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masu
May 28th 2006, 10:26 AM
Question: Let Sn be a random walk with mean=0 and S0=x. Suppose that
P(a-c<= ST<=b+d)=1, where a<x<b, c>=0, and d>=0
(a) Show that (a-c)*P(ST<=a) + b*P(ST>=b) <= x <= a*P(ST<=a) + (b+d)*P(ST>=b)

Sol: P(a-c<= ST <=b+d) =1
EST = x + ET*mean =x (since mean=0)
= sum (y*P(ST=y)) from y=a-c to b+d

and since mean=0, 0=mean=sum(y*P(ST=y)) from y=a+1 to b-1 .....

(p.s. the "mean" above is the mean of random variable Xi where the Sn=x+X1+X2+...+Xn, n>=1)

Is there anybody can tell me why? I can not get it

Thanks