y*p
发帖数: 47 | 1 A trader either wins or loses $100,000 on each trade, and has the probabilit
y of winning at .55. The
trader starts with a capital of $1,000,000. What is the probabiity for that
trader of being ruined
evenually? | J*****n
发帖数: 4859 | 2
probabilit
that
Markov 链
解迭代方程：
a_n=0.45*a_{n-1}+0.55*a_{n+1} (1)
a_0=1
a_{inf}=0
目标求a_10
(1)可以化为：11(a_{n}-a_{n-1})=9(a_{n-1}-a{n-2})
设D_n=a_{n}-a{n-1},则D_n=9*D_{n-1}/11
a_{inf}=a_0+(a_1-a_0)+(a_2-a_1)+.....
i.e. 0=1+D_1+D_2+.....
则D_1=-2/11,a_10=D_1+D_2+....+D_10+a_0=(9/11)^10
【在 y*p 的大作中提到】
: A trader either wins or loses $100,000 on each trade, and has the probabilit
: y of winning at .55. The
: trader starts with a capital of $1,000,000. What is the probabiity for that
: trader of being ruined
: evenually?
| s*********l
发帖数: 103 | 3 In a random walk with P(x->x+1) = p, the probability of ever
reaching the position z > 0 from the origin equals 1 if p>=q
and equals (p/q)^z if p
In this case, p = 0.45, q = 0.55, z = 10.
probabilit
that
【在 y*p 的大作中提到】
: A trader either wins or loses $100,000 on each trade, and has the probabilit
: y of winning at .55. The
: trader starts with a capital of $1,000,000. What is the probabiity for that
: trader of being ruined
: evenually?
| i*******e
发帖数: 29 | | o*******r
发帖数: 131 | 5 问题转化为
非对称random walk，往左走一步概率为P=0.55，往右走一部Q=0.45
开始点为0，左边边界为 -10， 右边为正无穷大。
问 P(先到达-10) 的概率
然后直接套公式得
1 - (Q/P)^infty | b***e
发帖数: 1419 | 6 Here is a simple version of the solution, which requires no Markov chain or
random walk formula.
It is easy to see that:
f(n) = p * f(n-1) + q * f(n+1)
where p is probability of losing and q is the probably of winning (p + q =
1).
Also, it is the case that:
f(n) = f(n-k) * f(k)
because we can always divide the money to 2 piles (arbitrarily), and to
lose all $n means to lose both piles of money (which are independent of
each other).
With these two equations, we can easily have:
f(n) = z^n
where z |
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