Help with Kilbaha 2007 exam

[Whatlol]

[IMG]http://img97.imageshack.us/img97/1339/kilbaha2007.png[/img]

I tried doing this question and seem to be getting the complement to the answer.

Ill display my working out so hopefully my error will be revealed (=



I assigned D to be driving to work, and C to be catching the train

0.25D + 0.35C =D
.: 0.35C = 0.75D
 and we know that D= (1-C)
0.35 = 0.75 - 0.75C
C = 0.75 / 1.1
   = 0.6818

The answer kilbaha has is 0.318

[kenhung123]

No idea what you did but you can simply do 0.75/(0.75+0.35) I'm afraid kilbaha is wrong.

[slothpomba]

What this is asking for is a steady state matrix.

A steady state matrix eventually will always take the same values regardless of the initial conditions, so the initial conditions do not even matter.

Your first error was including the initial conditions.

We shall assign X=0 for car and X=1 for Train.

Pr(X_n+1=1|X_n=0) = 0.75  a
[[(Probability he catches the train today given he took the car yesterday) a.k.a Pr(Train_n+1|Car_n) ]]

Pr(X_n+1=0|X_n=0) = 0.25 1-a
[[(Probability he takes the car today given he took the  car yesterday) a.k.a Pr(Car_n+1|Car_n) ]]  



Pr(X_n+1=0|X_n=1) = 0.35 b
[[(Probability he takes the car today given he took the train yesterday) a.k.a Pr(Car_n+1|Train_n) ]]

Pr(X_n+1=1|X_n=1) = 0.65 1-b
[[(Probability he catches the train today given he took the train yesterday) a.k.a Pr(Train_n+1|Train_n) ]]




fill it in



Steady State Formula:
Recall taking the train X=1

Steady State formula:

Pr(X_n=1) =




= 0.681818

Maybe they are wrong..

[Whatlol]

In my text book they wrote out this (except for a different question)

transition matrix x =

i dont think that is taking into initial conditions, that is saying how the system will just settle after a certain amount of time.

but regardless of that, in the solutions they took the same approach you did except instead of doing they did . Maybe it was a mistake

[slothpomba]

Yeah maybe they meant what proportion of time does he *not* take the train to work or they wrote train instead of car... or they just gave the wrong solution..

[Whatlol]

Yeah maybe they meant what proportion of time does he *not* take the train to work or they wrote train instead of car... or they just gave the wrong solution..

Yeah most likely, ill just leave it i guess.

Thanks for the clear explanation (=

[happyhappyland]

Kilbaha is right...



Thus

0.318

If you use 0.75 on top you are essentially using the prob of driving not train.

[kenhung123]

Err I think you got it the wrong way around 0.35 is the probability of driving and not train. 0.75 is the probability of train.

[Linkage1992]

Spot on. Since the train is on the bottom you have to do 0.35/ (0.75 + 0.35) to get the proportion of the time he catches the train.

you have to be careful...

EDIT: Damnit, I'm wrong. It is 0.75 on top. now i'm really confused...

[slothpomba]

Waaait im confused... i thought it was 0.75 on top.. i think ken was saying that too?

[kenhung123]

Yea I'm pretty damn sure its 0.75 on top..

[Whatlol]

Yea I'm pretty damn sure its 0.75 on top..

+1