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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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