This seems to be a pretty basic question for this forum, please bear with me.

Consider that I have $10,000$ measurements, say of the height of $10,000$ people. These height measurements vary from $140$ to $190$ cm. I now define three height-groups: short ($<150$ cm), medium ($150$ to $170$ cm) and tall ($>170$ cm). I can now calculate proportions of those groups in my set of $10,000$ people (eg: “$40\%$ of the people are short, $30\%$ are medium, $30\%$ are tall”).

Now, consider that there is a random error associated with each of the height measurements. By a separate experiment, I concluded that this error distribution is well-approximated by a normal distribution, with a mean of zero and a standard deviation of $10$ cm. That is, the measurers are unbiased, but do make some random errors.

Now, I would like to propagate this estimated error in height measurement to the proportions. That is, I would like to say something like “the percent of short men in $40 ± 3 \%$” ($±$ could be standard error). Is there a theoretical way to go about this problem, rather than resorting to a Monte Carlo simulation?

The original data of 10,000 measurements could be described in two ways:

  1. It is approximated by another normal distribution, of mean 165 cm and a standard deviation of 7.0 cm
  2. It is described in a programming language data structure context; it is in a R vector, say "origData". Here, I am expecting R code that will take this vector and other inputs (from the question) and give me the standard errors.
  • $\begingroup$ Do you have the original data? For instance, If I asked you how many people had a height of $140$ cm, could you tell me? Also, do you have a distribution of the heights? Because if everyone in your short group had a height of $149$ cm, your error would be much larger than if everyone was $140$ cm. $\endgroup$ – AlgorithmsX Jan 3 '17 at 4:20
  • $\begingroup$ Hi, thanks a lot for your response. I have edited the original question to address your comments, hope it will be sufficient. $\endgroup$ – user100102 Jan 3 '17 at 15:18
  • $\begingroup$ Is there something missing from my answer? $\endgroup$ – AlgorithmsX Jan 6 '17 at 14:21
  • $\begingroup$ I am sorry for the late reply. Thanks a lot for the suggestions below. I am a bit unsure of how to do the last part: "...compare this new distribution with your original distribution and calculate errors from there". For example, let us say that the new distribution says "39.5% men are short" (please see my original question). So, is my answer 40±0.5% or so? Also, I should admit that I have not had time to read up fully on your answer, so please excuse if I missed something obvious. $\endgroup$ – user100102 Jan 7 '17 at 20:33
  • $\begingroup$ Basically, you end up getting a new distribution. With enough people, your new distribution will be pretty accurate no matter your old distribution. I don't really know how this error works, but what you can do is calculate your old short vs. your new short and use the difference to figure out how off you originally were. $\endgroup$ – AlgorithmsX Jan 7 '17 at 22:52

For a continuous distribution of values, you can apply a Weierstrass Transform. This basically takes the value at every point and spreads it out according to the normal distribution. The Weierstrass Transform is a type of convolution.

For discrete distributions, you can use a Gaussian Blur, which is the discrete version of a Weierstrass Transform.

Either way, you end up with a new distribution. You can then compare this new distribution with your original distribution and calculate errors from there.


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