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How to calculate the conditional probability of an event?

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Myriam Moss
Myriam Moss on 23 Apr 2021
Answered: William on 25 Apr 2021
I have an array similar to this array = [A A B A C A B B B C C A A C]. I want to calculate p(C|A), p(C|B), p(C|C). How can I do this just having this information? I want to know what is the probability of C happening after a previous event A, B or C.

Accepted Answer

William
William on 23 Apr 2021
Hello Myriam. It is not clear whether A, B, and C here are text characters, or if they have numeric values. If we assume they have numerical values, like A=1, B=2 and C=3, then you could use
y = diff(array);
P_AC = sum(y==2);
P_BC = sum(y==1);
P_CC = sum(y==0);
  1 Comment
Myriam Moss
Myriam Moss on 24 Apr 2021
Hi William. Thank you. They are characters.
I'm new to matlab sorry. Could you explain to me your logic, please?

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More Answers (2)

William
William on 25 Apr 2021
Actually, I believe that p(C|A) would be:
y = strfind(array, 'A');
N_A = length(y);
p_CA = N_AC/N_A;
There is one further thing to consider, though. It may be true that the very last element of array is an 'A'. I don't think this should be counted in N_A because we don't know whether it would have been followed by a 'C' or not. So, if the last element of array is 'A', we should reduce N_A by 1.
y = strfind(array, 'A');
N_A = length(y);
if y(end)==length(array) || y(end)==length(array)-1 % The string might end
N_A = N_A - 1; % with an 'A' or an 'A '
end
p_CA = N_AC/N_A;

William
William on 25 Apr 2021
Myriam,
If A, B and C were variables with the values 1, 2 and 3, then in your example:
array = [1, 1, 2, 1, 3, 1, 2, 2, 2, 3, 3, 1, 1, 3]
The diff() function returns the difference between each value and the next value, so
diff(array) = [0, 1, -2, 2, -2, 1, 0, 0, 1, 0, -2, 0, 2]
Every time an A is followed by a C, the difference is 2. Every time a B is followed by a C, the difference is 1. So, I was suggesting that you count the number of times A is followed by C by counting the number of times that the value 2 appears in diff(array) with a statement like c = sum(diff(array) == 2). Unfortunately, I see now that this does not work correctly for the number of times B is followed by C, because this results in a value of 1 in diff(array), and a value of 1 is also produced when an A is followed by a B.
Since you have said that A, B and C are characters, I assume that you mean that:
array = 'A A B A C A B B B C C A A C';
In this case, maybe a better solution would be:
y = strfind(array, 'A C');
N_AC = length(y);
y = strfind(array, 'B C');
N_BC = length(y);
  1 Comment
Myriam Moss
Myriam Moss on 25 Apr 2021
Thank you William! Now I have the number of times C appears after A and B.
If I define
y = strfind(array, 'C');
N_C = length(y);
If I want p(C|A), for example, I should do:
p_CA = N_AC/N_C , do you agree? :)

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