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Math Help - PDF, CDF, Mean, Var Question

  1. #1
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    PDF, CDF, Mean, Var Question

    PDF question

    The question is from a past paper with no solutions, so I am trying to complete them and confirm with you guys if this is correct.

    a) Finding k, I solved for
    1 = k \left[  \int_{-1}^{0} (x+2)\ dx + \int_{0}^{1} (1-x^2)\ dx\   \right]
    for which I get k=\frac{6}{13}

    b) the CDF, I just took the integrals of all the terms so it should look like

    0 if x < -1
    \frac{x^2}{2} + 2x if -1 \leq x \leq 0
    x - \frac{x^3}{3} if 0 \leq x \leq 1
    0 if x > 1

    c) Expected value of X
    I want to confirm this before calculating the variance, because I am getting a negative mean of  -\frac{5}{26}

    I am attempting to calculate it as follows:
    E[X] = \frac{6}{13} \left[ \int_{-1}^{0} x(x+2)\ dx + \int_{0}^{1} x(1-x^2)\ dx    \right]

    e) Find the probability that X>0
    It is greater than 0 if it is between [0,1], so I've taken the integral:
    \frac{6}{13}  \int_{0}^{1} (1-x^2)\ dx = \frac{4}{13}

    Apologies the question is long, but they are related.
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  2. #2
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    First, you went to all that trouble to find 6/13. Why did you discard it for the CDF?

    CDF needs a LOT of work and there appears to be quite a bit of understanding missing. You studied indefinite integrals for a reason. You seem to have forgotten what they are.

    \frac{6}{13}\cdot\int x+2\;dx\;=\;\frac{6}{13}\cdot\left(\frac{x^{2}}{2} + 2x\right) + C. You absolutely must NOT ignore that arbitrary constant.

    Your task is to find the value of C such that \frac{6}{13}\cdot\left(\frac{(-1)^{2}}{2} + 2(-1)\right) + C = 0 You tell me why.

    Similarly, you must find the value of D, such that \frac{6}{13}\cdot\left((1) - \frac{(1)^{3}}{3}\right) + D = 1 Again, where did that come from?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    You tell me why.
    Because the cdf is a non-decreasing function between [0,1]. Thank you!
    Throughout the semester so far, we've only had cdf -> pdf questions, so this was new to me.

    Sorry about the missing k, I was meant to draw that up in front of the cdf per the original question.

    The revised CDF should be with the same limits as above:
    0
    \frac{6}{13}(\frac{x^2}{2} + 2x) + \frac{9}{13}
    \frac{6}{13}(x-\frac{x^3}{3}) + \frac{9}{13}
    0

    Would this be correct?
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  4. #4
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    Better. [-1,0] still is not correct. Why would the two constants be the same? They might be the same, but I would not expect it.

    Unless there is a probability mass in the middle, you should also demonstrate continuity, [-1,0] ends where [0,1] starts.
    Last edited by TKHunny; June 8th 2011 at 01:32 PM.
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  5. #5
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    Re: PDF, CDF, Mean, Var Question

    Thank you TKHunny, for the tips!
    I have had to review the definitions of a CDF, and understand why I have to demonstrate continuity.

    It turned out the two constants are the same for this question.
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  6. #6
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    Re: PDF, CDF, Mean, Var Question

    Right. Since the only questionable continuity is at x = 0, equality is requried. Good work.
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