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Thread: single varibale cummulative dist. func.

  1. #1
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    single varibale cummulative dist. func.

    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    \frac{3}{4} x^2(1-x^2) & \mbox{for} & -1 < x < 1 \\
    0 & \mbox{for} & \mbox{other}
    \end{array}\right. $

    find the cumulative distribution function for $\displaystyle P(X \leq x)$

    so far I get

    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1 \\
    \frac{3}{4}x - \frac{1}{4}x^3 & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    but the given solution is:

    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1 \\
    \frac{3}{4}x - \frac{1}{4}x^3 + \color{red} \frac{2}{3} & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    where those the $\displaystyle \frac{2}{3}$ come from?
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  2. #2
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    Quote Originally Posted by lllll View Post
    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    \frac{3}{4} x^2(1-x^2) & \mbox{for} & -1 < x < 1 \\
    0 & \mbox{for} & \mbox{other}
    \end{array}\right. $ Mr F asks: Is this the pdf? If so, why do you have f(x) below (*)?

    find the cumulative distribution function for $\displaystyle P(X \leq x)$

    so far I get

    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1 \\
    \frac{3}{4}x - \frac{1}{4}x^3 & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $ * ??

    but the given solution is:

    $\displaystyle f(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1 \\
    \frac{3}{4}x - \frac{1}{4}x^3 + \color{red} \frac{2}{3} & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    where those the $\displaystyle \frac{2}{3}$ come from?
    Regardless of your answer, note that to get the cdf you integrate the pdf from -1 to x .....

    Substituting -1 into the antiderivative of the pdf will account for any constant you get in the cdf .....
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  3. #3
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    those are typos and I didn't copy it accordingly, what's shown for the cdf is:

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    but after integrating out you get:

    $\displaystyle \frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

    so either I'm missing something or there is in fact an error in the solution...
    Last edited by CaptainBlack; Apr 12th 2008 at 11:10 PM.
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  4. #4
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    Quote Originally Posted by lllll View Post
    those are typos and I didn't copy it accordingly, what's shown for the cdf is:

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    but after integrating out you get:

    $\displaystyle \frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

    so either I'm missing something or there is in fact an error in the solution...

    What happed to the other $\displaystyle x^2$ in the density?

    RonL
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  5. #5
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    Quote Originally Posted by lllll View Post
    those are typos and I didn't copy it accordingly, what's shown for the cdf is:

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    $\displaystyle F(x) = \left\{ \begin{array}{rcl}
    0 & \mbox{for} & x \leq -1\\
    \frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
    1 & \mbox{for} & x \geq 1
    \end{array}\right. $

    but after integrating out you get:

    $\displaystyle \frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

    so either I'm missing something or there is in fact an error in the solution...
    $\displaystyle \int^{x}_{1} \frac{3}{4}(1-u^2) \ du = \frac{3}{4}x - \frac{1}{4}x^3 + \frac{1}{2}$ is correct ...... But the pdf and hence integral is up in the air I think ....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle \int^{x}_{1} \frac{3}{4}(1-u^2) \ du = \frac{3}{4}x - \frac{1}{4}x^3 + \frac{1}{2}$ is correct ...... But the pdf and hence integral is up in the air I think ....
    Well I don't see any reason for us to worry about the algebra so I let Maxima do the work:

    Code:
    (%i5) integrate((3/4)*(1-z^2), z, -1, x);
    
    (%o5) (3*(2/3-(x^3-3*x)/3))/4
    
    (%i6) radcan(%);
    
    (%o6) -(x^3-3*x-2)/4
    
    (%i7)
    and as this is $\displaystyle 1$ at $\displaystyle x=1$, we have that $\displaystyle (3/4)(1-x^2)$ for $\displaystyle x \in (-1,1)$ and zero otherwise is a density, but the $\displaystyle (3/4)x^2 (1-x^2)$ of the original post is not (it integrates up to $\displaystyle 1/5$)

    RonL

    (note Maxima unlike Mathematica is freeware - and does not just produce plain ASCII output as above either)
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Well I don't see any reason for us to worry about the algebra so I let Maxima do the work:

    Code:
    (%i5) integrate((3/4)*(1-z^2), z, -1, x);
     
    (%o5) (3*(2/3-(x^3-3*x)/3))/4
     
    (%i6) radcan(%);
     
    (%o6) -(x^3-3*x-2)/4
     
    (%i7)
    and as this is $\displaystyle 1$ at $\displaystyle x=1$, we have that $\displaystyle (3/4)(1-x^2)$ for $\displaystyle x \in (-1,1)$ and zero otherwise is a density, but the $\displaystyle (3/4)x^2 (1-x^2)$ of the original post is not (it integrates up to $\displaystyle 1/5$)

    RonL

    (note Maxima unlike Mathematica is freeware - and does not just produce plain ASCII output as above either)
    Good thinking, 99.
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