# Using the Cox axioms to derive unknown probabilities from known probabilities

To strengthen my understanding of fundamental probability theory, I am working my way through Professor Aaron Hertzmann's Introduction to Bayesian Learning course notes. Section 3.8 of these course notes includes the following exercise.

### Original Problem

Derive a formula for $P(\mathbf{A})$, assuming you know $P(\mathbf{A}|\mathbf{B}_1,\mathbf{C})$ and $P(\mathbf{A}|\mathbf{B}_2,\mathbf{C})$ and $P(\mathbf{B}_1|\mathbf{C})+P(\mathbf{B}_2|\mathbf{C})=1$.

### My Attempt

So, my goal is to express $P(\mathbf{A})$ in terms of known quantities.

\begin{align} P(\mathbf{B}_1|\mathbf{C}) &+ P(\mathbf{B}_2|\mathbf{C}) &=& 1 && \text{}\\ P(\mathbf{B}_1|\mathbf{C})P(\mathbf{C}) &+ P(\mathbf{B}_2|\mathbf{C})P(\mathbf{C}) &=& P(\mathbf{C}) && \text{Algebra}\\ P(\mathbf{B}_1,\mathbf{C}) &+ P(\mathbf{B}_2,\mathbf{C}) &=& P(\mathbf{C}) && \text{Product Rule}\\ P(\mathbf{B}_1,\mathbf{C}|\mathbf{A}) &+ P(\mathbf{B}_2,\mathbf{C}|\mathbf{A}) &=& P(\mathbf{C}|\mathbf{A}) && \text{Condition on } \mathbf{A}\\ \frac{P(\mathbf{A}|\mathbf{B}_1,\mathbf{C})P(\mathbf{B}_1,\mathbf{C})}{P(\mathbf{A})} &+ \frac{P(\mathbf{A}|\mathbf{B}_2,\mathbf{C})P(\mathbf{B}_2,\mathbf{C})}{P(\mathbf{A})} &=& P(\mathbf{C}|\mathbf{A}) && \text{Bayes Theorem} \\ \end{align}

\begin{align} &\frac{P(\mathbf{A}|\mathbf{B}_1,\mathbf{C})P(\mathbf{B}_1,\mathbf{C}) + P(\mathbf{A}|\mathbf{B}_2,\mathbf{C})P(\mathbf{B}_2,\mathbf{C})}{P(\mathbf{A})} &=& P(\mathbf{C}|\mathbf{A}) && \text{Algebra}~~~~~~~~~~~~~~\\ &\frac{P(\mathbf{A}|\mathbf{B}_1,\mathbf{C})P(\mathbf{B}_1,\mathbf{C}) + P(\mathbf{A}|\mathbf{B}_2,\mathbf{C})P(\mathbf{B}_2,\mathbf{C})}{P(\mathbf{C}|\mathbf{A})} &=& P(\mathbf{A}) && \text{Algebra}~~~~~~~~~~~~~~\\ \end{align}

Here is where I get stuck, because I don't know $P(\mathbf{C}|\mathbf{A})$, $P(\mathbf{B}_1,\mathbf{C})$, or $P(\mathbf{B}_2,\mathbf{C})$. It is not clear to me how I can express these probabilities in terms of the quantities I know.

These course notes are very much a work in progress. The author frequently writes TO-DO notes to himself about things to add and remove. Is it possible that this exercise is actually under-determined? Or am I simply going about this derivation the wrong way?

### Similar Questions

When searching StackExchange, I found this similar question. However, it was still not directly applicable to my problem.

### Tagging as Homework

Although this question is not homework for me, I am tagging it as homework anyway because: (1) it is very introductory in nature; (2) a similar question could conceivably be homework for someone else; and (3) I would be happy with hints.

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You can get $P(A\mid C)$ from the total probability law but cannot remove the conditioning on $C$. –  Dilip Sarwate Dec 20 '11 at 20:17
Yep, a direct application of the law of total probability using the conditional probability $P( \cdot | C)$ gives you $P(A|C) = P(A|B_1,C) P(B_1|C) + P(A|B_2,C)P(B_2|C)$. So you need to know $P(C)$ to answer. –  Elvis Dec 21 '11 at 10:01