Discrete Time Markov Chains: Steady State Distributions

Suppose that bases (letters) in DNA sequences can be modeled as a Markov chain with state space (A, C, G, T) and transition matrix

P = rbind(
  c(0.36, 0.15, 0.20, 0.29),
  c(0.41, 0.16, 0.04, 0.39),
  c(0.30, 0.21, 0.14, 0.35),
  c(0.31, 0.19, 0.18, 0.32)
)

	A	C	G	T
A	0.36	0.15	0.20	0.29
C	0.41	0.16	0.04	0.39
G	0.30	0.21	0.14	0.35
T	0.31	0.19	0.18	0.32

1. One of the following is the unique stationary distribution. Identify which one it is and explain your reasoning conceptually without doing any calculations.

$$\begin{array}{rl}{\mathit{\pi }}_1& =[0.1742,0.3430,0.3266,0.1562]\\ {\mathit{\pi }}_2& =[0.3430,0.1742,0.1562,0.3266]\\ {\mathit{\pi }}_3& =[0.4935,0.0093,0.0057,0.4915]\end{array}$$

2. Now use software to find the stationary distribution.

3. Find the probability that C is followed two letters later by T.

4. Find the probability that A is followed two letters later by T.

5. Find the probability that C is followed 100 letters later by T.

6. Find the probability that A is followed 100 letters later by T.

7. Find the probability that a three letter sequences spell “CAT”.