本次英国代写主要为随机过程的限时测试
1. 令 (Sn)n 0 是 S0 = 0 的随机游走，使得
P(Sn+1 = k + 1jSn = k) = 0:8
P(Sn+1 = k 1jSn = k) = 0:2 ：
令T为Sn = a或Sn = b的第一个时间，其中a和b为
正整数。
(a) 证明 Sn 3n
5和（
1个
4）Sn是mar，那S2
n +
6n
5 Sn 9n2
25是超级市场。
（b）找出P（ST = a）和P（ST = b）。
(c) 找出 E[T]。
2. 让 (Sn)n 0 是一个随机游走，使得
P（Sn + 1 = k + 2jSn = k）= 0：5
P（Sn + 1 = k 1jSn = k）= 0：5：
令（Fn）为S产生的分数。确定2 R，使得
Sn 是鞅。暗示：
E[锡+1
jFn] = 1
2个
锡+2
+ 1
2个
锡 1
：
3.与练习2中的情况相同。假设S0 = a> 0并设为
S 达到 0 的第一次。找到 P( < 1)，即概率
步行在一段时间内达到零。
4.令M为一个平方可积mar。流程不断增加
引用到M（用hMi表示）是唯一可预测的过程，例如
那M2
hMi是a，hMi0 =0。令M为平方
具有独立增量的可积mar，即Mn + 1 Mn为
不依赖于-代数Fn =（M0;：：：; Mn）。套装2
0 = V ar（M0）
并且对于 n 1; 2个
n ＝ V ar（Mn Mn 1）。求出V ar（Mn）。什么是 hMi？
5.令Sn为简单的随机游动：S0 = 0; Sn = U1 + : : : + Un，其中
Uis 是 iid 使得 0 < P(U1 = 1) = p < 1； P（U1 = 1）= 1 p = q。
假设 F0 = f;;
g 和 Fn = (U1; : : : ; Un)。
考虑符号功能
1. Let (Sn)n0 be a random walk with S0 = 0 such that
P(Sn+1 = k + 1jSn = k) = 0:8
P(Sn+1 = k 1jSn = k) = 0:2 :
Let T be the rst time that Sn = a or Sn = b where a and b are
positive integers.
(a) Prove that Sn 3n
5 and (
1
4 )Sn are martingales and that S2
n +
6n
5 Sn 9n2
25 is a supermartingale.
(b) Find P(ST = a) and P(ST = b).
(c) Find E[T].
2. Let (Sn)n0 be a random walk such that
P(Sn+1 = k + 2jSn = k) = 0:5
P(Sn+1 = k 1jSn = k) = 0:5 :
Let (Fn) be the ltration generated by S. Determine 2 R such that
Sn is a martingale. Hint:
E[Sn+1
jFn] = 1
2
Sn+2
+ 1
2
Sn 1
:
3. Same situation as in Exercise 2. Assume S0 = a > 0 and let be the
rst time that S reaches 0. Find P( < 1), i.e. the probability that
the walk reaches zero in a nite time.
4. Let M be a square integrable martingale. The increasing process asso-
ciated to M, denoted with hMi, is the unique predictable process such
that M2
hMi is a martingale and hMi0 = 0. Let M be a square
integrable martingale with independent increments, i.e. Mn+1 Mn is
independent of the -algebra Fn = (M0; : : : ;Mn). Set 2
0 = V ar(M0)
and for n 1; 2
n = V ar(Mn Mn 1). Find V ar(Mn). What is hMi?
5. Let Sn be a simple random walk: S0 = 0; Sn = U1 + : : : + Un, where
Uis are iid such that 0 < P(U1 = 1) = p < 1; P(U1 = 1) = 1 p = q.
Suppose F0 = f;;
g and Fn = (U1; : : : ; Un).
Consider the sign function