Multiple State Model : transition probability when state 0 can be re-entered

Question

In a multiple state model, where transition intensities are constant throughout the year : force of transition from state 0 to 1 = 0.05, force of transition from state 1 to 0 = 0.08, force of transition from state 1 to 2 = 0.15

I am asked to verify that tpx00 (the probability of being in state 0 at time t if starting in state 0 at time 0) = (10e^-0.03t + e^-0.25t) / 11

I am confused because there can be infinite transition from 0 to 1, back to 0, then back to 1 , over and over again as long as the final transition is to state 0. So how do I account for this and derive the above equation?

Any help would be immensely appreciated!!!

Answer 1

If i correctly understand the situation, $2$ is an absorbing state, and we have a Markov chain with matrix $$ A=\begin{bmatrix} \frac{19}{20} & \frac{1}{20} & 0 \\ \frac{2}{25} & \frac{77}{100} & \frac{3}{20} \\ 0 & 0 & 1 \end{bmatrix}\ . $$ The solution with a program like sage is as follows:

sage: A = matrix( QQ, 3,3,[ 0.95, 0.05, 0, 0.08, 0.77, 0.15, 0, 0, 1 ] )
sage: var('t');
sage: v = vector( QQ, [1,0,0] )
sage: v * A^t * v
10/11*97^t/100^t + 1/11*3^t/4^t

Why?

The given matrix is diagonalizable, its diagonal / Jordan form is the following one, sage again (in this century, life is too short else):

sage: J, S = A.jordan_form(transformation=True)
sage: J
[     1|     0|     0]
[------+------+------]
[     0|97/100|     0]
[------+------+------]
[     0|     0|   3/4]
sage: S
[  1   1   1]
[  1 2/5  -4]
[  1   0   0]
sage: A == S*J*S.inverse()
True

This represents $A$ as follows: $$ A = SJS^{-1}\ ,$$ where $S$ is the above invertible matrix, and $J$ is diagonal with eigenvalues $1$, and $97/100=0.97$, and $3/4=0.75$. After $t$ steps, the transition matrix (from the initial state) is $$ \begin{aligned} A^t &= (SJS^{-1})^t \\ &=\underbrace{(SJS^{-1})\ (SJS^{-1})\ \dots\ (SJS^{-1})}_{t\text{ times}} \\ &=SJ\ S^{-1}S\ J\ S^{-1}S\ \dots\ S^{-1}S\ J \ S^{-1} \\ &=SJ\ \ J\ \ \dots\ \ J \ S^{-1} \\ &=SJ^tS^{-1} \\ &= \begin{bmatrix} 1 & 1 & 1 \\ 1 & \frac{2}{5} & -4 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{97}{100} & 0 \\ 0 & 0 & \frac{3}{4} \end{bmatrix}^t \begin{bmatrix} 0 & 0 & 1 \\ \frac{10}{11} & \frac{5}{22} & -\frac{25}{22} \\ \frac{1}{11} & -\frac{5}{22} & \frac{3}{22} \end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 & 1 \\ 1 & \frac{2}{5} & -4 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1^t & 0 & 0 \\ 0 & \left(\frac{97}{100}\right)^t & 0 \\ 0 & 0 & \left(\frac{3}{4}\right)^t \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ \frac{10}{11} & \frac{5}{22} & -\frac{25}{22} \\ \frac{1}{11} & -\frac{5}{22} & \frac{3}{22} \end{bmatrix} \end{aligned} $$ We only need now the entry on the position $(1,1)$ of the above matrix $A^t$, written explicitly. (Here the upper index $t$ is a power.)

This entry is as computed initially in four lines,

sage: (A^t)[0,0]
10/11*97^t/100^t + 1/11*3^t/4^t
sage: latex(_)

$$ \frac{10 \cdot 97^{t}}{11 \cdot 100^{t}} + \frac{3^{t}}{11 \cdot 4^{t}} $$ which is an exact value, and now we should approximate e.g. $$ \left(\frac{97}{100}\right)^t = 0.97^t=(1-0.03)^t\approx(\exp(-0.03))^t=\exp(-0.03t)\ . $$