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Extracting sigma from sum of distributions with moving mean

  1. Sep 12, 2014 #1
    I have an experiment in which I want to extract the distribution function of a process. I expect it to be Gaussian. Each data point measured is an entire distribution, f(x), but I am forced to average over many points such that the result of the experiment is the sum of many measurements of f(x). If A and σ are believed to be constants but the mean, μ, varies a little for each point, the resulting sum of distributions appears broader as if σ is larger. My question: If I believe I know the deviation of the mean μ, can this affect be subtracted out so that I am left with the actual value of σ?
     
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  3. Sep 12, 2014 #2

    mfb

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    I'm not sure what you mean with "Each data point measured is an entire distribution". If every data point is a point, but comes from a distribution with fixed σ and variable μ, then it is possible. The measured standard deviation of your data points will be the quadratic sum of σ and the standard deviation of μ:
    $$\sigma_{total} = \sqrt{\sigma^2 + \sigma_\mu^2}$$
     
  4. Sep 12, 2014 #3
    Let me clarify. I'm running an experiment for which each result (what I called point) is a trace that gives a distribution. Based on the nature of the experiment, the standard deviation should be the same for each trace. However, there is a random shift of the mean that can obviously be measured. In order to achieve sufficient signal/noise, I must average over many traces such that this shifting of the mean causes a widening of the summation. I wasn't completely sure how to subtract out this effect. I wasn't able to find the solution in my old stats book that I had to dust off. I believe you answered the question, but I was also hoping to see this relationship proven. Thanks for the help.
     
  5. Sep 12, 2014 #4

    mfb

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    If you can measure the shift for each data point, then you can do better - you can just shift your measurement to correct for it (and take the uncertainty of this correction as uncertainty on μ). If σμ is small compared to σ this won't make a significant difference, but if the two are comparable it can be a huge improvement.
     
  6. Sep 12, 2014 #5
    Sorry again for the poor explanation. Because of the nature of the experiment, I can't measure the mean directly from each trace, but I can infer the deviation of the mean indirectly from other data.
     
  7. Sep 13, 2014 #6

    FactChecker

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    If you can do repeat measurements within the same distribution, before the mean changes, then you can estimate the variance for each distribution and draw some conclusions. Otherwise, you can not statistically separate the variation of distribution mean from the variation for a single distribution with a fixed mean. If you know why and how the mean changes, maybe you can back that out from the data before applying statistical methods. Otherwise, I don't think you can do anything to estimate the variance at a fixed mean.
     
  8. Sep 15, 2014 #7
    Follow up question. I get that if I know σ_μ than I can back out σ. Now if the distribution is bivariate such that the trace I see is actually g(x,y=0) and I also have some variation in the mean of y, is there a simple way to relate the mean of y to the resulting sigma_x? In other words, I will be averaging over g(x,y=+/-dy) and the trace (distribution) I see will appear broader. If I assume sigX=sigY, it seems as though I could back out σ again. Thanks again.
     
  9. Sep 15, 2014 #8

    mfb

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    How are y and x related?

    It would help to know more about the setup and your datasets...
     
  10. Sep 15, 2014 #9
    I'd avoided explaining for fear it might devolve off topic, but here goes. My question is not actually directly related to statistics but I have reason to assume a normal distribution so it seemed like a good place to pose the question. I have a 2D distribution of particles injected into a gas medium that I can image using a gated camera and a laser that provides excitation (LIF images). I want to study the rate of diffusion (take the image at different times). The result should be symmetric and gaussian. However, the limitations are that I need to accumulate over many shots/injections in order to see a distribution for a given moment in time after the particles enter. Also, I know from using a medium with higher signal that while sigma is not expected to change at one point in time, the centerline/mean varies from shot to shot. I can measure this under other circumstances and I'd like to back out the effect from the intensity/population distributions that I see. Imagine trying to measure the spread of bird shot by looking at a target but needing to subtract out changes in centerline trajectory caused by imperfect rifleman.

    So if I only needed to worry about the dimension in the direction of the beam (x-direction), I'd just use the equation you provided to subtract out the deviation of the mean, which I measure under other circumstances. However, since there is variation in the y-direction (perpendicular to beam) as well, I will be "missing" the centerline and getting a broadening effect there as well. It would be easy for me to generate a 2D normal distribution, assign random values for sigma within a confidence interval of my determination and take the average of a single line over many thousands of points and perform this repeatedly until I arrive at the answer/distribution that I see. However, I'd prefer to just find an actual solution. Any help is, again, greatly appreciated.
     
  11. Sep 15, 2014 #10

    mfb

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    For diffusion, x and y are independent, yes you can ignore y then (or look at it separately).
     
  12. Sep 16, 2014 #11
    hmmm

    I understand that the rate of diffusion is independent for x and y, but if I'm measuring distribution over many "shots" and it's moving in y than I'll be probing a different point at each measurement. The distribution will appear broader when y ≠ 0, correct?
     
  13. Sep 16, 2014 #12

    mfb

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    Why should the x-distribution change if your y moves?
     
  14. Sep 16, 2014 #13
    Because the function f(x,y) such that f(x,y=const) is different for all y? I'm just looking at the equation for a bivariate normal distribution.
     
  15. Sep 16, 2014 #14

    mfb

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    Why should the x-dependence be different?
    Independent variables mean you can find g, h such that f(x,y)=g(x)*h(y). You don't have to care about h to find g.
     
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