Full credit will only be given to correct answers with a clear explanation of how they are obtained. Use additional paper as necessary.

1. If only the S/W property of nucleotide bases is considered, one can generate a binary Markov chain with transition probability matrix
   
   

   P  =	(	0.6	0.4	)
   		0.4	0.6

   
   
   Find the stationary distribution (f(s), f(w)) of this Markov chain. Enter your answers in the table below.
   
   

   x	f(x)
   S
   W

   
   
   Solution: Using the fact that when at the stationary distribution, (f(S), f(W)) * P = (f(S), f(W)) we get the equations
   f(S) = (0.6)*f(S) + (0.4)*f(W)
   f(W) = (0.4)*f(S) = (0.6)*f(W)
   We know that f(S) + f(W) = 1.
   Simultaneously solve f(S) + f(W) = 1
   and either
   f(S) = (0.6)*f(S) + (0.4)*f(W)
   or
   f(W) = (0.4)*f(S) = (0.6)*f(W)
   to get
   f(S) = 0.5 and f(W) = 0.5.
2. Write down the system of linear equations you'll need to solve in order to find the stationary distribution of the Markov nucleotide sequence with the transition probability matrix in problem 1 of Exercise 2A. You do not have to solve the system unless you have plenty of time or a calculator that does matrix calculations.
   
   Solution: Recall the matrix from problem 1 of Exercise 2a:
   
   
   
   

   P  =	(	0.2	0.3	0.1		)
   		0.4		0.1	0.2
   		0.5	0.1		0.2
   			0.2	0.3	0.4

   
   Once the stationary distribution has been reached, we know that
   (f(A), f(C), f(G), f(T)) * P = (f(A), f(C), f(G), f(T))
   
   Using matrix multiplication, we get the equations
   
   f(A) = 0.2 * f(A) + 0.4 * f(C) + 0.5 * f(G) + 0.1 f(T)
   f(C) = 0.3 * f(A) + 0.3 * f(C) + 0.1 * f(G) + 0.2 f(T)
   f(G) = 0.1 * f(A) + 0.1 * f(C) + 0.2 * f(G) + 0.3 f(T)
   f(T) = 0.4 * f(A) + 0.2 * f(C) + 0.2 * f(G) + 0.4 f(T)
   
   The system to solve is comprised of any 3 of the above 4 equations along with the equation
   1 = f(A) + f(C) + f(G) + f(T)