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May 13th, 2016, 01:01 PM   #1
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Markov Chains

There are 3 test, J,K,L. One of these tests is taken each day.

A test is randomly chosen each day, and it can either have a good (G) or bad (B) score.

The probability that the score is good given a test is

P(J) = 3/10
P(K) = 3/5
P(L) = 9/10

If the test score is bad, we will always use test J tomorrow. If the test score is good, the test tomorrow could be either J,K or L, each with 1/3 probability.

There are two questions:
-------------------------

1. Calculate approx proportion of total number of days, the results are good if this testing procedure has been running for a very long time.

I calculated this to be 3/7 by solving for $\displaystyle \pi_{G}$ using the following:

$\displaystyle [\pi_{G} \pi_{B}] = [\pi_{G} \pi_{B}] [\begin{smallmatrix} 0.6&0.4\\ 0.3&0.7 \end{smallmatrix}] $

2. The testing procedure has been running for a very long time, calculate the proportions of time in which tests J, K, L were used.


I dont really know where to begin for this question, I cant see any example in my notes. There is a hint in the question which is the following:

Consider the fact that for i ={ J,K,L}

P ( Test i used on day n) = P (Test i used on day n and results are G on day n-1) + P (Test i used on day n and results are B on day n-1)

Then apply appropriate conditioning and limiting operations


Any ideas?
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May 13th, 2016, 06:54 PM   #2
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Part of the question is circular because the good or bad score influences what the next test is, but without knowing what percent of the time each test was, given, how can you calculator the probability of a good score? Test J is most likely, and Test K and Test L are equally likely. Therefore I'm going to start with made up percentages of 50% for Test J, 25% for Test K, and 25% for Test L. This results in a probability of a good score being 0.525 (52.5%). That results in Test J being given 65% of the time, Test K being given 17.5% of the game, and Test L being given 17.5% of the time. That completes iteration 1 (is that term mathematically appropriate here?). I'm not going to show my arithmetic, but I'll state my formulas. When the percent of good scores is known, call it the variable a, and make a variable b that = 1 - a. The probability of Test J being given is b + a/3. The probabilities for Test K and Test L are each a/3. The probability of a good score when the percent of the time each test is given is known = P(J)*probability of Test J being given + P(K)*probability of Test K being given + P(L)*probability of Test L being given. For iteration 2, use the 65%, 17.5%, and 17.5% probabilities to calculate the probability of a good score. This time that probability is .4575 (45.75%). That results in Test J being given 69.5% of the time, Test K being given 15.25% of the game, and Test L being given 15.25% of the time. If you're correct and the probability of a good score is 3/7 (42.9% to the nearest tenth of a percent), doing iterations will make the probability eventually get very close to 42.9% and not change much. I did 2 iterations, and my guess is that fewer than 10 will be necessary before they produce very close results. Some people could probability write a computer program to do this quickly. I don't like or remember matrices, so I can't comment on your 3/7 probability.
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May 14th, 2016, 06:23 AM   #3
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I did more work:

Iteration 3:
P(good score) = .43725
P(next test is J) = .7085
P(next test is K) = .14575
P(next test is L) = .14575

Iteration 4:
P(good score) = .431175
P(next test is J) = .71255
P(next test is K) = .143725
P(next test is L) = .143725

Iteration 5:
P(good score) = .4293525
P(next test is J) = .713765
P(next test is K) = .1431175
P(next test is L) = .1431175

Iteration 6:
P(good score) = .42880575
P(next test is J) = .7141925
P(next test is K) = .14293525
P(next test is L) = .14293525

I'll stop here. It seems like P(good score) = 3/7, P(next test is J) = 5/7, P(next test is K) = 1/7, and P(next test is L) = 1/7. Given P(good score) = 3/7, it's easy to know that P(bad score) = 4/7. P(next test is J) = 4/7 + 1/7. I should have thought of that last night.
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May 14th, 2016, 08:50 AM   #4
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Thanks EvanJ for all the info. So if I understand your second reply correctly, we don't need to use iterations to calculate the proportion of time that each test is used if we know what the P(good score) is ( which we do). We can instead just use

P(next test is X) = P(Bad score) * P(next test is X) + P(Good score)*P(next test is X)

i.e

P(J) = 4/7 * 1 + 3/7 * 1/3 = 5/7
P(K) = 4/7 * 0 + 3/7 * 1/3 = 1/7
P(L) = 4/7 * 0 + 3/7 * 1/3 = 1/7


To check this result we can add up 5/7 + 1/7 + 1/7 = 1.
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May 14th, 2016, 04:06 PM   #5
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Quote:
Originally Posted by calypso View Post
Thanks EvanJ for all the info. So if I understand your second reply correctly, we don't need to use iterations to calculate the proportion of time that each test is used if we know what the P(good score) is ( which we do). We can instead just use

P(next test is X) = P(Bad score) * P(next test is X) + P(Good score)*P(next test is X)

i.e

P(J) = 4/7 * 1 + 3/7 * 1/3 = 5/7
P(K) = 4/7 * 0 + 3/7 * 1/3 = 1/7
P(L) = 4/7 * 0 + 3/7 * 1/3 = 1/7


To check this result we can add up 5/7 + 1/7 + 1/7 = 1.
That's correct.
Thanks from calypso
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May 27th, 2016, 01:32 AM   #6
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Hey calypso.

Just out of curiosity for these "long time" questions - have you covered the nature of a stationary distribution and the theory behind obtaining it?
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