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X_{t+1} = M X_t e^{-|R|}, where \;\; M \~ N(1,.01) and R \~ N(0, .01)
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X_{t+1} = M X_t e^{-|R|}, where   M \~ N(1,.01) and R \~ N(0, .01)
 
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Revision as of 15:07, 3 March 2013

Markov Processes

(alec green)

I attempted to model a robot circulating about the origin with small noise in each step. You can directly plug in the Matlab code below. It generated the following images, showing the putative robot spiraling 'inward' or 'outward'.

The (first-order, stationary, discrete time, continuous state-space) Markov process representing this simple 'walk' is as follows:

$ X_{t+1} = M X_t e^{-|R|}, where M \~ N(1,.01) and R \~ N(0, .01) $


Inward Spiral Outward Spiral


Question: Is the noise early in the walk process has a greater effect on the final position of the robot than noise later in the walk process, suggesting a butterfly effect? It seems that the second image appears to suggest otherwise, since the robot spirals out early but later spirals inward ending at the red square.


% set params
num_steps = 10000;
std_rot = .01;
std_noise_mag = .01;

% generate random initial state with complex magnitude 1
walk = zeros(1,num_steps);
walk(1) = exp(i*rand);
% generate next states (probabilistic modifications on current state)
for j=2:length(walk),
  walk(j) = walk(j-1)*normrnd(1,std_noise_mag)*(i^(-abs(normrnd(0,std_rot))));
end;
plot(walk); grid on; hold on;

% plot starting (green) and end (red) markers
plot(walk(1), 'gs','LineWidth', 2);
plot(walk(length(walk)), 'rs','LineWidth', 2);

% generate black circle to compare with "robot" circulation
circ = zeros(1,1000);
circ(1) = i;
for j=2:length(circ),
  circ(j) = circ(j-1)*i^(2*pi/length(circ));
end;
plot(circ,':k','LineWidth', 2); hold off;

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang