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Consider a simple random walk on the four vertex graph (Square shape with A,B,C,D being vertex)

Assume that the payoff function is: $$f(A) =2,\\ f(B) = 4,\\ f(C) = 5,\\ f(D) = 3.$$ Assume that there is no cost associate with moving, but there is a discount factor $a$. What is the largest possible value of a so that the optimal stopping strategy is to stop at every vertex, i.e., so that $S_2 = \{A,B,C,D\}$?

I have really hard time solving this problem. Please help me.

5xum
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James
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1 Answers1

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Solve $f(A)=(1-a)\cdot\frac12\cdot(f(B)+f(D))$.

Did
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  • Can you please help elaborate on how you arrived at the equation above? I could not derive it. – ghjk Nov 21 '20 at 05:33
  • I meant, how come you have the factor $1-\alpha$ rather than only $\alpha$? – ghjk Nov 21 '20 at 10:49