Subject CT4 April 2013 Question 10 (iii)

[George Philip]

In part (iii) of this question, we are asked to solve 3px(1,2) in the 2 state (1,2) model
What we have found is that:
mu(1,2) = 0.2
mu(2,1) = 0.8

Using this, I have calculated P(1,1) = exp(-0.2) &
P(1,2) = 1 - P(1,1) = 1 - exp(-0.2)
P(2,2) = exp(-0.8) &
P(2,1) = 1 - P(2,2) = 1 - exp(-0.8)
The probability matrix will be of the form:
P =
0.8187 0.1813
0.55 0.45

therefore won't 3px(1,2) = P^3[1,2] = 0.242827 ?

But it doesn't match with the answer given.

Could someone please explain what I am doing wrong?
Thanks

 

[Andrew Martin]

Hi George

Looks like you may be mixing up Markov chains with Markov jump processes. Here we have continuous rather than discrete time. I think you may also be mixing up occupancy probabilities with more general transition probabilities.

Here we have the following occupancy probabilites:

\( p_{\bar{1,1}}(t) = exp(-0.2 * t) \)
\( p_{\bar{2,2}}(t) = exp(-0.8 * t) \)

However, we want \( p_{1,2}(3) \).

To find a general formula for \( p_{1,2}(t) \), we can solve its forward differential equation.

Hope this helps!

Andy