문제

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!

도움이 되었습니까?

해결책

%% 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));

다른 팁

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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