Given two Gaussians with the same variance $\sigma$ and means $\mu_1$ and $\mu_2$, where each Gaussian represents a class $C_1$ and $C_2$ with the same prior probabilities, i.e. $p(C_1) = p_(C_2)$, we should derive the decision boundary $x^*$ as a function of $\mu_1$ and $\mu_2$.
I know that it holds for the decision boundary $p(x|C_1) = p(x|C_2)$. We know further that $p(x|C_1) = \mathcal{N}(x|\mu_1, \sigma)$ and $p(x|C_2) = \mathcal{N}(x|\mu_2, \sigma)$.
So we have
$ \begin{align} &\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x - \mu_1)^2}{2\sigma^2}) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x - \mu_2)^2}{2\sigma^2})\\ \Longleftrightarrow & -\frac{(x - \mu_1)^2}{2\sigma^2} = -\frac{(x - \mu_2)^2}{2\sigma^2}\\ \Longleftrightarrow & x^2 - 2x\mu_1 + \mu_1^2 = x^2 - 2x\mu_2 + \mu_2^2\\ \Longleftrightarrow & x(2\mu_2 - 2\mu_1) = -(\mu_1^2 + \mu_2^2)\\ \Longleftrightarrow & x = -\frac{\mu_1^2 + \mu_2^2}{2\mu_2 - 2\mu_1} \end{align} $
Is this approach and the solution for the decision boundary right?
