Q&A Bank 2.13(iv)

[rinishj28]

To find the distribution of the employees can we multiply the matrix by itself 5 times to get the 5-year probabilities ?

 

[Tim.Sullivan]

Hi

You'll be more likely to receive help if you copy the question in your post - many people who would be able to assist don't have easy access to the notes (Q&A bank in this case).

Sorry to point that out - but I know that you have had zero responses to some of your other posts for, I suspect, the same reason!

Thanks

Tim

 

[Calum]

And the short answer - probably yes, since that would be the normal approach, however five times is a lot so I suspect there is a shortcut you can use.

 

[suraj]

To find the distribution of the employees can we multiply the matrix by itself 5 times to get the 5-year probabilities ?

For that you've calculate 1-year probabilities and then construct a 1 year transition matrix.

You can't use Jump chain matrix or Generator Matrix for this.

Instead calculate 5-year probabilities directly like this:

\(P_{11}(5)\) and \(P_{22}(5)\) are nothing but \(e^{-2.5}\) and \(e^{-1}\) respectively because holding times are exponentially distributed.

and \(P_{12}(5)\) =

\(\int_0^5\) P(Holding state 1 till t) * Transition to state 2 * P(Holding state 2 for remaining time(5-t) ) dt

= \[\int_0^5 e^{-0.5t} \times 0.25 \times e^{-0.2(5 - t)}~dt \]

 

[rinishj28]

For that you've calculate 1-year probabilities and then construct a 1 year transition matrix.

You can't use Jump chain matrix or Generator Matrix for this.

Instead calculate 5-year probabilities directly like this:

\(P_{11}(5)\) and \(P_{22}(5)\) are nothing but \(e^{-2.5}\) and \(e^{-1}\) respectively because holding times are exponentially distributed.

and \(P_{12}(5)\) =

\(\int_0^5\) P(Holding state 1 till t) * Transition to state 2 * P(Holding state 2 for remaining time(5-t) ) dt

= \[\int_0^5 e^{-0.5t} \times 0.25 \times e^{-0.2(5 - t)}~dt \]

Hi Suraj

I think that's where I'm stuck

What is the difference between the jump chain matrix and the normal matrix?

 

[suraj]

Hi Suraj

I think that's where I'm stuck

What is the difference between the jump chain matrix and the normal matrix?

Transition matrix consists of probabilities of moving between states in a given time period (like 1 year in NCD model)

While in Jump Chain, we only examine the process at the times of transition & time spent in each state is not taken into account.
You'll notice that the main diagonal entries in a Jump Chain matrix are always 0. That's because we're talking about probabilities of going to other states, when a process leaves a particular state.

We can use Jump Chain matrix for calculations where we're asked to find probabilities like -
Say, Probability that process goes to state X before it goes to state Y, and process is currently in some other state.
i.e. where time is not considered.

Have a look at Q8 of last session's UK paper for example.