Finding Markov Chain of $Y_n = M_n - S_n$?
up vote
1
down vote
favorite
Letting $X_n$ be i.i.d taking the value $1$ with probability $p$, and $-1$ with probably $1-p$,how do I show that $Y_n = M_n - S_n$, where $$M_n = max{0, S_1, S_2,ldots,S_n}$$ and $$S_n = X_1+ldots+X_n$$ is a Markov Chain? I can see that $M_n$ itself is not a Markov chain, but do not know how to show $Y_n = M_n - S_n$ is?
probability random-variables markov-chains
edited Nov 13 at 14:10
Snookie
3559
asked Nov 13 at 13:36
jessg12345
64
add a comment |
up vote
1
down vote
favorite
Letting $X_n$ be i.i.d taking the value $1$ with probability $p$, and $-1$ with probably $1-p$,how do I show that $Y_n = M_n - S_n$, where $$M_n = max{0, S_1, S_2,ldots,S_n}$$ and $$S_n = X_1+ldots+X_n$$ is a Markov Chain? I can see that $M_n$ itself is not a Markov chain, but do not know how to show $Y_n = M_n - S_n$ is?
probability random-variables markov-chains
edited Nov 13 at 14:10
Snookie
3559
asked Nov 13 at 13:36
jessg12345
64
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Letting $X_n$ be i.i.d taking the value $1$ with probability $p$, and $-1$ with probably $1-p$,how do I show that $Y_n = M_n - S_n$, where $$M_n = max{0, S_1, S_2,ldots,S_n}$$ and $$S_n = X_1+ldots+X_n$$ is a Markov Chain? I can see that $M_n$ itself is not a Markov chain, but do not know how to show $Y_n = M_n - S_n$ is?
probability random-variables markov-chains
edited Nov 13 at 14:10
Snookie
3559
asked Nov 13 at 13:36
jessg12345
64
Letting $X_n$ be i.i.d taking the value $1$ with probability $p$, and $-1$ with probably $1-p$,how do I show that $Y_n = M_n - S_n$, where $$M_n = max{0, S_1, S_2,ldots,S_n}$$ and $$S_n = X_1+ldots+X_n$$ is a Markov Chain? I can see that $M_n$ itself is not a Markov chain, but do not know how to show $Y_n = M_n - S_n$ is?
probability random-variables markov-chains
probability random-variables markov-chains
edited Nov 13 at 14:10
Snookie
3559
asked Nov 13 at 13:36
jessg12345
64
edited Nov 13 at 14:10
Snookie
3559
asked Nov 13 at 13:36
jessg12345
64
edited Nov 13 at 14:10
Snookie
3559
edited Nov 13 at 14:10
Snookie
3559
edited Nov 13 at 14:10
Snookie
3559
3559
asked Nov 13 at 13:36
jessg12345
64
asked Nov 13 at 13:36
jessg12345
64
asked Nov 13 at 13:36
jessg12345
64
64
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45
add a comment |
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
You need to look at the transition probability. Observe first that since $M_ngeq S_n$, you have $Y_ngeq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.
answered Nov 13 at 14:04
P. Quinton
1,261213
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You need to look at the transition probability. Observe first that since $M_ngeq S_n$, you have $Y_ngeq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.
answered Nov 13 at 14:04
P. Quinton
1,261213
add a comment |
up vote
0
down vote
You need to look at the transition probability. Observe first that since $M_ngeq S_n$, you have $Y_ngeq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.
answered Nov 13 at 14:04
P. Quinton
1,261213
add a comment |
up vote
0
down vote
up vote
0
down vote
You need to look at the transition probability. Observe first that since $M_ngeq S_n$, you have $Y_ngeq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.
answered Nov 13 at 14:04
P. Quinton
1,261213
You need to look at the transition probability. Observe first that since $M_ngeq S_n$, you have $Y_ngeq 0$. If $Y_n = 0$ then there is a $1/2$ probability that $Y_{n+1}=0$ and a $1/2$ probability that $Y_{n+1}=1$. if $Y_n>0$ then there is a $1/2$ probability that $Y_{n+1}=Y_n pm 1$, this is because $M_{n+1}=M_n$ in this case. These are the transition probability for the normal one sided random walk.
answered Nov 13 at 14:04
P. Quinton
1,261213
answered Nov 13 at 14:04
P. Quinton
1,261213
answered Nov 13 at 14:04
P. Quinton
1,261213
answered Nov 13 at 14:04
P. Quinton
1,261213
1,261213
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2996737%2ffinding-markov-chain-of-y-n-m-n-s-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
You may verify that from one of the equivalent definitions. Please show us your work and where you're stuck, and avoid asking no-clue questions.
– GNUSupporter 8964民主女神 地下教會
Nov 13 at 13:45