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.