Hi everyone. I did a study on the above topic and found the observations worth sharing. Here I talk of drawing a transition graph of a non markov model by introducing additional states and turning it into a markov one without just general reasoning and giving it a mathematical touch. Given below is an algorithm to do the same. 1. Find the maximum number of states any existing state can be extended into. 2. Extend every state into these many number of states. 3. Connect the states. 4. Remove the "unwanted states" from the model, i.e., which do not have any arrow pointed towards them. (I call them "unwanted" because no state wants to reach to them.) 5. Combine the new states into their source state iff in any direction (forward or backward) they do not point to different states alongside if in any direction, one points to some state and the other does not at all. 6. The resultant model forms a markov chain. Given below is an example to illustrate the same. Taking the model of an NCD policy having 4 discount levels namely, 0%, 25%, 40% and 60%; where in the case of a claim during the current year, the discount status moves up one step if the year is claim free, moves down one step if it's not and moves down two steps if the previous year was not claim free. Here, the model is not a markov one since our future discount status would not only be dependent on our present year's information but also on our previous year's information. So, according to the first step written above, we determine that any of the discount status can be extended into maximum two. Following the second step we extend every discount level into two namely 0+,0-,25+,25-,40+,40-,60+ and 60- with a (+) sign indicating that the previous year was claim free and a (-) sign indicating the opposite. I represent the states as 0+,0-,1+,1-,2+,2-,3+,3- for ease. Now, according to the third step we'll have to connect these states. So now, 0+ reaches 1+, 1+ reaches 2+, 2+ reaches 3+, 3+ reaches 3+ suggesting that the present year is claim free and previous year too was claim free. Also, 1+ reaches 0-, 2+ reaches 1-, 3+ reaches 2-, 0+ reaches 0- suggesting that present year is not claim free but previous year was. Nevertheless, 3- reaches 1-, 2- reaches 0-, 1- reaches 0- and 0- reaches 0- suggesting that both present year and previous year are not claim free. Following the fourth step, we'll have to remove unwanted states, which here are 0+ and 3- since none of the new states reaches them. Now comes the step where we have to combine new states into their source state. Since 0 is now reduced to 0- and 3 to 3+ ( we removed 0+ and 3- in the last step), 0- again becomes 0 and 3+ becomes 3. Coming to 1+ and 1-.. in the left both point to 0- and satisfy the condition of not pointing to different states in same direction. Also to the right, 1+ points to 2+ and 1- does not reach any state at all. So, now we should combine 1+ and 1- into their source state. Coming to 2+ and 2-.. to the left, 2+ reaches 1- and 2- reaches 0-. Hence, they cannot be combined given they reach different states in the same direction. The resultant model with new state space {0,1,2+,2-,3} forms a markov chain. Let me know if anyone of you finds any loopholes to this or if you think there's an improvement possible. I'll be grateful if any one of you finds this of any help. And if you do, then please let me know. Thank You, Akansha
Very Nice! I love CT4 too! Can understand why people say it's one of the hardest CT's though. If you haven't got a Maths degree it must be a nightmare. What is the purpose of your study Akansha? Peter
Hi Peter, Although CT4 requires somewhat more effort than other CT subjects, it's quite interesting. And that's the reason I could work out something like that. In the course notes, a Non Markov chain's transformation to a Markov chain is taught by general reasoning only, which I could never follow. It took me some time to form answers using the general reasoning method. Also, I was not always sure if I was correct. So, I thought of having a mathematical transformation of the same. And yes, then I derived the above algorithm. I shared it with my college teachers, who liked it. Mr. John Potter too approved this. And now I present this here for all the CT4 students too. I'm pleased that you liked it. Thanks for the appreciation. Akansha
Hi Akansha Very nice algorithm, I want to bring your notice to the below two phrases of your write-up. 1. “1- does not reach any state at all.” which I think is not correct because if the process is at 1- state and there is claim free year then process will go from 1- state to 2+ state. Though I agree with your conclusion that we can combine 1+ and 1- because 1- and 1+ both go to same original state (0 on the left side and 2 on the right side). 2. “where in the case of a claim during the current year” which I think kept by mistake while rephrasing your algorithm. Hope I did not misunderstand these phrases.
Hi , Please check the Solution of IAI october 2014, question 4 , how come the value of the a, b,c calculated thanks in advance
hi Akansha, Appreciate your work. I have tried to apply this to other sums, but not able to find soln. For your Reference kindly check sums...Q X 1.10 of assignment X and Q.6 April 2006 UK exam.
