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Math Help - PDF Problem

  1. #1
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    PDF Problem

    Heres a problem im stuck on.

    "Random variable X has the probability density function f(x) = 1/3 for -1 < x < 2 and zero elsewhere. Find the probability density function of

    Would it be
    for 0<y<1 , 0 elsewhere
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  2. #2
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    Quote Originally Posted by Xan21 View Post
    Heres a problem im stuck on.

    "Random variable X has the probability density function f(x) = 1/3 for -1 < x < 2 and zero elsewhere. Find the probability density function of

    Would it be
    for 0<y<1 , 0 elsewhere
    Not quite. According to your pdf, Pr(Y > 1) = 0 ...... Is that correct ....?

    If -1 < x < 2 it's clear that 0 < x^2 < 4 => 0 < y < 4. Therefore f(y) = 0 for y > 4 or y < 0.

    So you have to find the pdf for Y when 1 \leq y < 4. For such values of y, the cumulative density function is given by F(y) = \Pr(Y < y) = \Pr(X^2 < y) = \Pr(-1 < X < \sqrt{y}).
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  3. #3
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    Sorry i don't understand this

    Hi,
    I have a question why for 0<y<4.

    Why the Pr(X^2 < y)= Pr(-1<X<y^1/2) and not = Pr(-(y^1/2)<X<y^1/2) ?

    Thanks.
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  4. #4
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    Quote Originally Posted by embria View Post
    Hi,
    I have a question why for 0<y<4.

    Why the Pr(X^2 < y)= Pr(-1<X<y^1/2) and not = Pr(-(y^1/2)<X<y^1/2) ?

    Thanks.
    Yes it is \Pr( - \sqrt{y} \leq X \leq \sqrt{y}). But note:

    1. \Pr( - \sqrt{y} \leq X \leq \sqrt{y}) = \Pr( - \sqrt{y} \leq X \leq -1) + \Pr( -1 \leq X \leq \sqrt{y}) .

    2. It's given that -1 < x < 2. So the pdf for X is zero if x < -1. So \Pr( - \sqrt{y} \leq X \leq -1) = 0 when 1 \leq y < 4.

    3. So \Pr( -1 \leq X \leq \sqrt{y}) is the only calculation required when 1 \leq y < 4.
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