A physical device can be in three states: A, B, C. The device operates as follow

Question

A physical device can be in three states: A, B, C. The device operates as follows (all time units are in hours):

The device spends an exponentially distributed amount of time in state A (with mean of 12 minutes) and then with probability 0.6 goes to state B, and with prob. 0.4 goes to state C.

When in state B, the device moves to state C after an Exp(3) amount of time.

When in state C, the device goes to state A at rate 1/hour, and to state B at rate 2/hour.

Let Xt represent the device state at time t, and suppose X0 = ‘A?. Compute:

(a) Probability the device is in state `A' after 30 minutes.

(b) Probability the device is in state `A' after 30 minutes given that it was in state `B' after 5 minutes and in state `C' after 10 minutes.

(c) The long-run proportion of time the device spends in state `A'.

Explanation / Answer

The key is to understand how to convert between different descriptions of exponential distributions and Markov chains.

For exponential distributions, the following are equivalent:

A)Sleep for an Exp(?)Exp?(?) time before waking up.

B)Sleep for an exponentially distributed time with mean ??1??1 before waking up.

C)Transition from sleep to waking up at a rate of ??.

For a Markov chain, there are two ways to view transitions: the "call center" model and the "direct line" model. The following are equivalent:

Wait for Exp(?)Exp?(?) time for the phone to ring. Then get a call from Alice with probability pApA, from Bob with probability pBpB, from Charlie with probability pCpC, etc.

Wait for Exp(?pA)Exp?(?pA) time for a phone call from Alice, or for Exp(?pB)Exp?(?pB) time for a phone call from Bob, or for Exp(?pC)Exp?(?pC) time for a phone call from Charlie, etc. (Whichever comes first.)

In other words, the possible transitions from a state can be thought of as having several competing exponential distributions going at once, and picking the smallest, or as having a single exponential distribution going, and picking which way to go once it's done.

The driving property behind this equivalence is that for exponential distributions, we have

min{Exp(?1),Exp(?2),…,Exp(?n)}=Exp(?1+?2+?+?n).

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