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[ 60% 40% ]
Small[ 25% 75% ]
Am I correct in saying that the probability matrix can be re-written as....
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?
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%.
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:
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) = 
at time 1 (3 years hence):
x(1) = [ 0.6 0.25 ]  = [0.6]
[ 0.4 0.75 ]  [0.4]
at time 2 (6 years hence):
x(2) = [ 0.6 0.25 ]^2  = [ 0.46 0.3375 ]  = [0.46]
[ 0.4 0.75 ]  [ 0.54 0.6625 ]  [0.54]
(Which has the values you had already mentioned)
Yeah, thank you very much. I had seen that some people did that, whilst other examples didn't.
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]
(This is just an off-the-cuff idea, may not be rigorous)
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 [ 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."
Just for Fun, I worked out that matrix "P" where
P^3 = [0.6 0.25]
It is (approximately):
So, what is P^6 ?
And P^8 is:
And P^9 is
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
It's funny because there are two ways of doing it, viz:
Which one do you think sounds better?
Here's another theory I have:
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. ???
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):
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.
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!
No update as of yet....
Still working within the problem.
Hopefully other members will have some interesting ideas.
maths_buff/kris was the owner of rover and has now gone bust and i blame rod
I come back stronger than a powered-up Pac-Man
I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large
Fatboy Slim is a Legend
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".