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Uncertainty of binomial distribution?

  1. Aug 26, 2009 #1
    Hi, I'd like to know how to find the uncertainty of a function that has two binomial distribution s, something like

    Signal = N(yes) - N(no)

    Set p = 0.6 for yes. My problem is that I do not know how to find the uncertainty for N(yes) and N(no), and do not know how to find the uncertainty for the signal. Any help would be great! Thanks
  2. jcsd
  3. Aug 26, 2009 #2


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    One measure of uncertainty is the standard deviation, which is the square root of the variance. See http://en.wikipedia.org/wiki/Binomial_distribution" [Broken], since the error of a difference is larger than the error of either of the components. Does this help?
    Last edited by a moderator: May 4, 2017
  4. Aug 26, 2009 #3
    Everything made sense except the propagation of the error. I did not understand the equation given, it had a term COV_{ab} which I have never seen before, and reading the page on covariance does not help as I understand very little of it. Could you give me an equation or explain the term COV please?
  5. Aug 26, 2009 #4


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    That is a complication, but first, let's check if your two binomial distributions are independent. If they are, then the covariance is zero. (The covariance is a measure of whether the two distributions are correlated.)
  6. Aug 26, 2009 #5
    The first is of events that are same-sign (ie ++ or --), while the other is the number of events that are opposite sign (+-), called N(ss) and N(os) respectively. I've taken the uncertainty of these two as:

    SQRT( N(os).p_os(1 - p_os) ) and SQRT( N(ss).p_ss(1 - p_ss) )

    I would like to know now the uncertainty now of a function that is N(os) - N(ss)
  7. Aug 26, 2009 #6


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    OK, that doesn't look independent. I haven't run into this type of problem before, so unfortunately you'll need somebody else to stop by and help.
  8. Aug 26, 2009 #7
    Ok, Thank you for helping though!
  9. Aug 26, 2009 #8
    Okay, well you really only have one binomial distribution here. N(yes) - N(no) = N(yes) - (N(total) - N(yes)) = 2N(yes) - N(total)

    This is a linear transformation of a single binomial distribution where n = N(total). Specifically if X is distributed as B(n,p) then your variable is 2X - n.

    When you say "uncertainty" you also refer to the "uncertainty of the signal" which makes me think you are talking about the Shannon entropy of the variable. The Shannon entropy of 2X-n is the same as the Shannon entropy of X (assuming that n is a constant and not a random variable), and the entropy of X can be looked up since it is binomial. According to http://en.wikipedia.org/wiki/Binomial_distribution, the Shannon entropy of X is [tex]\frac{1}{2} \ln \left( 2 \pi n e p (1-p) \right) + O \left( \frac{1}{n} \right) [/tex]
  10. Aug 26, 2009 #9
    Thank you, I had not thought of considering it as a function of only the n(yes) or n(no), that helps a lot. The uncertainty is to do with a measurement of the signal, ie 112 events +/- 10.6 events, originally this was posted in the physics section. But I think I can solve it with the treatment you have told me about, I am grateful to you,

  11. Aug 26, 2009 #10
    Actually I got the same problem on what the uncertainty of the n_total is as it too is a sum of binomial distributions
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