1
$\begingroup$

enter image description here

My attempt:
(a) I solved that $a=\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$

(b) Here is where I'm running into trouble. I'm plugging the distributions into $\ln{\frac{P(X|C_0)P(C_0)}{P(X|C_1)P(C_1)}}$ and I get $a=\ln{\frac{P(C_0)}{P(C_1)}}+\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$.
I can see that $b=\ln{\frac{P(C_0)}{P(C_1)}}$ and $w^Tx=\frac{1}{2}(x-\mu_1)^T\Sigma^{-1}(x-\mu_1)-\frac{1}{2}(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)$.
I'm not sure how to simplify $w^Tx$ so that I can solve for $w$. Or is there something that I did wrong?

$\endgroup$
2
$\begingroup$

If you expand the terms, you can see that the quadratic terms cancel out.

\begin{align} a &= \ln \frac{P(C_0)}{P(C_1)} + \frac12(x - \mu_1)^T\Sigma^{-1}(x - \mu_1) - \frac12(x-\mu_0)^T\Sigma^{-1}(x-\mu_0)\\ &=\ln \frac{P(C_0)}{P(C_1)} + \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_1+\mu_1^T\Sigma^{-1}\mu_1\right]\\& - \frac12\left[x^T\Sigma^{-1}x-2x^T\Sigma^{-1}\mu_0+\mu_0^T\Sigma^{-1}\mu_0\right]\\ &= (\mu_0-\mu_1)^T\Sigma^{-1}x+\ln \frac{P(C_0)}{P(C_1)} +\frac12\left[\mu_1^T\Sigma^{-1}\mu_1-\mu_0^T\Sigma^{-1}\mu_0\right] \\ \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.