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Math Help - Extrapolation

  1. #1
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    Extrapolation

    Hello,

    Lets say i'm measuring the probability of something happening at different stages.

    For x = 1 I know the probability +- 0% (exact)
    For x = 2 x = 3 and x = 4 I know the probability +- 8%

    How well would i be able to say what the probability is at x = 5? and x = 6? And even further? (maximum to x = 10)
    Also, I expect the probability to decreases exponentially.

    If someone could help me with this that would be really awesome.
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  2. #2
    Junior Member Mathhead200's Avatar
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    Re: Extrapolation

    If the probability is decreasing exponentially, why is it constant at x = 2,3,4 ?
    Assuming it starts strictly decreasing at x = 4, i.e. P(x = 4) > P(x = 5), then:


    Let n = x - 4, and P_n = ar^n where n \geq 0 and 0 < r < 1
    Since P(x = 4) = P_{n = 0} = 0.08, it follows that a = 0.08
    So P_n = (0.08)r^n

    We also need \sum_{x=1}^{\infty}{P(x)} = 1 because we are dealing with probabilities.
    \sum_{x=1}^{\infty}{P(x)}
    P(x = 1) + P(x = 2) + P(x = 3) + \sum_{n=0}^{\infty}{P_n}
    0 + 0.08 + 0.08 + \sum_{n=0}^{\infty}{P_n}
    0.16 + \sum_{n=0}^{\infty}{P_n}

    So equivalently we need \sum_{n=0}^{\infty}{P_n} = 1 - 0.16 = 0.84

    But P_n is a geometric series, so \sum_{n=0}^{\infty}{P_n} = \frac{a}{1 - r} = \frac{0.08}{1 - r} = 0.84
    \frac{0.08}{1 - r} = 0.84
    0.08 = (0.84)(1 - r)
    0.08 = 0.84 - (0.84)r
    -0.76 = -(0.84)r
    r = \frac{-0.76}{-0.84} = \frac{19}{21} \approx 0.9048

    So we calculated P_n = (0.08)\left(\frac{19}{21}\right)^n

    Finally P(x) = \begin{array}{ll}  0  &  ;\; x = 1  \\  0.08  &  ;\; x =2,3  \\  (0.08)\left( \frac{19}{21} \right)^{x - 4}  &  ;\; x \geq 4  \end{array}
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  3. #3
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    Re: Extrapolation

    One inexact way to do it would be to find an equation that describes the probability for 1,2,3 and 4 using the upper bound for your measurements. When you have this approximate equation find the probabilities at 5 to 10, these would be the upper limit of what the probabilities are. Do the same for the lower bound of your measurements. Finding an equation that fits the data should be simple enough because you already have an idea that it is exponential.
    This method will tend to give exaggerated upper and lower limits of the probabilities.

    Quote Originally Posted by Mathhead200 View Post
    If the probability is decreasing exponentially, why is it constant at x = 2,3,4 ?
    I'm pretty sure the +-8% is that he knows the value of the probability lies in the range p\pm8% and there is something special about 0 that means the probability has to be zero and so the uncertainty range is zero.
    Last edited by Shakarri; April 9th 2013 at 10:53 AM.
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  4. #4
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    Re: Extrapolation

    Yeah I meant that at x = 1 I know the probabillity exactly. It is 100% at x = 1
    For x = 2, x = 3 and x = 4 I would know the probabillity within a range of +- 8%

    for example:
    x = 1, 100%
    x = 2, 75% +- 8%
    x = 3, 50% +- 8%
    x = 4, 20% +- 8%

    These are just numbers I made up, the reason i'm asking this because I want to know if its worth to measure them since it will take me alot of time to do so. And the higher the x gets the longer it will take. The guess at exponential decrease is just based on the few times I did it without actually measuring anything and on what others tell me.
    Last edited by StevenDR; April 9th 2013 at 12:20 PM.
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  5. #5
    Junior Member Mathhead200's Avatar
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    Re: Extrapolation

    Well you can use the same ideas that I used (when I misunderstood your question.) Look at different functions assuming it's a geometric series. BTW, how can P(x = 1) = 100\%; then P(x \neq 0) = 0\%...?
    Do you mean, P(x >= 1) = 100\%,\; P(x >= 2) = p_2 \pm 8\%,\; \ldots
    In this case: P(x = 1) = 100\% - (p_2 \pm 8\%)
    I would think about it this way, as computing an infinite sum (of the geometric series) only helps if you assume: \sum_{x=1}^{\infty}{P(x)} = 100\%
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