Problem 4. Random walk with killing. A random walker has initial location r, an
ID: 3311149 • Letter: P
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Problem 4. Random walk with killing. A random walker has initial location r, an integer between 1 and A + B. At each time n1,2,3,... he tosses a fair coin; if it comes up Heads, he moves one step +1 to the right; if Tails, he moves one step -1 to the left. But before each step, the random walk can be accidentally killed, ending the random walk; this happens with probability 0r i. (Thus, before each left-right move there is an independent r-coin toss to decide whether or not the random walk ends.) (A) Let u(x) be the probability that the random walker eventually reaches either endpoint 0 or A B without being killed first. Show that u(x) satisfies a second-order difference equation of the form (DE) a(z) = au(x-1) + au(x + 1) for x = 1, 2, . . . , A + B-1, What is a? What boundary conditions must the function u(x) satisfy?Explanation / Answer
a) The probability function can be modeled as after n-tosses. Lets's assume that the random walk has not ended. Within n tosses, we will assume i tosses were heads and the rest heads. Thus, the cumulative movement is +1*i+(-1)*(n-i)=i+i-n or 2i-n. The possible combinations of heads and tails can be written as nCi. Thus the total movement is
nCi*(2i-n)*(1/2)^i*(1/2)*(n-i) which can be simplified as nCi*(2i-n)*(1/2)^n. This has to be multiplied with (1-r)^n as the random walk is not killed.
Thus the total function becomes nCi*(2i-n)*[(1-r)/2]^n for a movement of 2i-n after n tosses.Thus, if the initial location is x, for the particle to reach A+B, 2i-n=A+B-x and for the particle to reach 0, 2i-n=-x. Thus, he value scan be substituted in the function to check for second order condition which gets satisfied.
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