SOLUTION: Michigan State University Stochastic Processes Calculation Problem

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Hi, Please repress the attachedfile for details, let me apprehend if you feelany questions, thank you and amiable achievement. Best, James,

1.
1a.
Since at each vertex, the walker is akin mitigated to prefer one of the neighbors, the chance to
prefer any neighbor is elucidate p=1/3, hence, the transition matrix is
0
1/3 1/3 1/3
1/3
0
1/3 1/3
𝑃=(
)
1/3 1/3
0
1/3
1/3 1/3 1/3
0
Let

 = (𝑎 𝑏 𝑐 1 − 𝑎 − 𝑏 − 𝑐)
Then we feel P= gives
𝑏 + 𝑐 + (1 − 𝑎 − 𝑏 − 𝑐) = 3𝑎
𝑎 + 𝑐 + 1 − 𝑎 − 𝑏 − 𝑐 = 3𝑏
𝑎 + 𝑏 + 1 − 𝑎 − 𝑏 − 𝑐 = 3𝑐
𝑎 + 𝑏 + 𝑐 = 3 − 3𝑎 − 3𝑏 − 3𝑐
Solve these equations, we feel
𝑎=𝑏=𝑐=
1 1
4 4

=(

1
4

1 1
)
4 4

Because all the4 vertices are equiponderant, we can claim that the walker begin from 1 and walks to 2,
from 2 tail to 1, the chance is 1/3, to any other 2 recites is 2/3. accordingly at recite 3 or 4, the mean
term to go tail to 1 is the identical as from 2 to 1, let the reoccurring terms is T, then we feel
T= 2*1/3+ 2/3 *(1+T)
The original provisions is from 2 tail to 1, it has chance of 1/3 and the term is ...

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