Compute probability of different states in a binary distribution

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Hello,
I have table array that is 300X1 consisting of a binary sequence [0,1,0,0,1,1,1,1,0,0,0,0,1...]
I would like to compute the probability that every second order value exists. Meaning how many 00, 01, 10, and 11 exist in the sequence. Do you have any Suggestions on how to go about it?.
Thank you
  2 comentarios
David Hill
David Hill el 1 de Oct. de 2019
Does 0 0 0 0 count as two 00 sequences and 1 1 1 1 0 counts as two 11 sequences or one 11 and one 10? In other words what are the rules for counting the sequences? Do you always start counting from the first element?
Danae Parissi
Danae Parissi el 1 de Oct. de 2019
I start counting from the first element and try to find the duads in a serial way. Meaning that in the sequence 01110000100110
there is 01 11 00 00 10 01 10
so there is 2 diplets of 01, one of 11, 2 of 00 and 2 of 10.

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Dimitris Kalogiros
Dimitris Kalogiros el 1 de Oct. de 2019
clc; clearvars
% input data
N=300; % always even
x=randi([0, 1], N, 1);
% two bits words
% 00->0 , 01->1 , 10->2 , 11->3
xmsb=x(1:2:end);
xlsb=x(2:2:end);
y=2*xmsb+xlsb;
% calculation of probability
[P,edges] = histcounts(y, 'Normalization', 'probability');
yIntervalCenters=(1/2)*(edges(1:end-1)+edges(2:end));
figure; stem(yIntervalCenters, P); ylabel('probability'); grid on;
  5 comentarios
Danae Parissi
Danae Parissi el 2 de Oct. de 2019
Could you help me to calculate the propability within a subset (let say within the first 30 values of the sequence?) and then the other 30 values and so on and so forth?
Dimitris Kalogiros
Dimitris Kalogiros el 2 de Oct. de 2019
You have just to apply this piece of code on chunks of the input data.
I'm giving an example:
clc; clearvars;
close all;
% input data
N=300; % always even
data=randi([0, 1], N, 1);
setLength=30;
for k=0:setLength:N-setLength
% extract an interval of data
x=data(k+1:1:k+setLength);
% two bits words
% 00->0 , 01->1 , 10->2 , 11->3
xmsb=x(1:2:end);
xlsb=x(2:2:end);
y=2*xmsb+xlsb;
% calculation of probability
edges=-0.5:1:3.5;
[P,~] = histcounts(y, edges, 'Normalization', 'probability');
yIntervalCenters=(1/2)*(edges(1:end-1)+edges(2:end));
figure(1);
stem(yIntervalCenters, P, 'LineWidth', 2); ylabel('probability');
grid on; ylim([0, 1]);
pause(1);
end
I've included some extra graphical tricks

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