How can I derive the mathematical relationship between P(A|B,C) with P(A,B), P(A,C) and, P(B,C) if any.

Question

I am trying to model a process mathematically, where I have three events. $(X=x)$, $(U=1)$, and $(T=t)$. I am interested in finding $\mathbb{P}(U=1\mid X=x,T=t)$.

From my model, I know that the event $(U=1)$ is dependent on $(X=x)$ and $(T=t)$, but $(X=x)$ and $(T=t)$ are not dependent on each other.

I hope to arrive at the relation $\mathbb{P}(U=1\mid X=x,T=t)=\mathbb{P}(U=1\mid X=x)\cdot \mathbb{P}(U=1\mid T=t)$.

Whether the above relation turns out to be true or not, I would like to know the relation between $\mathbb{P}(A,B,C)$ with $\mathbb{P}(A,B)$, $\mathbb{P}(A,C)$ and $\mathbb{P}(B,C)$, if any. So, I can take it from there and find a way forward.

Answer 1

I assume that by $P(A,B)$ you mean $P(A\bigcap B)$. Then the only formula we can get is :

$P(A,B,C)=P(A)\,P(B\mid A)\,P(C\mid A,B)$

Answer 2

I know that the event (U=1) is dependent on (X=x) and (T=t), but (X=x) and (T=t) are not dependent on each other.

Then we have: $$\mathsf P(U\,{=}\,1,X\,{=}\,x,T\,{=}\,t) ~=~ \mathsf P(U\,{=}\,1\mid X\,{=}\,x,T\,{=}\,t)\,\mathsf P(X\,{=}\,x)\,\mathsf P(T\,{=}\,t)$$

Generally: $\mathsf P(A,B,C) = \mathsf P(A\mid B,C)\,\mathsf P(B, C)$ .

And when $B,C$ are independent: $\mathsf P(B,C)=\mathsf P(B)\,\mathsf P(C)$.

I hope to arrive at the relation $\mathsf P(U\,{=}\,1\mid X\,{=}\,x,T\,{=}\,t)~=~\mathsf P(U\,{=}\,1\mid X\,{=}\,x)\cdot\mathsf P(U\,{=}\,1\mid T\,{=}\,t)$.

That's not a general result.

Answer 3

Let $B$ and $C$ be independent events. Let $\mathbb P(A,B) = q$, $\mathbb P(A,C) = r,$ and $\mathbb P(B,C) = s.$ Let $m = \min\{q,r,s\}.$ Note that $m$ defined in this way is the largest possible value of $\mathbb P(A,B,C).$

Suppose $m > 0$ and consider the following two joint distributions of $A, B, C,$ where the values of $t,$ $u,$ $v,$ and $w$ are chosen so as to make the events $B$ and $C$ independent in the distribution $f_1$:

\begin{array}{c|c|c|} & f_1 & f_2 \\ \hline \mathbb P(A,B,C) & 0 & m \\ \mathbb P(A^\complement,B,C) & s & s-m \\ \mathbb P(A,B^\complement,C) & r & r-m \\ \mathbb P(A^\complement,B^\complement,C) & t & t+m \\ \mathbb P(A,B,C^\complement) & q & q-m \\ \mathbb P(A^\complement,B,C^\complement) & u & u+m \\ \mathbb P(A,B^\complement,C^\complement) & v & v \\ \mathbb P(A^\complement,B^\complement,C^\complement) & w & w \end{array}

The probabilities of the four cases for $B$ and $C$ in $f_2$ are the same as in $f_1,$ so $B$ and $C$ are independent in $f_2$ as well.

But $\mathbb P(A\mid B,C) = 0$ for the joint distribution $f_1,$ whereas $\mathbb P(A\mid B,C) = m/s > 0$ for $f_2.$

By taking a distribution $kf_1 + (1-k)f_2$ you can make $\mathbb P(A\mid B,C)$ take on any value you like between $0$ and $m/s.$

So all you can conclude about $\mathbb P(A\mid B,C)$ from the three probabilities $\mathbb P(A,B) = q$, $\mathbb P(A,C) = r,$ and $\mathbb P(B,C) = s$ and the knowledge that $B$ and $C$ are independent is:

• If $m = \min\{q,r,s\} > 0$ then $\mathbb P(A\mid B,C) \leq m/s.$

• If $\min\{q,r\} = 0$ and $s > 0$ then $\mathbb P(A\mid B,C) = 0.$

• If $s = 0$ then $\mathbb P(A\mid B,C)$ is undefined.

Note also that in either $f_1$ or $f_2$ we find that $\mathbb P(A\mid B) = q/(q + s + u)$ and $\mathbb P(A\mid C) = r/(r + s + t),$ so knowing $\mathbb P(A\mid B)$ and $\mathbb P(A\mid C)$ will not gain any more information about $\mathbb P(A\mid B,C)$.

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Regarding the conditions under which $\mathbb P(A\mid B,C) = \mathbb P(A\mid B) \mathbb P(A\mid C)$:

If $A,$ $B,$ and $C$ are a set of mutually independent events, then $$\mathbb P(A\mid B,C) = \mathbb P(A\mid B) = \mathbb P(A\mid C) = \mathbb P(A),$$ so we have the requested relation only when $P(A) = 0$ or $P(A) = 1.$ In order for the probabilities to be non-trivial, $A$ must have a particular dependence on $B$ and $C.$

From the previous calculations we have $$ \mathbb P(A\mid B) \mathbb P(A\mid C) = \frac{qr}{(q + s + u)(r + s + t)} $$ while $\mathbb P(A\mid B,C) = \mathbb P(A,B,C)/s,$ so we need $$ \mathbb P(A,B,C) = \frac{qrs}{(q + s + u)(r + s + t)}. $$

I am finding it difficult to see a useful path toward your desired conclusion.