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Q&A Bank Part 1 QU 1.14 (iv)

S

Study Break

Member
The question asks you to calculate

P(R10=7|X10=4) where R10 is the residual holding time at time 10.

The solution says that this is the probability of remaining in state 4 for 6 time units and then leaving.

Can someone explain this to me please? Why is it 6 time units?

Many thanks.
 
Hi
BAsically, you have to leave state 4 at the time of the 7th transition from now (in order to have stayed in State 4 for a duration of 7 periods in total). So for the next 6 periods you stay in state 4 (prob 0.8) and in the 7th period you jump out of it (prob 0.2).

If it helps, you are currently in state 4. If you jumped out of this state at the next jump, the prob would be 0.2, and the holding time would be 1. If you stayed in at the first transition but then left at the next, then the prob would be 0.8x0.2 and the holding time would be 2. Extending this to a holding time of 7 gives you the answer.
Hope that helps
Robert
 
Thanks again.

I was reading it wrongly, by thinking I was in state 4 at time 10 and wanted to be in state 7 at time 10.

I can see it now.

Thanks.
 
The one thing I don't get Mr. Chadburn is that aren't you suppose to stay in state 4 for 7 time units? In the book, the defn. if holding time is the time between now and your next jump. So is it that the BEGINNING of the 7th period is when you jump or the end of it?

Thanks.
 
The end.

(At the end of the 7th period from now is when you jump out of it. So for 6 possible transition points you stay put, and leave at the 7th.)

Robert
 
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