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Thread: pdf for continuous random variable

  1. #1
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    pdf for continuous random variable

    Hello again everyone,

    I'm having some trouble understanding an example for a probability density function of a cont. RV.
    The part I'm not getting is the integrals. I need an elaboration on what is going on with the calculations
    so I can see what is happening. I assume that it has something to do with derivatives but I have never
    seen or don't remember what the $\displaystyle \mid^2_{-1}$ etc.. or what the $\displaystyle \int$ symbol means.
    Thanks for your help.

    Here is the example:

    [Question]
    Suppose that the error in the reaction temp., in Celsius, for a controlled lab. experiment is a cont.
    RV "X" having the pdf:

    $\displaystyle f(x)=\left\{\begin{array}{cc}\frac{x^2}{3},&-1 < x < 2\\0, & \mbox{ elsewhere }\end{array}\right.$


    1. Verify that $\displaystyle f(x)$ is a density function
    2. Find $\displaystyle P(0 < X \leq 1)$


    [SOLN]

    1. $\displaystyle f(x) \geq 0$, by def.
      $\displaystyle \int^\infty_{-\infty}f(x)dx = \int^2_{-1} \frac{x^2}{3}dx = \frac{x^3}{9}\mid^2_{-1} = \frac{8}{9} + \frac{1}{9} = 1$

    2. $\displaystyle P(0 < X \leq 1) = \int^1_{0}\frac{x^2}{3}dx = \frac{x^3}{9}\mid^1_{0} = \frac{1}{9}$
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  2. #2
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    Re: pdf for continuous random variable

    so you don't know what an integral is?
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  3. #3
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    Re: pdf for continuous random variable

    I guess not...
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  4. #4
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    Re: pdf for continuous random variable

    Quote Originally Posted by mathgex View Post
    I guess not...
    that's going to make it pretty hard to comprehend and work with continuous probability densities.

    Do you have the background to hit the web and read the 30 second version?
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  5. #5
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    Re: pdf for continuous random variable

    I suppose that is what I will have to do. I have to master this topic by Sunday...Pray for me lol.
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  6. #6
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    Re: pdf for continuous random variable

    Ok so I spent a great deal of time reading up on integrals and derivatives and while I was
    writing this reply I finally realized what I was missing..

    $\displaystyle \frac{8}{9} + \frac{1}{9}$, I was expecting to see $\displaystyle \frac{8}{9} - \frac{1}{9}$ since $\displaystyle (-1)^3 = -1$
    and I didn't understand why. Now I know that its + because you subtract F(-1) from F(2): $\displaystyle F(2)-F(-1)$ and in this case the
    $\displaystyle \frac{-1}{9}$ was turned positive via the double negative $\displaystyle \frac{8}{9} - (\frac{-1}{9})$
    Last edited by mathgex; Nov 8th 2015 at 10:36 PM.
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