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Question regarding probability of observation

  1. Aug 4, 2013 #1
    Hi Everyone,

    I am a newbie in probability theory and following is my question:

    Consider we have two binary random variables A and B. B is dependent on A. So we have two conditional probability tables P(A) and P(B|A) with the following parameters :

    A P(A)
    F 0.3
    T 0.7

    A P(B=T|A)
    F 0.4
    T 0.6

    Suppose that A=T is observed. So, now the probability of A being True is 1.0 instead of 0.3 and P(A=F) = 0.0 instead of 0.7. Observing A=T the probability of B=T is going to be 0.4, by just looking up the corresponding tuple in B's CPT. Consider a different scenario where we now observe only B=T. My first question is,

    1) is P(B=T) going to be 1.0, since we have observed it, comparing to the first scenario of observing A=T? I know that if nothing is observed then P(B=T) is calculated as P(A=T)xP(B=T|A=T)+P(A=F)xP(B=T|A=F), which is the marginal probability of B=T.

    My second question which follows from the first one is,
    2) if P(B=T) = 1.0 when B=T has been observed, then while calculating the posterior probability of A=T given B=T i.e. P(A=T|B=T) why is it that we don't put P(B=T)=1.0 in the denominator of the following Baye's rule

    P(A=T|B=T) = P(A=T) x P(B=T|A=T) / P(B=T)

    Why do we use the marginal value of P(B=T) [when nothing is observed] computed by the expression
    P(A=T) x P(B=T|A=T) + P(A=F) x P(B=T|A=F)?

    Thanks in advance.
  2. jcsd
  3. Aug 5, 2013 #2

    Simon Bridge

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    OK - denote ##\small A## meaning ##\small A\leftarrow T## and ##\small \lnot A## meaning ##\small A \leftarrow F##.
    $$\small P(A)=0.7\\
    \small P(\lnot A)=0.3\\
    \small P(B|A)=0.6\\
    \small P(B|\lnot A)=0.4$$
    So, if ##\small A## then ##\small P(B)=0.6##
    We want to ask, if ##\small B## then what is ##\small P(A)## ?

    Construct a tree and figure out how many situations could lead to B (=T).
    That should allow you to confirm or refute your assertions.
  4. Aug 5, 2013 #3
    Thank you Simon for your reply. If A=T then is P(A) = 1.0? I can understand that if A=T then P(B) = 0.6. I am just thinking what the probability of an observed event should be, not the unobserved event given some observation. This is actually my first question.
  5. Aug 5, 2013 #4

    Simon Bridge

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    My response was in two parts.
    The first part hoped to tidy up the notation in a way that may help you think about the problem.
    The second part hoped to point you in the direction of finding the answers to your questions.

    If you observed A, then P(A) no longer has any meaning by itself.
    Subsequent observations may see A or not depending on the nature of the system.

    If you tossed a (biased) coin, and left it there, then subsequent measurements will be the same as the first one. eg. if A="the coin shows heads side up when I look at it" and the result was A, then you can say that P(A)=1 for subsequent measurements (observations of the coin without tossing it).

    i.e. P(A|A)=1.

    You use coin A to select which of two possible B-coins to pick.
    You toss the indicated one ...

    But you wanted to consider the consequences of doing the math in reverse.
    To understand how that works, you need to think what the reverse process is.
    The reverse experiment would be that someone else follows the procedure and I see only the result on the B coin ... what does this tell me about the probable states of the A coin?

    If you were thinking of a different experiment, then please describe it.
    Last edited: Aug 5, 2013
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