Let $\{X_t\}_{t=1}^4$ be a Markov chain with $t$ denoting the time index. Simplify the following factorisation. $$\Pr(X_4)\Pr(X_3|X_4)\Pr(X_2|X_3,X_4)\Pr(X_1|X_2,X_3,X_4)$$
I really don't know how Markov chains work, so some direction on how to factorise this would be great.
Apply the definition of conditional probability repeatedly, and you'll get $$ \Pr(X_4)\Pr(X_3|X_4)\Pr(X_2|X_3,X_4)\Pr(X_1|X_2,X_3,X_4)=\Pr(X_1,X_2,X_3,X_4) \tag1 $$ Now reverse the subscripts in (1) to get another identity: $$ \Pr(X_1)\Pr(X_2|X_1)\Pr(X_3|X_2,X_1)\Pr(X_4|X_3,X_2,X_1)=\Pr(X_4,X_3,X_2,X_1)\tag2 $$ But the RHS of (1) is the same as the RHS of (2). And, by the Markov property, the LHS of (2) simplifies to $$ \Pr(X_1)\Pr(X_2|X_1)\Pr(X_3|X_2)\Pr(X_4|X_3) $$