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Finding error on half life.

  1. Oct 6, 2012 #1
    How may the error on the half life (using Poisson) be calculated? Barlow states that the error on the log would be 1/√N. Why is that?
     
  2. jcsd
  3. Oct 6, 2012 #2

    chiro

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    Hey peripatein.

    If you assume that the distribution for the half-life decay has a Poisson distribution with a parameter of lambda = 1/N, then the variance is also 1/N.

    The standard deviation of a distribution is the square root of the variance and this gives 1/SQRT(N).

    I'm assuming that this distribution is for some kind of decay events or something similar but if it's not, then could you tell us what the actual events refer to (if not decay events).

    For information about Poisson look at the following link (and note the mean and variance):

    http://en.wikipedia.org/wiki/Poisson_distribution
     
  4. Oct 7, 2012 #3
    Hi Chiro,
    Thank you your reply!
    Radioactive decay is indeed the case. I am familiar with the properties of the Poisson distribution, and obviously if the mean number of events (counts) were 1/N by all means would then sigma be 1/SQRT(N).
    My question is why would lambda be 1/N?
     
  5. Oct 7, 2012 #4

    mfb

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    The estimate for lambda should be the mean decay time of all events IIRC.
    I did a toy likelihood analysis for that problem (a few observed decay times) some months ago, I can try to find the code. There is an analytic expression for the likelihood, and if you consider log(lambda) it is a parabola I think.
     
  6. Oct 7, 2012 #5
    Right, mfb, estimate for lambda being the mean decay time of all events, why/how is that equal to 1/N? Can you demonstrate it mathematically, please?
     
  7. Oct 7, 2012 #6

    mfb

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    What is 1/N? If N is the number of events, 1/N is a dimensionless number, which cannot be the value of lambda. It is related to the relative uncertainty, but that is a different thing.
     
  8. Oct 7, 2012 #7
    Let me try and clarify the case -
    I have five Geiger counter readings taken at five short (1 minute) hourly intervals. If I plot lnR against count rate to find the decay constant, why would the error on the ln of the half life, using Poisson distribution, be 1/SQRT(N) (thus according to Barlow)?
     
  9. Oct 7, 2012 #8

    chiro

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    I should have asked this before, but what does the N actually represent in this half-life decay context?
     
  10. Oct 7, 2012 #9
    N is the number of counts.
     
  11. Oct 8, 2012 #10

    chiro

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    So is 1/N the average rate in terms of these counts (i.e. N events per unit time)?
     
  12. Oct 8, 2012 #11
    I am not sure why that would be. Barlow, in his explanation of the answer, merely states that after fitting a straight line of time against log of count rate, 1/SQRT(N) would be the error on that log using Poisson statistics. This is nearly an exact quote. I understand that the standard error on the mean should be sigma/SQRT(N) but that is still not 1/SQRT(N). I have arrived at the correct answer by inferring, from Barlow's explanation, that (ln tau) = 1/SQRT(N) and then applied propagation of errors to find the standard deviation of tau. But why is (ln tau) = 1/SQRT(N)? Could someone please demonstrate that to me mathematically?
     
  13. Oct 8, 2012 #12

    chiro

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    Well the Poisson distribution is just a distribution and its derived as a binomial in the limiting case where you look at the rate of so many events "happening" within one specific time interval and in this model each of the smallest intervals is atomic (i.e. disjoint) and they don't overlap at all.

    What the distribution is used for is that you have some interval and the probability function gives the probability for having x events happen in one interval where P(X = x) gives that probability. You can have no events or you can have one or more events, but that is what a Poisson distribution actually represents.

    The standard error of the mean is sigma/SQRT(N), but in a Poisson distribution sigma = SQRT(mu) so in a theoretical sense the SE(mu) = SQRT(x_bar)/N.

    This has problems most notably over-dispersion since the distribution has only one parameter controlling both the mean and variance, but that is another story.

    The main point though is that this distribution represents the probability of x events happening in one interval and the mean of this distribution gives the "average" number of events happening in one interval. I'm guessing these events are the "decays" themselves, but if there is no connection between the decays and the above description of a Poisson distribution, then that's something you need to check out.
     
  14. Oct 8, 2012 #13

    mfb

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    Ok, that is a completely different setup now.

    If you count N decays in that 1-minute-interval, you expect an uncertainty of sqrt(N) for the number of decays in that interval, as the number of decays is (in good approximation) a Poisson distribution.
     
  15. Oct 8, 2012 #14
    But why then is the error on the log of the half life equal to 1/SQRT(N)? Can you please show it mathematically?
     
  16. Oct 8, 2012 #15

    mfb

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    In first order, the error on the logarithm is the relative error, as ##\Delta \log(x) \approx \frac{\Delta x}{x}## ("chain rule")
     
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