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Currently I am learning about Markov chains. In the solution of a problem I find the following statement.

$$ \mathbb{P}\left(X_4=2 \mid X_3 \neq 0, X_2 \neq 0, X_1 \neq 0, > X_0=2\right) $$ Using the definition of conditional probability, we can rewrite this as $$ \frac{\mathbb{P}\left(X_4=2, X_3 \neq 0, X_2 > \neq 0, X_1 \neq 0 \mid X_0=2\right)}{\mathbb{P}\left(X_3 \neq 0, X_2 > \neq 0, X_1 \neq 0 \mid X_0=2\right)} . $$

Unfortunately I don't see how the rule of conditional probability is applied here.

From my knowlegde the law of conditional probability is as follows:

$P(A \mid B)=\frac{P(A \cap B)}{P(B)}$

Following this rule I would say that:

$A: X_4=2$

$B: X_3 \neq 0, X_2 \neq 0, X_1 \neq 0, X_0=2$

Question: Why is the condition placed the way it is in the solution?

Tim
  • 684

1 Answers1

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Notice that the following is true:

$$\frac{P(A,B|C)}{P(B|C)} = \frac{\frac{P(A,B,C)}{P(C)}}{\frac{P(B,C)}{P(C)}} = \frac{P(A,B,C)}{P(B,C)}$$

so their form is equivalent to what you propose should be done.

Manifoldski
  • 848