Consider an irreducible finite Markov chain with states 0, 1, ..., N. (a) (20 pts) Starting in state i, what is the probability
that the process will visit state j at most 3 times in total? (b) (10 pts) Let xi = P {visit state N before state 0|start in i}. Find a set of linear equations which the xi satisfy, i = 0, 1, ..., N. (c) (10 pts) If P j jPi,j = i for i = 1, ..., N −1, show that xi = i/N is a solution to the equations in part (b).
1 answer:
Answer:
If for each j ∈ S, Recalling that P m ij is precisely the (ij) th component of the matrix P
m, we conclude that (4) can be expressed in matrix form byπj exists as defined in (3) and is independent of the initial
state i, and P j∈S πj = 1, then the probability distribution π = (π0, π1, . . .) on the state space S is called the limiting or stationary or steady-state distribution of the Markov chain.
see attached image 1.
That is, when we average the m-step transition matrices, each row converges to the vector of stationary probabilities π = (π0, π1, . . .). The i th row refers to the intial condition X0 = i in (4), and for each such fixed row i, the j th element of the averages converges to πj . A nice way of interpreting π: If you observe the state of the Markov chain at some random time way out in the future, then πj is the probability that the state is j. To see this: Let N (our random observation time) have a uniform distribution over the integers {1, 2, . . . n}, and be independent of the chain; P(N = m) = 1/n, m ∈ 2, . . . n}. Now assume that X0 = i and that n is very large. Then by conditioning on N = m we obtain (see image 2)
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The initial population is
P₀ = 94 million in 1993
The growth formula is
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.
Because P(5) = 99 million (in 1999),
In the year 2005, t = 12 years, and
Answer: 106 million (nearest million)
P₀ = 94 million in 1993
The growth formula is
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.
Because P(5) = 99 million (in 1999),
In the year 2005, t = 12 years, and
Answer: 106 million (nearest million)
Answer:
0% probability that at least 1,500 will agree to respond
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
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In this problem, we have that:
So
What is the probability that at least 1,500 will agree to respond
This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So
has a pvalue of 1.
1 - 1 = 0
0% probability that at least 1,500 will agree to respond