This is a 2-dimensional pattern recognition system that I am working on.

It is known that the distribution between the two classes are $1/2$ and $1/2$ respectively for class $\omega_1$ and class $\omega_2$. The feature vectors of the two classes are normally distributed around:

  • $\mu_1 = \begin{bmatrix}3\\6\end{bmatrix}$ with covariance matrix $\Sigma_1=\begin{bmatrix}1/2 & 0\\0 & 2\end{bmatrix}$ for class $\omega_1$
  • and around $\mu_2 = \begin{bmatrix}3\\-2\end{bmatrix}$ with covariance matrix $\Sigma_2=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$ for class $\omega_2$.

I am having some trouble with showing the ́posteriori probabilities. Does anyone know how to compute the posterior probabilities?

  • $\begingroup$ No ticks (yet) for the answer ? $\endgroup$ – Gilles Mar 2 '16 at 13:42


With Bayes' theorem you have:

$$p\left(\omega_i|\mathbf{x}\right) = \frac{p\left(\mathbf x|\omega_i\right)p(\omega_i)}{p\left(\mathbf{x}\right)}=\frac{p\left(\mathbf x|\omega_i\right)p(\omega_i)}{\sum_{i=1}^N p\left(\mathbf{x}|\omega_i\right)p(\omega_i)}$$

In your case $N = 2$, and your priors are equal: $p(\omega_1) = p(\omega_2) = \frac 12$. So, you have your posteriors as:

$$p\left(\omega_i|\mathbf{x}\right) = \frac{p\left(\mathbf x|\omega_i\right)}{\sum_{i=1}^2 p\left(\mathbf{x}|\omega_i\right)}=\frac{p\left(\mathbf x|\omega_i\right)}{p\left(\mathbf{x}|\omega_1\right)+p\left(\mathbf{x}|\omega_2\right)}$$

Your feature vector in each class $\omega_i$ are distributed according to:

$$p(\mathbf x|\omega_i)=\frac{1}{2\pi \left|\Sigma_i\right|^{1/2}}\exp\left[-\frac 12 (\mathbf x - \mu_i)^T \Sigma_i^{-1}(\mathbf x - \mu_i)\right]$$

And you plug all the information you have in your posteriors.

| cite | improve this answer | |

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.