matrix multiplication with probability

Hi, I want to assign a probability of multiplication for each value in a 3X3 matrix and multiply it by a random number. Can anyone please help me with this.

2 Comments

An example would help our understanding.
Thank you Jan for your answer, I sincerely appreciate it. but my requirement is for example
A = [1 0.5 1; 0.6 0.3 0.4; 0.4 0.5 1];
K= 1:1:10
a = rand;
C= [a 0 0]; here i want to assign probabilities to the 3 positions of 'C' where 'a' can be assigned i.e C = [10% 20% 30%] so during every iteration, a random 'a' is generated and it is placed in one of the position of 'C' based on the assigned probability and 'C' will get multiplied with 'A'

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 Accepted Answer

If p is a matrix the same size as A with the probability values you describe, do this:
SA = size(A);
a = rand(SA);
A = A + (A.*a-A).*(rand(SA)<=p);
At each element of A it is multiplied by the random number in ’a’ with probability in the corresponding element in p. Otherwise the element is unchanged.
It is possible you meant 'a' to be a single random scalar value rather than a matrix of them. If so, change the code to:
a = rand;
A = A + ((a-1)*A).*(rand(size(A))<=p);

3 Comments

Thank you very much, Mr.Roger Stafford. what if i want a vector 'C' of 3 rowsX1 column. The 'a' should be positioned in any one of the element of the 'C' based on my pre-defined probability of P= [10%;20%30%]. once the random 'a' is positioned in any one of the elements of 'C' that 'C' will be multiplied with 'A' matrix i.e B= A*C. Can you please help me for this.
I will help with your comment request by showing you how to produce an index with the requested probabilities. You can then use that index in dealing appropriately with your matrix A.
P = [.1;.2;.3];
CP = cumsum(P);
r = rand;
ix = sum(CP<=r)+1;
Given the values in P, the index ‘ix’ will assume the values 1, 2, 3, or 4 with respective probabilities .1, .2, .3, and 1-(.1+.2+.3)=.4 . If you wish to do something in the case ix=4, you can write
if ix<=3
% Handle the ix = 1,2,3 cases
else
% Handle the ix = 4 case
end
Otherwise, just omit the ‘else’ part.
I got it. thank you very much!.

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More Answers (1)

Maybe:
A = [1 0.5 1; 0.6 0.3 0.4; 0.4 0.5 1];
for k = 1:10
a = rand;
B = A * a;
...
end

1 Comment

Thank you Jan for your answer, I sincerely appreciate it. but my requirement is for example
A = [1 0.5 1; 0.6 0.3 0.4; 0.4 0.5 1];
K= 1:1:10
a = rand;
C= [a 0 0]; here i want to assign probabilities to the 3 positions of 'C' where 'a' can be assigned i.e C = [10% 20% 30%] so during every iteration, a random 'a' is generated and it is placed in one of the position of 'C' based on the assigned probability and 'C' will get multiplied with 'A'

Sign in to comment.

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