MHB How Do I Create a Transition Matrix for This Markov Chain Scenario?

spence1
Messages
1
Reaction score
0
I just discovered this website and want to thank everyone who is willing to contribute some of their time to help me. I appreciate it more than you know

First off, assume that state 1 is Chinese and that state 2 is Greek, and state 3 is Italian.

A student never eats the same kind of food for 2 consecutive weeks. If she eats a Chinese restaurant one week, then she is equally likely to have Greek as Italian food the next week. If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week. If she eats a Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week.

I feel like I am on the right track, but I'm having trouble translating the words to notation (the most important part).
Chinese could be represented by x, Greek by y, and Italian by z, correct? And that has to add up to 1?

"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...
So yeah. I kinda get the idea, I kinda don't. I think
 
Last edited:
Physics news on Phys.org
spence said:
"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...

She eats restaurants?
 
spence said:
I just discovered this website and want to thank everyone who is willing to contribute some of their time to help me. I appreciate it more than you know

First off, assume that state 1 is Chinese and that state 2 is Greek, and state 3 is Italian.

A student never eats the same kind of food for 2 consecutive weeks. If she eats a Chinese restaurant one week, then she is equally likely to have Greek as Italian food the next week. If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week. If she eats a Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week.

I feel like I am on the right track, but I'm having trouble translating the words to notation (the most important part).
Chinese could be represented by x, Greek by y, and Italian by z, correct? And that has to add up to 1?

"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...
So yeah. I kinda get the idea, I kinda don't. I think
In this problem there are three states: Chinese, Greek and Italian. The transition matrix tells you the probability of changing from one state to another. So it will be a $3\times3$ matrix. The rows and columns of the matrix will correspond to the states. So row 1 corresponds to Chinese, row 2 to Greek, and row 3 to Italian, and similarly for the columns. The $(i,j)$-element of the matrix give the probability of changing from state $j$ to state $i$. So for example the $(1,1)$-element of the matrix is the probability of changing from Chinese to Chinese. But the student never eats the same kind of food for 2 consecutive weeks. That tells you that the $(1,1)$-element of the matrix is $0$.

The $(2,1)$-element is the probability of changing from Chinese to Greek, and the $(3,1)$-element is the probability of changing from Chinese to Italian. Those probabilities must add up to $1$, and you are told that they are both equally llikely. So they must both be $\frac12$. The first column of the matrix is therefore $\begin{bmatrix}0 \\ \frac12 \\ \frac12 \end{bmatrix}$.

Now you have to do the same thing for the other two columns.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top