Schwarzschild90 said:
Computing U(l) using MATLAB I get the following ruin probabilities: (0.8000, 0.6000, 0.4000, 0.2000). I.e. the same values that you do, but I have not built into my matrix the absorbing boundary conditions.
It's comforting to know my approach works. Also, thank you for computing the matrix.
Taking the first and last row equal to zero returns a computational error when using the matrix T to compute U(l). Also, I can no longer take the inverse of T. How do I work around this? (I'm trying to implement the absorbing boundary conditions, i.e. all zeros in the first and last row)
Matlab code:
clear
M=6;
X = ones(M,1);
A = diag(-2*X,0);
clear X
X = ones(M-1,1);
B = diag(X,1);
C = diag(X,-1);
T = A+B+C;
T(1,:)=0;
T(M,:)=0;
clear X B C;
T = \begin{pmatrix}<br />
0 & 0 & 0 & 0 & 0 & 0 \\<br />
1 & -2 & 1 & 0 & 0 & 0 \\<br />
0 & 1 & -2 & 1 & 0 & 0 \\<br />
0 & 0 & 1 & -2 & 1 & 0 \\<br />
0 & 0 & 0 & 1 & -2 & 1 \\<br />
0 & 0 & 0 & 0 & 0 & 0 \\<br />
\end{pmatrix}
Your T is a 4x4 submatrix of my T. Your T does have an inverse, but mine does not. As I said before, I don't really see any point in computing the inverse, because there are much easier, standard ways of solving such problems.
One final point to clear up: the ruin probabilities do not depend in any way on the value of ##\tau## you choose. The value of ##\tau## affects the speed at which the limits are attained, but not the actual limits themselves. This is easy to see. From the DE ##dP(t)/dt = \tau T P(t)## it follows that for small ##\Delta t > 0## we have for states ##X(t) = 0,1,2, \ldots, 5## and transition rate matrix ##T## that
$$ \Pr \{ X(t + \Delta t) = j| X(t) = i \} = \begin{cases}
1- 2 \tau \Delta t,& i \neq 0,5, j = i \\
1 \tau \Delta t, & i \neq 0,5, j = i \pm 1\\
1 , & i = 0, 5, j = i \\
0, & i = 0, 5, j \neq i
\end{cases}
$$
Thus, for states i = 1,2,3,4 we have ##\Pr \{ X(t+\Delta t) \neq i | X(t) = i \} = 2 \tau \Delta t##. That means that the probability of leaving state ##i## in the time interval from ##t## to ##t+\Delta t## is ##\tau 2 \Delta t## to first order in small ##\Delta t.## That is why the holding time in state ##i## (i.e., the time spent in state ##i## is exponentially-distributed with rate parameter ##2 \tau## and mean ##\frac{1}{2\tau}##. However, we have
$$\Pr \{ X(t + \Delta t) =i \pm 1 | X(t + \Delta t) \neq i\: \&\: X(t) = i \} = \frac{1 \tau \Delta a}{ 2 \tau \Delta t} = \frac{1}{2} $$
and
$$\Pr \{ X(t + \Delta t) \neq i \pm 1 | X(t + \Delta t) \neq i\: \&\: X(t) = i \} = 0 $$.
Note that these transition probabilities (for the next "position") are independent of the value of ##\tau##. That is why the exact same 1-step discrete-time matrix ##P## is obtained without any reference to the value of ##\tau##.
Another way to see this is that (using my ##T## rather than yous) we have
$$P(t) \equiv (P_{ij}(t) ) = \exp(t \tau T), $$
which can be written as ##P(t) = \exp( \sigma T)##, where ##\sigma = t \tau##. The ruin probabilities are ##R(i) = \lim_{t \to \infty} P_{i0}(t)## (because ##P_{i0}(\infty)## is the probability we are in state 0 at time ##\infty##, and that means that we must have hit state 0 at some finite time---and stay there forever more. Anyway, ##\lim_{t \to \infty} \cdots = \lim_{\sigma \to \infty} \cdots##, so the ruin probabilities can be computed as though ##\tau = 1##.