/2)-((sqrt(y-1))^3)/32+1/2
    this isn't a proper CDF when i checked graphing it...

    Any help would be appreciated

    Edit: yeah, my mistake, i want the CDF of Y
    Last edited by skirk34; April 7th 2009 at 03:16 AM. Reason: clearing up question
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  2. #2
    MHF Contributor matheagle's Avatar
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    I assume you want the CDF of Y, not X.
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  3. #3
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    Quote Originally Posted by skirk34 View Post
    I have this question i cant nut out

    f(x)=3/32(4-x^2) I(-2,2)(x)
    hence F(x)=(3/8)x-(x^3)/32+1/2

    let Y=(X^2)+1, find the CDF of Y

    Heres my (probably wrong method or something) working out:
    Fy(y)=P(Y<y)=P((x^2+1)<y)=P(X<sqrt(Y-1))=Fx(sqrt(y-1))

    therefore Fy(y)=3/8(sqrt(y-1))^(1/2)-((sqrt(y-1))^3)/32+1/2
    this isn't a proper CDF when i checked graphing it...

    Any help would be appreciated

    Edit: yeah, my mistake, i want the CDF of Y
    What does the stuff in red mean?
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  4. #4
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    It is an indicator function (i think thats its name), it equals one when -2<x<2 and 0 otherwise
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  5. #5
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    Quote Originally Posted by skirk34 View Post
    It is an indicator function (i think thats its name), it equals one when -2<x<2 and 0 otherwise
    F_y (y) = \Pr(Y \leq y) = \Pr(X^2 + 1 \leq y) = \Pr(-\sqrt{y - 1} \leq X \leq \sqrt{y - 1}) for 1 \leq y \leq 5

    .
    .
    .

    = \frac{(13 - y) \sqrt{y - 1}}{16} for 1 \leq y \leq 5.

    Note that F_y (y) = 0 for y \leq 1 and F_y (y) = 1 for y \geq 5.
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