# A Markov chain {Xn, n ≥ 0} with states 1, 2,3 has the transition probability matrix with an initial distribution (1/2,0,1/2), what is P(X1=3|X2=1)

A Markov chain {Xn, n ≥ 0} with states 1, 2,3 has the transition probability matrix P

$$\begin{bmatrix}0&0.4&0.6\\1&0&0\\0.3&0.3&0.4\end{bmatrix}$$

with an initial distribution A (0.5,0,0.5), what is $$P(X_1=3|X_2=1)$$?

(I know a Markov chain property is the future, given the present, is independent of the past. the question here look like given future, what is the probability of the past? I am wondering

$$P(X_1=3|X_2=1)=P(X_1=3)=A_3=0.5$$

or $$P(X_1=3|X_2=1)=P_{13}=0.6$$ or else?

• Please add the self-study tag to your question Feb 16, 2021 at 7:17

1. the transition matrix $$\mathsf P$$ is made of the probabilities $$\mathbb P(X_t=j|X_{t-1}=i)$$ as $$(i,j)$$ entries,
2. the probability $$\mathbb P(X_{t-1}=j|X_{t}=i)$$ can be written as$$\mathbb P(X_{t-1}=j|X_{t}=i)=\dfrac{\mathbb P(X_t=i|X_{t-1}=j)\mathbb P(X_{t-1}=j)}{\mathbb P(X_t=i)}$$by Bayes' theorem,
3. the probability $$\mathbb P(X_t=i)$$ can be written as$$\mathbb P(X_t=i)=\sum_{j=1}^3 \mathbb P(X_t=i|X_{t-1}=j)\mathbb P(X_{t-1}=j)$$by the law of total probability,
the reverse probabilities $$\mathbb P(X_{1}=j|X_{2}=i)$$ can be derived from the entries of the homework question.