Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

|
Options

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]

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?