If I define a Markov chain,

$$

A\to B\to C

\\P(A,B,C)=P(A)P(B|A)P(C|B)

$$

Can I derive an expression for $P(C|A)$?

I feel like this should be $P(C|B)P(B|A)$, but I am not seeing how to prove it.

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# Probability – Understanding Markov Chain and Conditional Probabilities

###### Related Question

conditional probabilitymarkov-processprobability

If I define a Markov chain,

$$

A\to B\to C

\\P(A,B,C)=P(A)P(B|A)P(C|B)

$$

Can I derive an expression for $P(C|A)$?

I feel like this should be $P(C|B)P(B|A)$, but I am not seeing how to prove it.

## Best Answer

Your direction is correct, but you need to sum that expression over all possible $b$:

$$P(C=c|A=a)=\sum_b P(B=b|A=a)P(C=c|B=b)$$

It should cover all the paths from $A=a$ to $C=c$.