# SOLUTION: Michigan State University Stochastic Processes Calculation Problem

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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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