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Finding the distribution

  1. Aug 22, 2009 #1

    I'm not too sure if this is the correct location for my post, but it's the best fit I can see!

    The cdf of the continuous random variable X is
    [tex]F(x)=\left\{\begin{array}{cc}0&\mbox{ if }x< 0\\
    {1\over 4} x^2 & \mbox{ if } 0 \leq x \leq 2\\
    1 &\mbox{ if } x >2\end{array}\right.[/tex]

    Q1-Obtain the pdf of X
    Q2-If Y = 2 - X, derive the pdf of the random variable Y

    A1-I think the cdf is given by [tex]f(x) = F'(x)=\left\{\begin{array}{cc}{1\over 2}x &\mbox{ if } 0 \leq x \leq 2 \\
    0 &\mbox{ elsewhere } \end{array}\right.[/tex]
    Is that correct?

    A2-For the pdf of Y: [tex]G(Y) = P(Y \leq y) = P(2 - x \leq y) = P(x \geq 2-y)
    [/tex] but I'm not sure how to proceed??
  2. jcsd
  3. Aug 22, 2009 #2


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    A1 - you're right.

    A2 [tex]G(y) = P(Y \leq y) = P(2 - X \leq y) = P(X \geq 2-y) = 1-P(X < 2-y)=1-F(2-y)[/tex]
    G'(y)=F'(2-y) for the pdf
  4. Aug 22, 2009 #3
    Q1-how would I evaluate:
    [tex]G(y) = 1-F(2-y)\
    \mbox{to give the cdf? Do I just substitute 2-y into }{1 \over 4} x^2 \mbox{ to give }
    G(Y) = 1 - {1 \over 4} (2-y)^2 = 4y - {1 \over 4}y^2\ for \ 0 \leq y \leq 2[/tex]
    [tex] \mbox{for the cdf of Y? and so for the pdf: } 4-{1 \over 2}y[/tex]

    Q2-Why are you able to make this step: [tex]1-P(X < 2-y)=1-F(2-y)[/tex]
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