If I have two normal distributions A (mean = 0, variance = 4) and B (mean = 0, variance = 16), how can I calculate the probability that an observation with magnitude 2 comes from A?

  • $\begingroup$ I assume your mixture has equal weights, since you don't specify otherwise? $\endgroup$ – Stephan Kolassa Feb 5 at 15:18
  • $\begingroup$ @StephanKolassa you are correct, apologies for leaving it out $\endgroup$ – Bob Doe Feb 5 at 15:23
  • $\begingroup$ Welcome to CV. This is a good question for a newer learner to work on. It is valuable to look at the different ways you can engage the question. What form of an answer is more meaningful to you? $\endgroup$ – EngrStudent Feb 5 at 16:10
  • $\begingroup$ @EngrStudent Thank you. An example would be great but a hint or two would suffice. $\endgroup$ – Bob Doe Feb 5 at 16:29
  • $\begingroup$ Bob, there are several families of answers with many answers per family. Computationally, you could do it in R, python, excel, or something else. You could simulate it a hundred thousand times and look at rates. Computationally-Analytically you could do one of Nick Cox's convolutions and get results that are equivalent to infinite samples. Analytics/Symbolic - you could do it a number of ways with "squiggly herding" in a structure of a mathematical proof. There are 100 ways to make an example, and hearing that gives you a start exploring the (wonderful and miraculous) landscape of stats. $\endgroup$ – EngrStudent Feb 5 at 17:09

Here's some Python code

from scipy.stats import norm
pA = norm.cdf(2.0, loc=0, scale=2) - norm.cdf(-2.0, loc=0, scale=2)
pB = norm.cdf(2.0, loc=0, scale=4) - norm.cdf(-2.0, loc=0, scale=4)
print('P(-2.0 < x < 2.0 | A) = {}'.format(pA))
print('P(-2.0 < x < 2.0 | B) = {}'.format(pB))
print('P(-2.0 < x < 2.0 | A) / P(-2.0 < x < 2.0 | B)= {}'.format(pA / pB))

And it's output:

P(-2.0 < x < 2.0 | A) = 0.6826894921370859
P(-2.0 < x < 2.0 | B) = 0.38292492254802624
P(-2.0 < x < 2.0 | A) / P(-2.0 < x < 2.0 | B)= 1.7828285701395248

I just ask, what's the probability I'd draw a data point within the specified value of 2.0 (two-sided). Do this for both, and take ratio. This means it's 1.78 more likely the data point belongs to A.


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