Class 1 data samples follow a zero-mean Laplacian distribution; class 2 data samples follow a Gaussian distribution, as shown below:

$$ \begin{align*} p(x|\omega_1) &= \dfrac{1}{2b}e^{-|x|/b} = \begin{cases} \dfrac{1}{2b}e^{-x/b} &\text{if $x\geq 0$}\\ \dfrac{1}{2b}e^{x/b} &\text{if $x < 0$} \end{cases}\\ p(x|\omega_2) &= \dfrac{1}{\sqrt{2\pi}\sigma_2}\text{exp}\left[\dfrac{-1}{2}\left(\dfrac{x-\mu_2}{\sigma_2}\right)^2\right] \end{align*} $$

$p(x|\omega_1)$ parameter

b =


$p(x|\omega_2)$ parameters

mu =


variance =


I need to calculate the Bayesian decision boundary based on the information above.

So far I know the value of the parameters for both PDFs. I think I am suppose to get the discrimant function for both PDFs and set it equal to each other.

$$ \begin{align*} g_i(x) &= \log{p(x|\omega_i)} + \log{P(\omega_i)}\\ g_1(x) &= g_2(x) \end{align*} $$

This is close to Combination of two Gaussians

but in this case, I have two different distributions.

  • $\begingroup$ The Laplacian distribution in your statement has only one parameter, so it's not clear what $b$ and $\mu$ mean. Also, you may need to be more clear what you mean by decision boundary. $\endgroup$ – conjectures Mar 27 '17 at 15:09
  • $\begingroup$ This appears to be a self-study question. If that is the case, please add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ – Tavrock Mar 27 '17 at 15:11
  • $\begingroup$ class 1 is a zero mean, you don't see the mu in the function.See the PDF of the en.wikipedia.org/wiki/Laplace_distribution $\endgroup$ – Pete Mar 27 '17 at 15:17

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