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I saw in many different context that any probability result that is true for unconditional probability remains true if everything is conditioned on some event. For instance, consider the following equation: $$P(B\cap C)=P(B)P(C|B).$$ Then by conditioning both sides on $A$, we get $$P(B\cap C|A)=P(B|A)P(C|B\cap A).$$ However, it is possible to find counterexamples as given here.

So I was wondering when this conditioning the both sides of equality operation is valid. I am, in particular, interested in the following equality: $$P(X_4, X_3, X_2, X_1)=P(X_4 \mid X_3, X_2, X_1)\cdot \mathrm P(X_3 \mid X_2, X_1)\cdot \mathrm P(X_2 \mid X_1)\cdot \mathrm P(X_1).$$ Is it possible to condition both sides on $A$ and get $$P(X_4, X_3, X_2, X_1|A)=P(X_4 \mid X_3, X_2, X_1,A)\cdot \mathrm P(X_3 \mid X_2, X_1,A)\cdot \mathrm P(X_2 \mid X_1,A)\cdot \mathrm P(X_1|A)$$ here?

Many thanks in advance.

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If $P(A) \neq 0$ then we always have $P(B\cap C|A)=P(B|A)P(C|B\cap A)$.

Here is how to prove it.

$$ P(B \cap C \cap A) =P(B \cap C | A) P(A) \tag{1}$$ \begin{align*} P(B \cap C \cap A) & = P(C \cap B \cap A) = \\& = P(C |B\cap A) P(B\cap A) = \\ & = P(C |B\cap A) P(B | A) P(A) \tag{2} \end{align*}

From $(1)$ and $(2)$,

$$ P(B \cap C | A) P(A) = P(C |B\cap A) P(B | A) P(A) $$

Since $P(A) \neq 0$, we have

$$ P(B \cap C | A) = P(C |B\cap A) P(B | A) $$

So, if $P(A) \neq 0$, by repeatedly applying the equality we have just proved, we have
$$ \mathrm P(X_4, X_3, X_2, X_1|A)= \mathrm P(X_4 \mid X_3, X_2, X_1,A)\cdot \mathrm P(X_3 \mid X_2, X_1,A)\cdot \mathrm P(X_2 \mid X_1,A)\cdot \mathrm P(X_1|A)$$

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