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The conditional probability formula is given by: $$\mathbb{P}(A|B)\mathbb{P}(B)=\mathbb{P}(A\cap B)$$

How do I use this result directly to prove the following? $$\mathbb{P}(X_0=i_0,X_1=i_1,...,X_t=i_t)=\mathbb{P}(X_0=i_0)\times\mathbb{P}(X_1=i_1|X_0=i_0)\times...\times\mathbb{P}(X_t=i_t|\{X_{t-1}=i_{t-1},...X_0=i_0\})$$

Note: I can see how this statement holds true when applying this case by case for $t=1,2,3,...$ (or also by induction), but I am asking how we can see this from the direct use of the probability formula

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  • $\begingroup$ I wouldn't know how to prove this without the use of induction. $\endgroup$ – asdq Jan 21 '18 at 20:20
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You don't need induction to apply the multiplication rule $n$ times, i.e., $$ \mathbb{P}\left(\bigcap_{i=1}^n A_i\right) = \mathbb{P}\left(A_1 \cap\left( \bigcap_{i=2}^n A_i \right)\right) = \mathbb{P}\left(A_1\right) \mathbb{P}\left(\bigcap_{i=2}^n A_i | A_1\right), $$ and then apply the rule alliteratively $n-1$ times to get $$ \mathbb{P}\left(\bigcap_{i=1}^n A_i\right) = \mathbb{P}\left( A_1 \right) \mathbb{P}\left( A_2 | A_1 \right)\cdots\mathbb{P}\left(A_n|\bigcap_{i=1}^n A_i\right) . $$

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