Assume we are trying to classify between 2 classes, each has a Gaussian conditional probability, with different means but same variance, i.e. $X | y=0 sim N(mu_0, Sigma); X | y=1 sim N(mu_1, Sigma)$.

Our decision rule would be $1 iff P(y = 1 | X) > P(y=0|X)$ (and vice versa for 0).

Using Bayes rule we can invert the conditional probabilities, and get: $iff frac{P(X|y=1)P(y=1)}{P(X)} > frac{P(X|y=0)P(y=0)}{P(X)}$.

We can next eliminate the denominator.

Now, if $P(y=1) = P(y=0)$ we could eliminate that as well, and the decision rule would be simplified to $P(X|y=1) > P(X|y=0)$, which basically is which $mu$ is $X$ closer to. Now intuitively this translates to $1 iff X > c = frac{mu_0 +mu_1}{2}$.

I have 2 questions:

Can we prove this intuition mathematically?

What happens if $P(y=1) neq P(y=0)$ ?

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#### Best Answer

- Yes you can. For example, $$ mathbb{P}left(X=x|y=1right)=mathbb{P}left(X=x|y=0right) $$ happens exactly when $x$ is between the means, which can be calculated directly from the definition. Essentially, you are finding the decision boundary for this image, ,

but it's even simpler because you are assuming that the standard deviations are equal. You'd need to show with a simple equation that the probability (area under the curve) on the right is equal to the area on the left if the boundary is in the middle.

- I'm not willing to do the math here, but you have to slightly alter the equation above to find what happens to the boundary. I can tell it doesn't change linearly, because, for example, if you believe the left bubble to be twice as likely as the right bubble, you're looking for a boundary so that the area under what's left of the left bubble is half that of what's right of the boundary under the right bubble.

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