How to prove that $\mathbb{P}(A \cap B \cap C)=\mathbb{P}(A\mid C)\times \mathbb{P}(B\mid C)\times \mathbb{P}(C)$

Question

I am trying to prove ${\mathbb{P}(\mathrm{A \cap B \cap C)}=\mathbb{P}(\mathrm{A }\mid C)\times \mathbb{P}(\mathrm{B}\mid C)\times \mathbb{P}(\mathrm{C})}$, where $\mathbb{P}$ denotes probability and $A, B, C$ are three events where $\mathbb{P}(\mathrm{C})>0$.

Here is how I try to prove :-

$\mathbb{P}(\mathrm{A \cap B \cap C)}\\=\mathbb{P}((\mathrm{A \cap B)\mid C)}\times \mathbb{P}(\mathrm{C})\ \longrightarrow (1)$

Now my question is -

1. Can we write $\mathbb{P}((\mathrm{A \cap B)\mid C)}=\mathbb{P}(\mathrm{A \mid C)}\times \mathbb{P}(\mathrm{B\mid C)}$ and if yes (or no), why can (or cannot) we do so?

2. If $1^{st}$ approach does not apply, then how do we plug-in the result of conditional probability $$\mathbb{P}(\mathrm{A \cap B)}=\mathbb{P}\mathrm{(B \mid A)}\times \mathbb{P}(\mathrm{A})$$ in (1) and how do we go to the desired final expression?

Alternatively trying to prove this in the following way, leads me to :-

$$\mathbb{P}(\mathrm{A \cap B \cap C)}=\mathbb{P}(\mathrm{C\mid A \cap B)}\times \mathbb{P}(\mathrm{A \cap B)}\\=\mathbb{P}(\mathrm{C\mid A \cap B)}\times\mathbb{P}\mathrm{(B \mid A)}\times \mathbb{P}(\mathrm{A})$$, something similar to the final desired expression.

If I am even close to prove this expression I still surely miss something for which I am stuck. All helps and explanations are welcome, appreciated and valuable.

I have read these three posts, yet cannot reach at the final expression.

Answer 1

Conditional probabilities are ordinary probabilities defined in conditional Universe, and any formula for conditional probabilities has analog formula for original probabilities (defined in original Universe).

You are asking can we write $$\mathbb{P}((\mathrm{A \cap B)\mid C)}=\mathbb{P}(\mathrm{A \mid C)}\times \mathbb{P}(\mathrm{B\mid C)}$$ The analog formula is $$\mathbb{P}(\mathrm{A \cap B)}=\mathbb{P}(\mathrm{A})\times \mathbb{P}(\mathrm{B)}$$ The last formula is correct if events A and B are independent (it is in fact definition of independence). You can guess that the first formula is definition of conditional independence (given C).

As for your original formula $${\mathbb{P}(\mathrm{A \cap B \cap C)}=\mathbb{P}(\mathrm{A }\mid C)\times \mathbb{P}(\mathrm{B}\mid C)\times \mathbb{P}(\mathrm{C})}$$

it is correct if 2 conditions are satisfied:

1. A and B are conditionally independent given C: $$\mathbb{P}((\mathrm{A \cap B)\mid C)}=\mathbb{P}(\mathrm{A \mid C)}\times \mathbb{P}(\mathrm{B\mid C)}$$
2. $\mathrm{A} \cap \mathrm{B}$ and C are independent: $$\mathbb{P}((\mathrm{A \cap B)\cap C)}=\mathbb{P}(\mathrm{A \cap B)}\times \mathbb{P}(\mathrm{ C})$$

Answer 2

This is a form of conditional independence and it does not always hold.

Counterexample: Say you are tossing a fair penny and a fair dime independently.

Let $A$ be the event "the penny comes up $H$".

Let $B$ be the event "the dime comes up $H$"

Let $C$ be the event "the two coins do not match."

Then $A\cap B\cap C=\emptyset$ so the left hand of your desired expression is $0$. But none of the factors on the right are $0$. $A,C$ are independent as are $B,C$ so the right hand, in this case is $\frac 12\times \frac 12\times \frac 12=\frac 18$.

For an even cheaper sort of counterexample, consider the case where $A,B$ are mutually exclusive (but $A\cap C$ and $B\cap C$ are both positive probability). Then the left hand is $0$ but the right hand is not.

On the positive side, note that the claim is true if $A,B,C$ are mutually independent events (my example with the coins concerns pairwise independent events). In that case, both sides reduce to $P(A)\times P(B)\times P(C)$ so equality does hold (under the very strong assumption of mutual independence).

Answer 3

As an addition to the answer from lulu: This is True if and only if $A$ and $B$ are stochastically independent given $C$. In this Case $$P(A \cap B\cap C) = P(A\cap B|C)P(C) = P(A|C)P(B|C)P(C)$$

example: Consider this two stage bet:

In the first round you bet on the coin flip. You get 1 if it comes up head and 0 otherwise. If you win in the first round you are allowed in 50% of cases to participate in a second round of Heads or Tail with the same payoffs. Consider the following events:

$A$ = you win at the first round

$B$ = you win at the second round

$C$ = you participate in the second round

Now $A$ and $B$ are clearly not independent since in case you loose the first round you are not allowed to participate in the second and can therefore not win there. However, given that you participate in the second round your chances of winning are independent of the first round since the Coin Flips are independent.