Solutions to Additional Problems

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Problem 1:

The transition matrix is:

P = \left[\begin{array}{cccc} .75 & .25 & 0 & 0 \\ .25 & .50 & .25 & 0 \\ .25 & 0 & .50 & .25 \\ .50 & 0 & 0 & .50 \end{array}\right].

The matrix satisfies the conditions of LN8, Theorem 1, so we expect there exists a unique stationary distribution π = (π0,π1,π2,π3) such that:

\pi_j = \sum_{i=0}^3 \pi_i P_{ij}, \quad j=0,1,2,3,

and

\sum_{i=0}^3 \pi_i = 1.

We can write this as a set of linear equations:

\begin{align} \pi_0 &= \frac{3}{4} \pi_0 + \frac{1}{4} \pi_1 + \frac{1}{4} \pi_2 + \frac{2}{4} \pi_3 \\ \pi_1 &= \frac{1}{4} \pi_0 + \frac{2}{4} \pi_1 \\ \pi_2 &= \frac{1}{4} \pi_1 + \frac{2}{4} \pi_2 \\ \pi_3 &= \frac{1}{4} \pi_2 + \frac{2}{4} \pi_3 \\ 1 &= \pi_0 + \pi_1 + \pi_2 + \pi_3 \end{align}

This is easy to solve by hand, and we get the solution:

\begin{align} \pi_0 &= \frac{8}{15} \approx 0.533 \\ \pi_1 &= \frac{4}{15} \approx 0.267 \\ \pi_2 &= \frac{2}{15} \approx 0.133 \\ \pi_3 &= \frac{1}{15} \approx 0.067 \end{align}

Problem 2:

Since Y1 has a N(1,4) distribution,

E[Y1] = 1.

Then, using Yt = Yt − 1 + εt,

E[Y2] = E[Y1] + E[ε2] = 1 + 0 = 1.

Proceeding recursively we see that for all t, we have E[Yt] = E[Yt − 1], so

E[Yt] = 1.

Using a similar argument,

V[Y1] = 4,

V[Y2] = V[Y1] + V[ε2] = 4 + 1 = 5,

and in general,

V[Yt] = 3 + t.

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