April 2011 question 4

[AaronD]

Hi,

I was looking at question 4 from April 2011 and can't seem to get my head around the transition probabilities.

For example.
p1,1 = 0.7
Is this due to the fact that the only way we can go from state 1 (Poor) to state 1 is if we move downwards either once or twice, with probs .2 and .1 respectively, Which we take from 1 to give 0.7? If this is the case I don't understand why this would hold.

Another example:
p2,2 = .4
I think it might have something to do with 1 - prob of not moving. I.e. 1-(.2+.2+.1+.1)=.6 But I can be sure.

Any guidance here would be appreciated.

Thanks in advance

 

[Shiksha]

Hi Aaron,

I think since the probability of moving upwards to State 2 (Satisfactory) and to State 3 (Good) is 0.2 and 0.1 respectively and the values in a row of a transition matrix must sum to 1, the probability of being in State 1 is (1-0.2-0.1 = 0.7)

As for being in State 2, again the probability of moving upwards to State 3 and State 4 is 0.2 and 0.1 respectively but the probability of moving downwards from State 2 to State 1 is 0.2 since the question says there is a 20 per cent chance of moving down one level so by the law of total probability (i.e. that all the values in a row must sum to 1), the probability of being in State 2 is (1-0.2-0.1-0.2 = 0.4)

I hope this helps.