how to correct randsrc in matlab

Hi, we have an in-built function called 'randsrc(m,n)', where ideally it should generate an m-by-n matrix, each of whose entries independently takes the value -1 with probability 1/2, and 1 with probability 1/2.

Problem: I have used

X = [-5   -1 3  5  7];
pmf = [0.1  0.25  0.15  0.3 0.2] ;
out=randsrc(100, 1, [X; pmf] )

and im not getting a the rigth probability .how can i fix it

Please help me in finding out the solution for this.

[Walter Roberson]

What difficulty are you observing?

When I test, I see the expected fractions of symbols, to within statistical bounds (e.g., my test was slightly low on 3 and slightly high on 7)

Are you wanting to produce results with exactly the given fractions? So for 0.1 you would want exactly 100*0.1 = 10 copies of -5, exactly 25 copies of -1, exactly 15 copies of 3, and so on?

[alex korhin]

i want the exactly probability that was asking

[Nguyen Tien Dat]

Hello Alex, I am facing the same issue with you when using this algorithm. Have you figured out how to resolve it by this algorithm or you tried another one?

[Steven Lord]

Build a "deck of cards" and draw from it.

X = [-5   -1 3  5  7];
pmf = [0.1  0.25  0.15  0.3 0.2] ;
totalNumber = 100;
numberOfEach = round(pmf*totalNumber)

numberOfEach = 1×5

10 25 15 30 20

if sum(numberOfEach) ~= totalNumber
    error("Unable to generate exactly " + totalNumber + ...
        " elements from probability vector")
end
deck = repelem(X, numberOfEach);
shuffledDeck = deck(randperm(totalNumber));

% Show the shuffled deck
disp(reshape(shuffledDeck, 10, 10))

     7    -1     3     7    -1     5     5    -1     5     5
     7     7    -5    -1     3     5     3    -5     3     3
     3     5    -1     7     3    -1    -1     5    -1     5
    -5     7     5     5     3     5     5     5    -5     5
     5    -1    -1     7     7     5     5     5     5    -1
    -5     5    -1    -1    -1     7    -1     5     5    -1
    -5     3    -5     3     3    -1     7     5     7     7
    -1     5     3     7     5     7    -1     7    -5    -1
     3     7     5     5     3    -1     3     5     7    -1
    -1    -5    -1     5     7     5     7    -5     7    -1

% Check
fprintf("There should be " + numberOfEach(3) + " copies of " + X(3) + ...
    " in the deck and there are " + nnz(deck == X(3)) + ".")

There should be 15 copies of 3 in the deck and there are 15.

Answer 1

[Walter Roberson]

N = 100;
X = [-5   -1 3  5  7];
pmf = [0.1  0.25  0.15  0.3 0.2];

%round() by itself could result in overallocation if you are not careful.
howmany = round(pmf * N); howmany(end) = N - sum(howmany(1:end-1));

[~, ~, bin] = histcounts(1:N,1+[0 cumsum(howmany)]);
bin = bin(randperm(length(bin))) .';
out = X(bin);

[Steven Lord]

Once you've computed howmany, you could use repelem instead of histcounts to create the sorted vector of replicates of elements of X. Then you just need to shuffle them using randperm.

[alex korhin]

i need random numbers

[Walter Roberson]

randperm uses rand() internally.