This is a question/answer from my first assignment (intro class on pattern recognition)

I don't understand how they used the p(x given w1) and p(x given w2) in the discriminant function. For example, how did they used the mean vector [0,0] and [d, e] in g(x)?

Can someone explain the missing steps from the original function to the simplified one?





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    $\begingroup$ Please add the self-study tag and detail why you cannot follow the answer. $\endgroup$ – Xi'an Oct 27 '17 at 5:17

I confirmed the result so let me give you a few tips:

First of all, you want to know which values for $\boldsymbol{x}$ are possible in the boundary. The boundary should satisfy $g_1(\boldsymbol{x}) = g_2(\boldsymbol{x})$, so this is equivalent to:

$$ -\frac{1}{2} \left( \begin{array}{cc} x_1 & x_{2} \\ \end{array} \right)\left( \begin{array}{cc} a & c \\ c & b \\ \end{array} \right)^{-1}\left( \begin{array}{c} x_1 \\ x_2 \\ \end{array} \right)-\frac{1}{2} \log \left| \begin{array}{cc} a & c \\ c & b \\ \end{array} \right| +\frac{1}{2} \left( \begin{array}{cc} x_1-d & x_2-e \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)^{-1}\left( \begin{array}{c} x_1-d \\ x_2-e \\ \end{array} \right)+\frac{1}{2} \log \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right|=0$$

where $\log\left|\Sigma\right|$ means the logarithm of the determinant $|\Sigma|$. Here you basically need to follow the usual rules of matrix multiplication. Notice that the second $\log$ will be $0$.

After you have performed the previous multiplications, remember to use the condition you're given: $a*b-c^2=1$. This will be helpful (repeatedly) to simplify the previous expression. In fact, you can see that the first $\log$ can also be removed from the expression immediately.

I hope that helps.


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