# How to Quantify Likelihood

We are given two normal populations A & B with means $\mu_1$, $\mu_2$ and standard deviations of $\sigma_1$, $\sigma_2$ respectively, as well as a subject X. Our purpose is to determine which population X is more likely to belong assuming that each population is equally likely to occur. The obvious place to start is to calculate a z-score and corresponding percentiles.

Let's say the z-score for X in regard to population A is :

$A_X = 1.3 \space\space \Rightarrow \space\space 90.32 \space$ (percentile)

And for population B, we have:

$B_X = -2.1 \space\space \Rightarrow \space\space 1.79 \space$ (percentile)

What is the proper method for quantifying which population X would most likely belong to?

1. Should we simply reference the absolute distance from the mean? (IMO, this is very dissatisfying)
2. Is it correct in saying that if X was from A, there would be a $\approx 5.4$ times greater chance in encountering a more extreme value than if the X was from B $(1 - 0.9032 = 0.0968 \space\space \Rightarrow \space\space \frac{0.0968}{0.0179} \approx 5.4)$?
• You are implicitly assuming that A and B are equally likely to occur, is this on purpose? If not, this "prior probability" is an important variable in your calculation
– Sam
Aug 30, 2017 at 15:27
• @Sam yes,I am assuming that A and B are equally likely to occur on purpose, Aug 30, 2017 at 15:31

One way to get a probability for this is by using Bayes rule. What you want is $P(A|X)$ and $P(B|X)$.

For probabilities, the following holds: $P(A|X) = P(X| A) P(A) / P(X)$

$P(X|A)$ is the probability of X occurring if X belongs to A - so you can calculate that from the probability density function of the normal distribution with mean $\mu_1$ and variance $\sigma_1^2$.

So e.g. in R, you'd just calculate dnorm(X, mu, sigma)

$p(A)$ is your prior probability, so 0.5

and $p(X) = p(X | A) p(A) + p(X | B) p(B)$ is your "normalizing constant", which ensures that your output is still a probability.

So let's say $P(X | A) = 0.9$ and $P(X | B) = 0.3$ (these are probability densities, so they don't have to add up to 1)

then $P(X | A) p(A) = 0.9 * 0.5$ and $P(X | B) p(B) = 0.3 * 0.5$, so $p(X) = 0.6$ and hence $p(A | X) = \frac{0.45}{0.6} = 3/4$

The reason why you take the probability density, rather than the cumulative density ("percentile"), is easy to illustrate in the multimodal case:

Let's assume that A is a multimodal Gaussian Let's say your you want to get $p(X = 0.5 | A)$. If you take the percentile, it would tell you it's about 50% - However as you can see, the probability of it occurring is very small. What you want for $p(X|A)$ is really the relative likelihood of occurrence, which is given by the probability density value (on the y-axis)

• Nice answer. Wouldn't P(X|A) simply be the percentile (i.e. .9032)? E.g. in R we have pnorm(X, mean = mu1, sd = sigma1 = 0.9031995, where mu1 and sigma1 are from the question. Aug 30, 2017 at 16:23