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Given a variable x, with dom(x) = {1,0}

p(x = 1) refers to the probability of variable x to be in state '1'

p(x = 0) refers to the probability of variable x to be in state '0' = 1 - p(x = 1)

  1. What does it mean by p(x)?

Also, given a distribution

p(a,b,c,d,e) = p(d|a,b)p(e|c)p(a)p(b)p(c)

dom(a) = dom(b) = dom(c) = dom(d) = dom (e) = { 0 , 1 }

If one is asked the p(a = 1) given an observation that b = 1.

We solve this (if i am not mistaken): p(a=1|b=1) = p(a=1,b=1)/p(b=1) = marginalizing over variables c,d,e p(a=1,b=1,c,d,e) / marginalizing over variables a,c,d,e p(a,b=1,c,d,e)

Similarly, how do you interpret the statement p(a|b=1) in this context?

Been wondering about this for weeks now. o.O

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up vote 1 down vote accepted

When you get an expression like $p(x)$ or $p(a|b=1)$, this represents the whole distribution of $x$, or of $a$ given a value of $b$, rather than a single number as in $p(x=1)$. The notation is, admittedly, not so clear. To be precise, you'd think of $p(x)$ as a function from $x$'s domain $\{0,1\}$ to the probability of each of $x$'s possible values.

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Does that[p(x)] mean computing the individual probabilities of x being in each state in its domain? – dumbo Aug 10 '12 at 7:33
If I'm following you correctly, yes. – Kevin Carlson Aug 10 '12 at 11:59

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