Conditional Probability on Intermediate Event

[3.141592654]

I have seen in class the following formula used:

P(A | C) = \sum_{B} P(A | B \cap C)*P(B | C)

I don't understand how this formula works? Can anyone help me understand how it can be derived and how I can understand it intuitively? Can a venn diagram be drawn to illustrate this formula?

 

[SW VandeCarr]

I have seen in class the following formula used:

P(A | C) = \sum_{B} P(A | B \cap C)*P(B | C)

I don't understand how this formula works? Can anyone help me understand how it can be derived and how I can understand it intuitively? Can a venn diagram be drawn to illustrate this formula?

It seems you need a B somewhere on the left side of the equation for it to make sense (ie P(A|B,C))

 

[3.141592654]

This is a formula that is seen, for example, in Uniformization:

P_{ij}(t) = P(X(t) = j|X(0) = i)

= \sum^{n=0}_{\infty} P(X(t) = j|X(0) = i, N(t) = n)P(N(t) = n|X(0) = i)

= \sum^{\infty}_{n=0} P^{n}_{ij} \frac{(e^{vt})(vt^n)}{n!}

Where P_{ij}(t) is the transition probability in a continuous time Markov Chain.

Going from the first to second line above is where I got the generalized equation I presented in my original post.

 

[chiro]

This is a formula that is seen, for example, in Uniformization:

P_{ij}(t) = P(X(t) = j|X(0) = i)

= \sum^{n=0}_{\infty} P(X(t) = j|X(0) = i, N(t) = n)P(N(t) = n|X(0) = i)

= \sum^{\infty}_{n=0} P^{n}_{ij} \frac{(e^{vt})(vt^n)}{n!}

Where P_{ij}(t) is the transition probability in a continuous time Markov Chain.

Going from the first to second line above is where I got the generalized equation I presented in my original post.

Hey 3.141592654.

You should try going from the fundamental matrix relationship for continuous time markov chains which is in the form:

dP/dt = PQ for a valid matrix A where P is your transition matrix. You should then get the general solution P(t) = e^(tQ) for t >= 0. You can use properties of operator algebras for any general expression in terms of calculating or you can use some algebra to find closed for expression for p(i,j)(t).

 

[Stephen Tashi]

I have seen in class the following formula used:

P(A | C) = \sum_{B} P(A | B \cap C)*P(B | C)

I don't understand how this formula works? Can anyone help me understand how it can be derived and how I can understand it intuitively? Can a venn diagram be drawn to illustrate this formula?

Do you understand the formula P(A) = \sum_{B} P(A| B) P(B)?

The formula you quoted above is the same formula.

(In both formulas, the variable B must range over a collection of mutually exclusive sets whose union contains A. In problems, this collection of sets is usually a "partition" of the entire probability space. )

The notation "|" for "given" cannot be captured by the visual appearance of a Venn diagram. The event denoted by "X|Y " and the event denoted by X \cap Y are the same set of points.

The notation P(X|Y) implies that we consider the "probability space" to be the set Y.

The notation P(X \cap Y) implies that we consider the whole probability space to be whatever it is in the statement of the problem, before any conditions are mentioned.

Any "law" of probability like P(A^c) = 1 - P(A) is understood to apply within some given probability space of "all possible outcomes". If we let S represent this space then we could write P(A^c|S) = 1 - P(A|S). Usually we leave the space S out of our notation.

The formula P(A) = \sum_{B} P(A|B) P(B) applied when the probability space is C becomes:

P(A|C) = \sum_{B} P( (A|B) | C) P(B|C)

This leaves only the problem of interpreting P( (A|B) | C). We have to argue that this is the same as P(A | B \cap C). It might be a tangle of words to do so, but I hope it is intuitively clear.