Suppose that a life insurance coverage waives premium upon disability of the insured. You model
the coverage as a homogeneous Markov chain with three states: active, disabled, and dead.
The annual transition matrix is:
$$M = \begin{bmatrix}
0.75 & 0.15 & 0.1\\
0.5 & 0.3 & 0.2\\
0 & 0 & 1
\end{bmatrix}$$
Currently, $90$% of insureds are active and $10$% are disabled.
(a) Compute the percentage of the current population of insureds that are (i) active (ii) disabled and (iii) dead, at the end of three years.
(b) Compute the probability that a currently disabled life will be active at the end of three years.
My attempt:
Note: I am using actuarial notation.
(a) After three years, the transition matrix is:
$$M^3 = \begin{bmatrix}
0.75 & 0.15 & 0.1\\
0.5 & 0.3 & 0.2\\
0 & 0 & 1
\end{bmatrix}^3 = \begin{bmatrix}
0.556875 & 0.142875 & 0.30025\\
0.47625 & 0.12825 & 0.3955\\
0 & 0 & 1
\end{bmatrix}$$
Therefore with $0$ as active, $1$ as disabled and $2$ as dead:
%active = $0.9(_3P_x^{00} + _3P_x^{10}) = 0.9(0.556875 + 0.47625) = 92.98$%
%disabled = $0.1(_3P_x^{01} + _3P_x^{11}) = 0.1(0.142875 + 0.12825) = 2.711$%
%dead = $_3P_x^{02} + _3P_x^{12} = 0.30025 + 0.3955 = 69.575$%
(b) P(disabled being active after three years) = $_3P_x^{10} = 0.47625$
Is what I have done correct? I am not sure, especially about (a). Any assistance is much appreciated.
Best Answer
Your 3-year transition matrix is correct, but your interpretation and use is not.
One way to see that your answer cannot possibly be correct is that the percentages do not add up to $100\%$.
Moreover, the first row, third column entry of $M^3$, $0.30025$, represents the probability that an insured who is active (state $0$) at the beginning of the $3$-year period, arrives at state $2$--dead--within that $3$-year period. As such, this means that active insureds have a slightly over $30\%$ chance of death over a $3$-year period, and disabled insureds have almost $40\%$ chance over that same period. As death is an absorbing state, this means roughly $(0.9)(0.3) + (0.1)(0.4) = 0.31$ or $31\%$ of the book of business will be dead in $3$ years. The exact quantity is given by the product of the row vector $\begin{bmatrix} 0.9 & 0.1 & 0 \end{bmatrix}$ with $M^3$; i.e., $$\begin{bmatrix} 0.9 & 0.1 & 0 \end{bmatrix} \begin{bmatrix} 0.556875 & 0.142875 & 0.30025 \\ 0.47625 & 0.12825 & 0.3955 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}0.548813 & 0.141413 & 0.309775\end{bmatrix}.$$ This gives the probability distribution of each state at the end of $3$ years.
Why is your computation incorrect? The reason is because it should read, for instance, $$(0.9) {}_3 p_x^{00} + \color{red}{(0.1)} {}_3 p_x^{10} + (0) {}_3 p_x^{20} = (0.9)(0.556875) + (0.1)(0.47625) + (0)(0) = 0.548813.$$ I have written out the third term which is $0$ in order to illustrate how this calculation corresponds to the matrix multiplication that was performed above--i.e., that what you are really doing here is taking the dot product of the row vector $\begin{bmatrix} 0.9 & 0.1 & 0 \end{bmatrix}$ representing the initial distribution of insureds, with the column vector $\begin{bmatrix} 0.556875 & 0.47625 & 0 \end{bmatrix}^\top$ representing the transition probability vector going from each of the three states to the active state.
The probabilities for the other two states are calculated accordingly, hence are omitted.
Your answer for part (b) is correct; this is simply the probability contained in the second row, first column of $M^3$, which is $0.47625$.