Given a draw from one of two overlapping normal distributions, what is the probability it came from one vs. the other?

Question

I'm developing some game theory models that involve normal distributions, and am not sure how to solve this probability problem:

Suppose there are two normal distributions: $$ X_1 \sim N(\mu_1, \sigma^2) $$ $$ X_2 \sim N(\mu_2, \sigma^2) $$

You know what both distributions are, but that's all you know. I take a draw (call it $s$) from one of the two distributions and show it to you. Given $s$, what is the probability I chose from $X_1$?

Thanks for any help!

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Here's of a picture of two overlapping normal distributions in case it's helpful to have a visual:

Answer 1

As Adrian already suggested you need to know the prior probability that $X$ came from each distribution. If $Y$ is an indicator telling us whether or not $X$ came from distribution one and $p_1$ and $p_2$ are the mixing (prior) probabilities then

$$ P(Y = 1 \mid X = x) = \frac{p_1 f_{X \mid Y=1}(x)}{p_1 f_{X \mid Y=1}(x) + p_2 f_{X \mid Y=0}(x)} . $$

All you've specified are the conditional densities $f_{X \mid Y = 1}$ and $f_{X \mid Y = 0}$ but this isn't enough to calculate the probability. You also need to know something about $p_1$ and $p_2$.

Answer 2

The probability is the ratio of the densities.

$ P(s \text{ from }X_1) =\frac{f_{X_1}(s)}{f_{X_2}(s)} $

Intuition. Look at where the distributions cross:

Anything to the left of this line is more likely to come from the distribution with the smaller mean, while anything above this line is more likely to come from the distribution with the larger mean. In the picture you've shown if the value $s$ actually does come from the distribution with the smaller mean, and it falls below $\mu_1$ then we are almost certain it comes from $X_1 \sim N(\mu_1,\sigma^2)$ since the pdf of $X_2$ is basically 0 at this point (i.e. $f_{X_2}(s)\approx 0, s<\mu_1)$

Answer 3

I'll try giving a very low-level answer. If there's no prior reason to believe that s was drawn from X₁ or X₂, i.e. the value of s is the only information you have, then the probability is directly obtainable from the relative values of the probability density functions evaluated at s. That is, $P(X_1 | s)\propto X_1(x=s) / X2(x=s)$. (The actual probability would require some normalization.)

For example, if s=0 in your example, it's much more likely that your draw was from the solid line distribution (value ~ 0.8) than the dashed line distribution (value ~ 0.1), and the relative probability is approximately 8:1. i.e. It's 8 times more likely that s=0 was drawn from the solid line distribution than the dashed line.