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Math Help - Continuous Random variable

  1. #1
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    Continuous Random variable

    Hi, I'm unsure where to go with this question:

    Let  f(a,b) (x) \begin{cases} = { \left ( \frac{bx}{a} \right )  for  \in   [0,a] \\ {0} otherwise \end{cases}

    If a > 0 for what values of b > 0 is f(a,b) the pdf of a continuous variable. Also for what values of a and b is the E(X) = 1/3


    So what I'm doing is  \int_{a}^{0} {bx/a} \, dx and then I get   -(a b)/2 . I was thinking to make this equal to 1, but don't no what to do next.
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  2. #2
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    Quote Originally Posted by axa121 View Post
    Hi, I'm unsure where to go with this question:

    Let  f(a,b) (x) \begin{cases} = { \left ( \frac{bx}{a} \right ) for \in [0,a] \\ {0} otherwise \end{cases}

    If a > 0 for what values of b > 0 is f(a,b) the pdf of a continuous variable. Also for what values of a and b is the E(X) = 1/3


    So what I'm doing is  \int_{a}^{0} {bx/a} \, dx and then I get  -(a b)/2 . I was thinking to make this equal to 1, but don't no what to do next.
    You require:


    1. \displaystyle \frac{bx}{a} \geq 0 for 0 \leq x \leq a.


    2. \displaystyle \int_0^a \frac{bx}{a} \, dx = 1.


    And for the third part, you require \displaystyle \int_0^a \frac{bx^2}{a} \, dx = \frac{1}{3}.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You require:


    1. \displaystyle \frac{bx}{a} \geq 0 for 0 \leq x \leq a.


    2. \displaystyle \int_0^a \frac{bx}{a} \, dx = 1.


    And for the third part, you require [tex]\displaystyle \int_0^a \frac{bx^2}{a} \, dx = \frac{1}{3}[\math].

    I got a = 1 and b = 1 for the expected value to be 1/3

    But do I need to solve the inqualities for the first bit. After integrating I have ab/2 = 1, but the first equation has [tex] \displaystyle \frac{bx}{a} \geq 0 [\math] which has an x in it.
    Last edited by axa121; November 11th 2010 at 02:56 PM.
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  4. #4
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    Im still stuck, after integrating for the first part I get ab = 2 how can I solve this?
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  5. #5
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    Quote Originally Posted by axa121 View Post
    Im still stuck, after integrating for the first part I get ab = 2 how can I solve this?
    You can't. All you can do is get a in terms of b. If you are expected to simultaneously satisfy the expected value being equal to 1/3, then you solve the following simultaneously:

    ab = 2

    ba^2 = 1

    (I have no idea where your answers in post #3 came from but they are not correct).
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  6. #6
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    Right so a = 1/2 and b = 4, this gives E(X) = 1/3. But the first part asks for what values of b > 0 is f(a,b)(x) the pdf of a continuous random variable.
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  7. #7
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    Quote Originally Posted by axa121 View Post
    Right so a = 1/2 and b = 4, this gives E(X) = 1/3. But the first part asks for what values of b > 0 is f(a,b)(x) the pdf of a continuous random variable.
    b = 2/a.
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  8. #8
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    I see, I was looking for one or two specific values. Thank you very much.
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