Pregunta

Lets say I want to simulate a particle state, which can be normal (0) or excited (1) in given frame. The particle is in excited state f % of time. If the particle is in excited state, it lasts for ~L frames (with poisson distribution). I want to simulate that state for N time points. So the input is for example:

N = 1000;
f = 0.3;
L = 5;

and the result will be something like

state(1:N) = [0 0 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 ... and so on]

with sum(state)/N close to 0.3

How to do that? Thanks!

¿Fue útil?

Solución

%% parameters
f = 0.3; % probability of state 1
L1 = 5;  % average time in state 1
N = 1e4;
s0 = 1; % init. state
%% run simulation
L0 = L1 * (1 / f - 1); % average time state 0 lasts
p01 = 1 / L0; % probability to switch from 0 to 1
p10 = 1 / L1; % probability to switch from 1 to 0
p00 = 1 - p01;
p11 = 1 - p10;
sm = [p00, p01; p10, p11];  % build stochastic matrix (state machine)
bins = [0, 1]; % possible states
states = zeros(N, 1);
assert(all(sum(sm, 2) == 1), 'not a stochastic matrix');
smc = cumsum(sm, 2); % cummulative matrix
xi = find(bins == s0);
for k = 1 : N
    yi = find(smc(xi, :) > rand, 1, 'first');
    states(k) = bins(yi);
    xi = yi;
end
%% check result
ds = [states(1); diff(states)];
idx_begin = find(ds == 1 & states == 1);
idx_end = find(ds == -1 & states == 0);
if idx_end(end) < idx_begin(end)
    idx_end = [idx_end; N + 1];
end
df = idx_end - idx_begin;
fprintf('prob(state = 1) = %g; avg. time(state = 1) = %g\n', sum(states) / N, mean(df));

Otros consejos

The average length of the excited state is 5. The average length of the normal state, should thus be around 12 to obtain.

The strategy can be something like this.

  • Start in state 0
  • Draw a random number a from a Poisson distribution with mean L*(1-f)/f
  • Fill the state array with a zeroes
  • Draw a random number b from a Poission distribution with mean L
  • Fill the state array witb b ones.
  • Repeat

Another option would be to think in terms of switching probabilities, where the 0->1 and 1->0 probabilities are unequal.

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