The equation provided is a linear equation in x.
Taking the logarithm of both sides and rearranging the terms results in a linear equation.
The coefficient of x^2 on the left hand side is 0.
Consider the general case where the two classes have different means and possibly different variances:
\displaystyle \displaystyle X|C=0 \displaystyle \sim \displaystyle \mathcal{N}(\mu _0,\sigma _0^2)
\displaystyle X|C=1 \displaystyle \sim \displaystyle \mathcal{N}(\mu _1,\sigma _1^2)
and the prior distribution of C is
\displaystyle \displaystyle \mathbf{P}(C=1) \displaystyle = \displaystyle p
\displaystyle \mathbf{P}(C=0) \displaystyle = \displaystyle 1-p.
To find the decision boundary, we need to solve the following equation for x:
\displaystyle \displaystyle \mathbf{P}(C=0) \mathbf{P}(X=x|C=0)=\mathbf{P}(C=1)\mathbf{P}(X=x|C=1).
It is easier to solve this by first taking the logarithm on both sides. Rearranging the terms, we get:
\displaystyle \displaystyle \ln \left[\mathbf{P}(C=0)\mathbf{P}(X=x|C=0)\right]-\ln \left[\mathbf{P}(C=1)\mathbf{P}(X=x|C=1)\right] \displaystyle = \displaystyle 0.
What is the degree of the equation? Assume \sigma _0^2\neq \sigma _1^2 and \mu _0\neq \mu _1.
quadratic equation in x.
linear equation in x
none of the above
correct
Report the coefficient of x^2 on the left hand side in the equation above.
(If applicable, type mu_0, mu_1 for \mu _0,\, \mu _1 respectively, sigma_0\mathbf{\wedge }2 for \sigma _0^2, and sigma_1\mathbf{\wedge }2 for \sigma _1^2 respectively.)
Coefficient of x^2:
1 answer