A problem on Markov Chains

[russel]

Hello there,
I'm stuck at a problem on markov chains... Could anyone help?
Here it is:
There are two machines that operate or don't during a day. Let $$X(n)$$ be the number of machines operating during the n-th day. Every machine is independently operating during the n-th day with probability $$\frac{1+X(n-1)}{4}$$. Show that the process $$\{ X(n):n=1,2,...\}$$ is markovian and find its transition matrix.

 

[chiro]

Hey russel and welcome to the forums.

You need to show us what you've tried and what you are thinking.

Markov chains primarily have to have valid probabilities and then need to satisfy 1st order conditional dependence.

So the first hint is that first order conditional independence says that P(X1 = a, X2 = b, X3 = c, blah blah, Xn = whatever | X0 = x) = P(X1 = a|X0 = x). In other words only the current state depends on what happened just before and no more.

What have you thought about for this expression?

 

[russel]

I tried to prove that P(X(n+1) | X(1),...,X(n))=P(X(n+1) | X(n)) but I can't. Must I use the P(A|B)=P(A,B)/P(B) ? Is it easier to prove that has independent increments and, therefore, it's markovian? The funny part is I found the transition matrix...
Could I just say that the probability X(n) depends only on X(n-1) (if you see the given equation), so it is markovian?

 

[chiro]

I tried to prove that P(X(n+1) | X(1),...,X(n))=P(X(n+1) | X(n)) but I can't. Must I use the P(A|B)=P(A,B)/P(B) ? Is it easier to prove that has independent increments and, therefore, it's markovian? The funny part is I found the transition matrix...
Could I just say that the probability X(n) depends only on X(n-1) (if you see the given equation), so it is markovian?

You can equate this long expression with your (X(n) - 1)/4 and then show that only X(n) matters which cancels all the rest of the terms.

Usually what happens in the independent cases is that if something is independent then P(C and D) = P(C)P(D). In many of these proofs you get an expression with you get P(E)P(D)/[P(C)P(D)] = P(E)/P(C) and P(E) = P(X1 = a, X0 = b) and P(C) = P(X0 = b).

In your case though you are given that it only depends on X(n) so you can just use this result directly, but in the future it's important to note that for many Markovian problems you will need to show independence like in the above scenario so that you get the right cancellation.

 

[blue_raver22]

Hello,

Markov chains are a useful tool for modeling systems that have a random or probabilistic nature. In this case, the system being modeled is the operation of two machines during a day. The variable $$X(n)$$ represents the number of machines operating on the n-th day.

To show that this process is Markovian, we need to demonstrate that the future state of the system (the number of machines operating on the next day) depends only on the current state of the system (the number of machines operating on the current day) and not on any previous states. In other words, the system has no memory and the future state is independent of the past states.

In this problem, the probability of a machine operating on the n-th day is dependent on the number of machines operating on the previous day. However, this does not violate the Markov property, as the current state (number of machines operating on the current day) is still the only factor determining the probability of the future state (number of machines operating on the next day).

To find the transition matrix, we need to determine the probabilities of transitioning from one state to another. In this case, there are only two possible states - 0 machines operating and 1 machine operating. Therefore, the transition matrix will be a 2x2 matrix.

Let's consider the first state, 0 machines operating. The probability of transitioning from this state to 0 machines operating on the next day is given by the probability that both machines do not operate on the n-th day. This can be calculated as $$P(X(n)=0|X(n-1)=0)=\frac{3}{4}$$.

Similarly, the probability of transitioning from the first state to 1 machine operating on the next day is given by the probability that one machine operates and one does not. This can be calculated as $$P(X(n)=1|X(n-1)=0)=\frac{1}{4}$$.

Now, let's consider the second state, 1 machine operating. The probability of transitioning from this state to 0 machines operating on the next day is given by the probability that one machine does not operate and one does. This can be calculated as $$P(X(n)=0|X(n-1)=1)=\frac{1}{4}$$.

Lastly, the probability of transitioning from the second state to 1 machine operating on the next day is given by the probability that