# pattern classification when the prior probabilities are not equal

In the case of 2 class classification, the decision boundary occurs when 2 discriminant functions are equal: $$g_1(x) = g_2(x)$$ $$g_i(x) = p(x|w_i)P(w_i)$$ $$p(x|w_i) = \mathcal{N}(x;\mu,\,\Sigma)$$

solving $$g_1(x) = g_2(x)$$ we get $w^T(x-x_0)$, where $w^T$ is $\mu_1 - \mu_2$ and $x_0$ is $1/2*(\mu_1 + \mu_2)-(\sigma^2(\mu_1 - \mu_2)/||\mu_1 - \mu_2||^2)*ln[P(w_1)/P(w_2)]$ If $P(w_1)=P(w_2)$ $x_0$ is halfway between the means, but if $P(w_1)\neq P(w_2)$ then $x_0$ shifts away from the more likely mean. My question is in the diagram the point to be classified is closer to the mean of class 2 than class 1, and the prior prob. of class 1 is 0.9 and prior prob. of class 2 is 0.1, in this case will the point be classified as class 1 or class 2?

You would still be classifying as the group with higher posterior probability. If $Y$ is the class membership indicator, $\pi_i$ the prior probability of class $i$, $f$ the density of the predictor $x$ and $i = 1, 2$ this equals
$$P(Y = i \mid x) = \frac{\pi_i f(x \mid Y = i)}{f(x)} .$$
Your classification rule (after plugging in suitable estimates for the prior probabilities and conditional densities) is then $\text{arg max}_{i \in \{1, 2 \}} \pi_i f(x \mid Y = i)$.
• The picture above is not weighting the densities according to the prior probabilities, it's only looking at the conditional distribution of $x$ given the class. Just imagine in your example that the proportion of oranges in the population is vanishingly small. Shouldn't this affect your decision to classify something as an orange? – dsaxton Jul 16 '15 at 20:09