i*******w
发帖数: 405 | 1 偶太笨，还是没相通该怎么解这个问题，哪位牛讲一下，谢谢。
一个青蛙，每次跳1m，往左跳1m概率是0.4，往右跳1m概率0.6，问往右第一次跳到9m的期望步数 | P*****f
发帖数: 2272 | 2 use recursion. Let E(x) be the expectation of steps when first hit the X
meter point on the right then we have
E(x) = 0.6(E(x-1)+1) + 0.4(E(x-1)+1+E(2)).
to solve the recursive, we need first get what is E(2).
let x = 2, then we have
E(2) = 1+E(1) + 0.4*E(2) => 0.6*E(2) = 1+E(1). (a)
Then analyze the E(1), we can easily get
E(1) = 0.6 + 0.4(1+E(2)) (b)
combine condition (a) and (b), we can eaily get E(1) =5, E(2) = 10
Then the recursion becomes E(x) = E(x-1) + 5. E(9) should be 45
偶太笨，还是没相通该怎么
【在 i*******w 的大作中提到】
: 偶太笨，还是没相通该怎么解这个问题，哪位牛讲一下，谢谢。
: 一个青蛙，每次跳1m，往左跳1m概率是0.4，往右跳1m概率0.6，问往右第一次跳到9m的期望步数
| s*******e
发帖数: 28 | 3
Good solution!
Another way is to use Martingale. Set X_i = 1 with p and -1 with q.
let S_n = \sum_1^n X_i - n*E[X_i] and because E[X_i] = p - q, where p \= q.
Then according to Optional stopping theorem, if n is finite, then
E[S_n] = E[S_0] = 0.
Therefore, as \sum_1^n X_i = 9, we get 0 = 9 - E[n] (p-q).
Then E[n] = 9/(p-q) = 9/(0.6-0.4) = 45.
Note before we use this theorem, the condition that stopping time is finite
must be justified. In this case it happens to be true.
If p=q, the stopping
【在 P*****f 的大作中提到】
: use recursion. Let E(x) be the expectation of steps when first hit the X
: meter point on the right then we have
: E(x) = 0.6(E(x-1)+1) + 0.4(E(x-1)+1+E(2)).
: to solve the recursive, we need first get what is E(2).
: let x = 2, then we have
: E(2) = 1+E(1) + 0.4*E(2) => 0.6*E(2) = 1+E(1). (a)
: Then analyze the E(1), we can easily get
: E(1) = 0.6 + 0.4(1+E(2)) (b)
: combine condition (a) and (b), we can eaily get E(1) =5, E(2) = 10
: Then the recursion becomes E(x) = E(x-1) + 5. E(9) should be 45
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