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Quant版 - a questin about Markov Chain
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话题: status话题: person话题: changes话题: find话题: questin
进入Quant版参与讨论
1 (共1页)
i*******e
发帖数: 8
1
We consider 3 types of the status X_t:
(i) a person is employed (E);
(ii) a person is unemployed (U);
(iii) a person is retired (R).
denote the employment status measured annually, Let assume that probability P(X_t+1|X_t
) is given, and initial status is E,
P(E|E) = 0.8; P(U|E) = 0.01; P(R|E)= 0.19;
P(E|U) = 0.5; P(U|U) = 0.375; P(R|U)= 0.125;
P(E|R) = 0; P(U|R) = 0; P(R|R)=1;
Find the the average number of status changes.
analytical solution?
b*****i
发帖数: 58
2
1. find the stationary distribution for U = (P(E), P(U), P(R))' by sovling T
'*U = U, where T is the transition matrix between status
2. calclulate the expected number of status changes, e.g., it is 1-0.8 = 0.2
if a status is E.
correct?

_t

【在 i*******e 的大作中提到】
: We consider 3 types of the status X_t:
: (i) a person is employed (E);
: (ii) a person is unemployed (U);
: (iii) a person is retired (R).
: denote the employment status measured annually, Let assume that probability P(X_t+1|X_t
: ) is given, and initial status is E,
: P(E|E) = 0.8; P(U|E) = 0.01; P(R|E)= 0.19;
: P(E|U) = 0.5; P(U|U) = 0.375; P(R|U)= 0.125;
: P(E|R) = 0; P(U|R) = 0; P(R|R)=1;
: Find the the average number of status changes.

n*w
发帖数: 41
3
Is there stationary state when solving T'U=U?

T
.2

【在 b*****i 的大作中提到】
: 1. find the stationary distribution for U = (P(E), P(U), P(R))' by sovling T
: '*U = U, where T is the transition matrix between status
: 2. calclulate the expected number of status changes, e.g., it is 1-0.8 = 0.2
: if a status is E.
: correct?
:
: _t

i******d
发帖数: 54
4
不是应该找到transition matrix的特征值为1的特征向量么?
i*******e
发帖数: 8
5
我好像漏了初始状态是E, 我做MC是7.5,但analytical 不知道

T
.2

【在 b*****i 的大作中提到】
: 1. find the stationary distribution for U = (P(E), P(U), P(R))' by sovling T
: '*U = U, where T is the transition matrix between status
: 2. calclulate the expected number of status changes, e.g., it is 1-0.8 = 0.2
: if a status is E.
: correct?
:
: _t

i*******e
发帖数: 8
6
能给 any detail ? thanks

【在 i******d 的大作中提到】
: 不是应该找到transition matrix的特征值为1的特征向量么?
l******n
发帖数: 9344
7
suppose the average number of status changes starting with E is x
.... with U is y
Since R is absorbing, it will stop at R.We have
x=.8(x+1)+.01(y+1)+.19*1
y=.5(x+1)+.375(y+1)+0.125*1
Solve for x and y


probability P(X_t+1|X_t

【在 i*******e 的大作中提到】
: We consider 3 types of the status X_t:
: (i) a person is employed (E);
: (ii) a person is unemployed (U);
: (iii) a person is retired (R).
: denote the employment status measured annually, Let assume that probability P(X_t+1|X_t
: ) is given, and initial status is E,
: P(E|E) = 0.8; P(U|E) = 0.01; P(R|E)= 0.19;
: P(E|U) = 0.5; P(U|U) = 0.375; P(R|U)= 0.125;
: P(E|R) = 0; P(U|R) = 0; P(R|R)=1;
: Find the the average number of status changes.

i*******e
发帖数: 8
8
thanks, 应该是这样吧,跟我mc的差不多,只是不明白为什么 x = .8(x+1)...而不是x=.8(x)...
这种情况下状态没变,为什么要加1呢? thanks

【在 l******n 的大作中提到】
: suppose the average number of status changes starting with E is x
: .... with U is y
: Since R is absorbing, it will stop at R.We have
: x=.8(x+1)+.01(y+1)+.19*1
: y=.5(x+1)+.375(y+1)+0.125*1
: Solve for x and y
:
:
: probability P(X_t+1|X_t

p****o
发帖数: 88
9
(x+1)是因为人为地加了一步造成的吧,x是natural process的话。

是x=.8(x)...

【在 i*******e 的大作中提到】
: thanks, 应该是这样吧,跟我mc的差不多,只是不明白为什么 x = .8(x+1)...而不是x=.8(x)...
: 这种情况下状态没变,为什么要加1呢? thanks

1 (共1页)
进入Quant版参与讨论
相关主题
请教一下关于Markov Chain和Brownian Movement的知识或书籍看看这道题(probability)
请问Markov Chain Monte Carlo和 Monte Carlo根本性的区别是什么?发个概率题
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问一到markov chain的题目请教一道最近面的Stochastic Process面试题
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面试的时候真会被问到Markov Chain么?[合集] 请问有人作financial risk management 方面的工作吗?
相关话题的讨论汇总
话题: status话题: person话题: changes话题: find话题: questin