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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!!!

Ck Law
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1 Answers1

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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. $$ \left(\frac{97}{100}\right)^t = 0.97^t=(1-0.03)^t\approx(\exp(-0.03))^t=\exp(-0.03t)\ . $$

dan_fulea
  • 32,856
  • Thanks for your prompt reply. However, this is actually a paper-and-pencil exam question. So I don't think we are expected to use programming. Is there any simpler way? My original thoughts were to use integration (works pretty well normally if there's no "re-entering" state 0 from state 1), or Kolmogorov forward equation. By using integration, I could not account for the possibility of many rounds of transition in between state 0 and 1. – Ck Law May 03 '18 at 11:02
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    Using computer was only for being complete, and having less to type. If this is an exam question, best mention it in the post, then there will always be further comments in the answers. Here, the exam solution would be not structural from my point of view. First of all we eliminate the state $2$ from the discussion. Consider only $B=\begin{bmatrix}0.95&0.05\0.08&0.77\end{bmatrix}$. The eigenvalues are $0.97$ and $0.75$. Eigenvectors for them are quickly found, paper+pencil, the diagonal form uses as base change the matrix with these eigenvectors as columns, $B^t$ is found, its entry (1,1). – dan_fulea May 03 '18 at 18:13