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- MathsIsFun
- 2005-06-09 21:44:23
Ha ... ha.
Actually I kinda like my solution. But because I invented it without a real deep understanding of Markov Chains I didn't want to say "trust me - this'll work".
- Mr T
- 2005-05-31 21:51:37
maths_buff/kris was the owner of rover and has now gone bust and i blame rod
- maths_buff
- 2005-05-28 18:27:04
No update as of yet....
Still working within the problem.
Hopefully other members will have some interesting ideas.
- maths_buff
- 2005-05-27 23:14:32
It matters and it doesn't matter; after six years it doesn't matter because on average everyone will have a new car. But it terms of after 8 years, that the questionable part. Perhaps 99% of the population buys their car in that year.
Remember that we're only worrying about current large car owners who'll have a large car in eight years time.
I will try and fiddle with it because as long as the elements in each row add to 1, I can indicate what I did is correct.
I'll wait and see what other forum members say.
Once again, MathsIsFun, thank you!
- MathsIsFun
- 2005-05-27 23:06:07
Indeed, if you are looking at one car owner, he will be on his "6 Year car" even after 8 years !
But we are looking at a large population here, I imagine, who are changing their cars every day.
Possible future for Large Car Owners (Year 0,3 and 6):
LLL
LLS
LSL
LSS
The question is for just one car owner, though. But it does not say how long the owner has had his present car.
So, he/she may exchange his car later today! Or not for 3 years.
- maths_buff
- 2005-05-27 22:59:57
Here's another theory I have:
Large Small
Large [ 0.60 0.40 ]
Small [ 0.25 0.75 ]
There are two possible ways for a large car owner to own a large car in six years:
A) Someone has a large car now, buys a large car after three years, buys another large car after six years
B) Someone has a large car now, buys a small car after three years, buys another large car after six years
Therefore the probability is P(LL)P(LL) + P(LS)P(SL) = 23/50 like I suspected.
Maybe it's wrong; maybe it's right. ???
- maths_buff
- 2005-05-27 22:51:20
Thanks, MathsIsFun!
It's funny because there are two ways of doing it, viz:
http://ceee.rice.edu/Books/LA/markov/
http://ceee.rice.edu/Books/LA/markov/
Which one do you think sounds better? 
- MathsIsFun
- 2005-05-27 22:35:47
Just for Fun, I worked out that matrix "P" where
P^3 = [0.6 0.25]
[0.4 0.75]
It is (approximately):
[0.818 0.114]
[0.182 0.886]
So, what is P^6 ?
[0.460 0.338]
[0.540 0.662]
And P^8 is:
[0.422 0.362]
[0.578 0.638]
And P^9 is
[0.411 0.369]
[0.589 0.631]
There is some drift due to calculation accuracy, but if you worked P out more accurately you may have something workable. But there may well be a rigorous way to do this rather than my "hey, lets use Excel and see what we get" approach
- maths_buff
- 2005-05-27 21:21:42
I can see your point MathsisFun, I can see it indeed.
But would that limit accuracy? Because I believe the transition matrix refers to three year intervals only. I will type the question word-for-word to clarify things.
"Market analysis in a certain region has established that, on average, a new car is purchased every three years. With respect to those changing cars, the buying patterns are described by the matrix:
Large Small
Large [ 60% 40% ]
small [ 25% 75% ]
a) Rewrite the matrix as a probability matrix
b) Find the probability that a person who now owns a large car will own a large car in eight years' time."
- MathsIsFun
- 2005-05-27 20:27:51
OK, well, we still have your "8 year" problem ...
... you could cheat and work out the probabilites at 6 and 9 and ratio in between.
Or, you could work out an equivalent matrix that works on 1 year intervals.
In other words, what is the Matrix "P" where:
P^3 = [0.6 0.25]
[0.4 0.75]
(This is just an off-the-cuff idea, may not be rigorous)
- maths_buff
- 2005-05-27 20:25:22
Yeah, thank you very much. I had seen that some people did that, whilst other examples didn't.
Cheers,
Kris
- MathsIsFun
- 2005-05-27 20:21:20
Hi, maths_buff.
Markov chains are not my specialty, I am hoping that Milos or one of the other members is better versed in these than I am.
Having said that, I think your probability matrix needs to be transposed:
L S
S = L [ 0.6 0.25 ]
S [ 0.4 0.75 ]
So, if someone owns a large car at time 0 we will have
x(0) = [1]
[0]
at time 1 (3 years hence):
x(1) = [ 0.6 0.25 ] [1] = [0.6]
[ 0.4 0.75 ] [0] [0.4]
at time 2 (6 years hence):
x(2) = [ 0.6 0.25 ]^2 [1] = [ 0.46 0.3375 ] [1] = [0.46]
[ 0.4 0.75 ] [0] [ 0.54 0.6625 ] [0] [0.54]
(Which has the values you had already mentioned)
- maths_buff
- 2005-05-27 19:59:00
After six years I know the respective probabilities are....
[ 23/50 27/50 ]
[ 27/80 53/80 ]
So after six years the probability of someone currently owning a large car and still owning one is 23/50 or 46%.
- maths_buff
- 2005-05-27 19:38:13
I am enquiring as to a series of Markov Chains questions I have.
I have come across a question that says market analysis has established that, on average, a new car is purchased every three years. Buying patterns are described by the matrix:
Large Small
Large[ 60% 40% ]
Small[ 25% 75% ]
Am I correct in saying that the probability matrix can be re-written as....
L S
S = L [ 0.6 0.4 ]
S [ 0.25 0.75 ]
In addition, how would I calculate the probability of someone owning a large car still owning a large car in eight years' time, considering that the problem itself deals with car purchases every three years on average?