o****e
发帖数: 80 | 1 1) a rabbit can either jump up 2 steps or it will jump down 1 step, each with
50% chance. What is the probability
that it will eventually get to -1?
2) A game, you have $2, I have $1, toss a unfair coin, you win probability is 0.4 and I win probability 0.6,
once bet $1, what is the probability I win in the final. How many expect toss before the game is over. | c*****y
发帖数: 4 | 2 两道题解题思路一样. 构造鞅.
详细见:
http://www.mitbbs.com/article_t/Quant/31249959.html
with
probability is 0.4 and I win probability 0.6,
toss before the game is over.
【在 o****e 的大作中提到】
: 1) a rabbit can either jump up 2 steps or it will jump down 1 step, each with
: 50% chance. What is the probability
: that it will eventually get to -1?
: 2) A game, you have $2, I have $1, toss a unfair coin, you win probability is 0.4 and I win probability 0.6,
: once bet $1, what is the probability I win in the final. How many expect toss before the game is over.
| o****e
发帖数: 80 | 3 谢谢你，我弄明白第一题了。 第二题的概率可以用gamblers ruin来解，但是
expected toss怎么解？
【在 c*****y 的大作中提到】
: 两道题解题思路一样. 构造鞅.
: 详细见:
: http://www.mitbbs.com/article_t/Quant/31249959.html
:
: with
: probability is 0.4 and I win probability 0.6,
: toss before the game is over.
| o****e
发帖数: 80 | 4 expected toss in second question:
my solution is like this,
Markov chain
u0=0 u3=0
u1=1+0.6*u2+0.4*u0
u2=1+0.6*u3+0.4*u1 → u1=40/19 u2=35/19
so the expected toss is 40/19?
可是为什么从第二个人算，expected toss 就是 35/19?
【在 o****e 的大作中提到】
: 谢谢你，我弄明白第一题了。 第二题的概率可以用gamblers ruin来解，但是
: expected toss怎么解？
| c*****y
发帖数: 4 | 5 from the point of view of the one owns 2:
suppose current state is 0, then ending state is -2 or 1. Let n(k) be the
expected times need from state k=-2, -1, 0, 1, then, in responding to your
calculation, we have
u0=n(1), u1=n(0), u2=n(-1), u3=n(-2). | o****e
发帖数: 80 | 6 就是说两个player的mc不同因为赢得概率不同，一个0。4， 一个0。6。
从p=0.6的player 1的角度来看，就是我刚才这样算得
Markov chain
u0=0 u3=0
u1=1+0.6*u2+0.4*u0
u2=1+0.6*u3+0.4*u1 → u1=40/19 u2=35/19
so the expected toss is u1=40/19
那么u2=35/19又代表什么呢？
your
【在 c*****y 的大作中提到】
: from the point of view of the one owns 2:
: suppose current state is 0, then ending state is -2 or 1. Let n(k) be the
: expected times need from state k=-2, -1, 0, 1, then, in responding to your
: calculation, we have
: u0=n(1), u1=n(0), u2=n(-1), u3=n(-2).
| s****n
发帖数: 1237 | 7 u2表示p1=0.6的$2，p2=0.4的$1开始的概率。和现在p1=0.4的$2，p2=0.6的$1开始不一
样。
如果p1=p2=0.5那u1应该=u2。
【在 o****e 的大作中提到】
: 就是说两个player的mc不同因为赢得概率不同，一个0。4， 一个0。6。
: 从p=0.6的player 1的角度来看，就是我刚才这样算得
: Markov chain
: u0=0 u3=0
: u1=1+0.6*u2+0.4*u0
: u2=1+0.6*u3+0.4*u1 → u1=40/19 u2=35/19
: so the expected toss is u1=40/19
: 那么u2=35/19又代表什么呢？
:
: your
| p*****k
发帖数: 318 | 8 ogtree, in addition to the excellent posts above, see
http://www.mitbbs.com/article_t/Quant/31207033.html
for the martingale approach to problem (2)
(a=2 & b=1 for your particular question)
once you get the ruin probability, simply use the martingale:
X(n)-0.2*n
and apply optional stopping to get the expected # of tosses:
E[X(n)-0.2*n]=0,
i.e., E[n]=5*E[X(n)]=5*[9/19*(+2)+10/19*(-1)]=40/19 | J*******g
发帖数: 267 | 9 yes, two diff mc from the perspective of two diff players, and diff initial
state
all factors included, you get the same answer
【在 o****e 的大作中提到】
: 就是说两个player的mc不同因为赢得概率不同，一个0。4， 一个0。6。
: 从p=0.6的player 1的角度来看，就是我刚才这样算得
: Markov chain
: u0=0 u3=0
: u1=1+0.6*u2+0.4*u0
: u2=1+0.6*u3+0.4*u1 → u1=40/19 u2=35/19
: so the expected toss is u1=40/19
: 那么u2=35/19又代表什么呢？
:
: your
| o****e
发帖数: 80 | 10 谢谢你，你真是好人那
【在 p*****k 的大作中提到】
: ogtree, in addition to the excellent posts above, see
: http://www.mitbbs.com/article_t/Quant/31207033.html
: for the martingale approach to problem (2)
: (a=2 & b=1 for your particular question)
: once you get the ruin probability, simply use the martingale:
: X(n)-0.2*n
: and apply optional stopping to get the expected # of tosses:
: E[X(n)-0.2*n]=0,
: i.e., E[n]=5*E[X(n)]=5*[9/19*(+2)+10/19*(-1)]=40/19
|
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