Imagine you start with two relatively large samples of sample sizes 15 and 18 respectively. We can assume they each come from a different normal distribution.

The first (15 observations) comes from distribution A and has a mean 10 and a standard deviation of 3

The second (18 observations) comes from distribution B has a mean 15 and a standard deviation of 3.5

(to be clear, the mean and standard deviations listed above are of observations, not the distributions themselves)

Then you are given a third and a fourth sample (sample C and sample D) of 3 observations each: let's say {12, 11, 8} and {14, 16, 13} You know that each new sample comes from one of the two distributions and that they both come from different distributions. How do you determine the probability that the sample C comes from distribution A and sample D comes from distribution B and vice versa?

Tries so far: I have written a bootstrapping program that simulates the scenario mentioned 100,000 times, but I would prefer an analytical solution. I have tried using the negative log likelihood of a T distribution similar to the solution mentioned below, but the answers I have gotten are substantially different than what I get in my program.

  • $\begingroup$ This sounds like a homework question. What is the exact phrasing? The idea that immediately came to mind does not answer the question as you’ve phrased it but does answer a related question. Also, please add the self-study tag, read its wiki, and edit your question to include what you’ve done so far. $\endgroup$ – Dave Jul 10 '20 at 5:56
  • $\begingroup$ @dave It's not a homework question. The topic came up organically in some data analysis I'm working on. I wrote the question like this because the actual problem I'm trying to solve is a bit more convoluted, and I wanted to reduce it to its most basic parts. $\endgroup$ – Jeremy Dorner Jul 10 '20 at 6:09
  • $\begingroup$ Distribution A and B are assumed to have potentially different variance? $\endgroup$ – Sextus Empiricus Jul 14 '20 at 18:00
  • $\begingroup$ Yes, the observation is that they have different standard deviations $\endgroup$ – Jeremy Dorner Jul 14 '20 at 22:24
  • $\begingroup$ If the moments of the two normal distributions are known then the initial data is irrelevant --- the problem is solved by a simple application of Bayes' theorem. Are you sure you intend for the initial data to be irrelevant? $\endgroup$ – Ben Jul 16 '20 at 7:03

One way would be using log likelihood, so the sum of log likelihoods of each unit, in this case obtained from a normal distribution. Computation in R

> -sum(dnorm(c(12,11,8),10,3,log=T))
[1] 6.552652
> -sum(dnorm(c(12,11,8),15,3.5,log=T))
[1] 9.535513

The goal is to minimize this value so the first case with mean 10 and SD of 3 is a better fit for the new data.

  • $\begingroup$ I like the approach, but if I understand correctly, this method does not account for the uncertainty of the mean and standard deviation of the original two samples $\endgroup$ – Jeremy Dorner Jul 10 '20 at 6:22
  • 1
    $\begingroup$ @JeremyDorner Regarding your edit, the mean and SD are unbiased estimators for the case of normal distributions, so they are your best "guesses" for the population. $\endgroup$ – user2974951 Jul 10 '20 at 6:45
  • $\begingroup$ I see. I was looking for the probability that each classification is correct, rather than the most likely classification. $\endgroup$ – Jeremy Dorner Jul 10 '20 at 6:57
  • 1
    $\begingroup$ @JeremyDorner Rough estimate 1-(6.552652/(6.552652+9.535513)) to get the probability of belonging in the first population. $\endgroup$ – user2974951 Jul 10 '20 at 7:18
  • $\begingroup$ Ah I got your point. Yeah that correctly answers the question as I asked, but would it still be valid if sample A and sample B have different sample sizes? I am not sure. I have updated the question to reflect this. My bad for not including it originally. $\endgroup$ – Jeremy Dorner Jul 10 '20 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.