- #1
hughwf
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Hello,
This question relates to Bayes law. I think my problem is I am not sure of the name of the thing I am trying to derive...
I have 2 variables a and b.
a = 1 or 0, b = 0...n
I have the data to calculate;
p(a = 1 and b) p(b)
for any b. Hence I can find p(a=1|b) = p(a = 1 and b)/p(b)
What I want is p(a=1|b), but 'given' that a = 1. I don't want this to be affected by p(b), hence I am not trying to find p(b|a=1).
To explain further what i mean; If event a = 1, what is the prob it will happen at a certain b, independant of the frequency of occurences of different b's.
So I normalise;
\sum_{b = 0}^n p(a=1|b).N = 1
Where N is a constant.
N = \frac{1}{\sum_{b = 0}^n p(a=1|b) }
p(a=1,b) = \frac{p(a=1|b)}{\sum_{b = 0}^n p(a=1|b)}
is that alright and does it have a name?
Many thanks in advance for any advice...
Hugh
This question relates to Bayes law. I think my problem is I am not sure of the name of the thing I am trying to derive...
I have 2 variables a and b.
a = 1 or 0, b = 0...n
I have the data to calculate;
p(a = 1 and b) p(b)
for any b. Hence I can find p(a=1|b) = p(a = 1 and b)/p(b)
What I want is p(a=1|b), but 'given' that a = 1. I don't want this to be affected by p(b), hence I am not trying to find p(b|a=1).
To explain further what i mean; If event a = 1, what is the prob it will happen at a certain b, independant of the frequency of occurences of different b's.
So I normalise;
\sum_{b = 0}^n p(a=1|b).N = 1
Where N is a constant.
N = \frac{1}{\sum_{b = 0}^n p(a=1|b) }
p(a=1,b) = \frac{p(a=1|b)}{\sum_{b = 0}^n p(a=1|b)}
is that alright and does it have a name?
Many thanks in advance for any advice...
Hugh
