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matrix multiplication with probability

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Santhosh Chandrasekar
Santhosh Chandrasekar am 11 Feb. 2018
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 Kommentare
Walter Roberson
Walter Roberson am 11 Feb. 2018
An example would help our understanding.
Santhosh Chandrasekar
Santhosh Chandrasekar am 11 Feb. 2018
Bearbeitet: Santhosh Chandrasekar am 12 Feb. 2018
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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Akzeptierte Antwort

Roger Stafford
Roger Stafford am 12 Feb. 2018
Bearbeitet: Roger Stafford am 12 Feb. 2018
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 Kommentare
Roger Stafford
Roger Stafford am 12 Feb. 2018
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.
Santhosh Chandrasekar
Santhosh Chandrasekar am 14 Feb. 2018
I got it. thank you very much!.

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Weitere Antworten (1)

Jan
Jan am 11 Feb. 2018
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 Kommentar
Santhosh Chandrasekar
Santhosh Chandrasekar am 12 Feb. 2018
Bearbeitet: Santhosh Chandrasekar am 12 Feb. 2018
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'

Melden Sie sich an, um zu kommentieren.

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