Latest revision as of 02:16, 8 July 2014

Homework 2, Statistical Pattern Recognition, Summer 2014 (Bochum)

a) Suppose that two classes have the following densities:

$ \rho(x|\omega_1) = \frac{1}{2\pi} e^{-\frac{x^2}{2}}, $

$ \rho(x|\omega_2) = \frac{1}{2\pi} e^{-\frac{(x-10)^2}{2}}, $

and that the two classes have equal priors:

$ Prob(\omega_1)=Prob(\omega_2)=0.5. \ $

For $ N=5,10,15,\ldots, 1000 $, generate N points from the mixture density

$ \rho(x) = 0.5 \rho(x|\omega_1)+0.5 \rho(x|\omega_2), \ $

keeping track of the true class each point was drawn from (the ground truth-- the label of each data point). Then use Baye's decision rule (assuming perfect knowledge of the parameters of the class density) to classify the points and record the (empirical) percentage of error. Repeat this experience 100 times for each N, and plot your results (percentage of error versus N) on a graph. Also plot the average of the error, over all 100 trials, for each N. Plot the standard derivation of the error, over all 100 trials, for each N, as well. How do the errors you obtained compare with Bayes' error? Discuss your results.

b) Repeat a) using the following class 2 density instead

$ \rho(x|\omega_2) = \frac{1}{2\pi} e^{-\frac{(x-1)^2}{2}}, $.

Discuss your results.

c) Repeat a) using the following class 2 density instead

$ \rho(x|\omega_2) = \frac{1}{2\pi} e^{-\frac{(x-10)^2}{20}}, $.

Discuss your results.

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