MHB Markov Chains - Finding a Transition Matrix for Probabilities

[Umar]

Hi! I have a question regarding making the transition matrix for the corresponding probabilities. The main problem I feel I have here is figuring out how to represent the probabilities in the question in the transition matrix. Like if something is 7 times more likely than something else.. Any help would be appreciated.

View attachment 6432

 

[I like Serena]

Hi! I have a question regarding making the transition matrix for the corresponding probabilities. The main problem I feel I have here is figuring out how to represent the probabilities in the question in the transition matrix. Like if something is 7 times more likely than something else.. Any help would be appreciated.

Hi Umar! ;)

If I'm not mistaken $a_{31}$ is the probability that the quantum state shifts from state 1 to state 3? (Wondering)

Then $a_{21}$ is the probability of the transition of state 1 to 2.
So if the latter is 7 times more likely then the first, that implies that $a_{21} = 7 a_{31}$.

The given statements result in:
$$A = \begin{bmatrix}
& a_{12} & a_{13} \\
7a_{31} && a_{13} \\
a_{31} &4a_{12} &
\end{bmatrix}$$
We can combine it with the fact that every row sums to $1$.

 

[Umar]

Hi Umar! ;)

If I'm not mistaken $a_{31}$ is the probability that the quantum state shifts from state 1 to state 3? (Wondering)

Then $a_{21}$ is the probability of the transition of state 1 to 2.
So if the latter is 7 times more likely then the first, that implies that $a_{21} = 7 a_{31}$.

The given statements result in:
$$A = \begin{bmatrix}
& a_{12} & a_{13} \\
7a_{31} && a_{13} \\
a_{31} &4a_{12} &
\end{bmatrix}$$
When we combine it with the fact that every row sums to $1$, we have a system of 3 equations and 3 unknowns...

Thanks for the reply! Your explanation really helped me to understand how to construct the transition matrix. So when I tried to solve the system for a31, a32 and a33, I got 1/29, 28/29 and 0 respectively. When I put this into my assignment website, I got 1/3... which I think was probably for getting the 0 right. But I don't get what mistake I could have made, because plugging in those values into the rows gives me the expected result of 1...

 

[I like Serena]

Thanks for the reply! Your explanation really helped me to understand how to construct the transition matrix. So when I tried to solve the system for a31, a32 and a33, I got 1/29, 28/29 and 0 respectively. When I put this into my assignment website, I got 1/3... which I think was probably for getting the 0 right. But I don't get what mistake I could have made, because plugging in those values into the rows gives me the expected result of 1...

Rereading the problem statement, I think we can deduce the following:
$$a_{11}=a_{22}=a_{33}=0$$
since it is given that the particle makes a transition to a different state at every step.
(And otherwise we wouldn't be able to solve the system, since we'd have 6 unknowns.)

Additionally, I think it's the column sum that has to add up to 1 instead of the row sum.
That's because if it's given that the particle is in state 1, then afterwards it's either in state 2 ($a_{21}$) or state 3 ($a_{31}$).
So together those should yield a probability of 1.

Finally, are we talking probability amplitudes or actual probabilities? (Wondering)
If they are probability amplitudes, perhaps we have to square them before comparing or adding?

 

[Umar]

Rereading the problem statement, I think we can deduce the following:
$$a_{11}=a_{22}=a_{33}=0$$
since it is given that the particle makes a transition to a different state at every step.
(And otherwise we wouldn't be able to solve the system, since we'd have 6 unknowns.)

Additionally, I think it's the column sum that has to add up to 1 instead of the row sum.
That's because if it's given that the particle is in state 1, then afterwards it's either in state 2 ($a_{21}$) or state 3 ($a_{31}$).
So together those should yield a probability of 1.

Finally, are we talking probability amplitudes or actual probabilities? (Wondering)
If they are probability amplitudes, perhaps we have to square them before comparing or adding?

Oh okay, I guess you're right with the columns having to add up to 1. I'll try that and report back what I get.

 

[Umar]

Rereading the problem statement, I think we can deduce the following:
$$a_{11}=a_{22}=a_{33}=0$$
since it is given that the particle makes a transition to a different state at every step.
(And otherwise we wouldn't be able to solve the system, since we'd have 6 unknowns.)

Additionally, I think it's the column sum that has to add up to 1 instead of the row sum.
That's because if it's given that the particle is in state 1, then afterwards it's either in state 2 ($a_{21}$) or state 3 ($a_{31}$).
So together those should yield a probability of 1.

Finally, are we talking probability amplitudes or actual probabilities? (Wondering)
If they are probability amplitudes, perhaps we have to square them before comparing or adding?

Hi! Just wanted to let you know that it worked, thanks for your help!