Check whether a stochastic process is Markov
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I have difficulty checking the Markov property of some stochastic process.
Let $W(t)$ be a standard Wiener process adapted to $mathscrF_s$ which is a filtration generated by this Wiener process. I want to check whether $X_t = cos(W_t)$ and $Y_t = W_t+W_t-2$ is Markovian.
The definition says, if a random process is Markovian, for any Borel-measurable function $f$ there exists another Borel-measurable function $g$ such that $E[f(X_t)|mathscrF_s] = g(X_s)$.
beginalign*
E[f(X_t)|mathscrF_s] &= E[f(cos(W_t))|mathscrF_s]\
&= E[f(cos(W_t-W_s+W_s))|mathscrF_s]\
endalign*
Let $h(w) = E[f(cos(W_t-W_s+w))|mathscrF_s]$, then:
beginalign*
h(w) &= int_-infty^infty f(cos(y+w))frac1sqrt2pi (t-s)e^-fracy^22(t-s) mathrmdy\
&= int_-infty^infty f(cos(y))frac1sqrt2pi (t-s)e^-frac(y-w)^22(t-s) mathrmdy
endalign*
So $g(x)=h(arccos(x))$? It seems strange since $W_s$ is not necessarily between $-1$ and $1$. I intuitively think it is markovian, due to the periodicity of $cos$. How to overcome this problem?
And for $Y_t$, I can only start with
beginalign*
E[f(W_t+W_t-2)|mathscrF_s] &= E[f(W_t+W_t-2-(W_s + W_s-2)+(W_s + W_s-2))|mathscrF_s]\
&= E[f((W_t-W_s)+(W_t-2- W_s-2)+(W_s + W_s-2))|mathscrF_s]
endalign*
I guess it is not Markovian since $W_t-2- W_s-2$ is not necessarily independent from $mathscrF_s$, then I cannot apply the same approach in $X_t$. But I am not sure whether I am right and how to get the result rigorously.
Thank you for any help!
stochastic-processes markov-chains conditional-expectation markov-process
asked Jan 31 at 1:10
Edward Wang
697411
add a comment |Â
up vote
0
down vote
favorite
I have difficulty checking the Markov property of some stochastic process.
Let $W(t)$ be a standard Wiener process adapted to $mathscrF_s$ which is a filtration generated by this Wiener process. I want to check whether $X_t = cos(W_t)$ and $Y_t = W_t+W_t-2$ is Markovian.
The definition says, if a random process is Markovian, for any Borel-measurable function $f$ there exists another Borel-measurable function $g$ such that $E[f(X_t)|mathscrF_s] = g(X_s)$.
beginalign*
E[f(X_t)|mathscrF_s] &= E[f(cos(W_t))|mathscrF_s]\
&= E[f(cos(W_t-W_s+W_s))|mathscrF_s]\
endalign*
Let $h(w) = E[f(cos(W_t-W_s+w))|mathscrF_s]$, then:
beginalign*
h(w) &= int_-infty^infty f(cos(y+w))frac1sqrt2pi (t-s)e^-fracy^22(t-s) mathrmdy\
&= int_-infty^infty f(cos(y))frac1sqrt2pi (t-s)e^-frac(y-w)^22(t-s) mathrmdy
endalign*
So $g(x)=h(arccos(x))$? It seems strange since $W_s$ is not necessarily between $-1$ and $1$. I intuitively think it is markovian, due to the periodicity of $cos$. How to overcome this problem?
And for $Y_t$, I can only start with
beginalign*
E[f(W_t+W_t-2)|mathscrF_s] &= E[f(W_t+W_t-2-(W_s + W_s-2)+(W_s + W_s-2))|mathscrF_s]\
&= E[f((W_t-W_s)+(W_t-2- W_s-2)+(W_s + W_s-2))|mathscrF_s]
endalign*
I guess it is not Markovian since $W_t-2- W_s-2$ is not necessarily independent from $mathscrF_s$, then I cannot apply the same approach in $X_t$. But I am not sure whether I am right and how to get the result rigorously.
Thank you for any help!
stochastic-processes markov-chains conditional-expectation markov-process
asked Jan 31 at 1:10
Edward Wang
697411
@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have difficulty checking the Markov property of some stochastic process.
Let $W(t)$ be a standard Wiener process adapted to $mathscrF_s$ which is a filtration generated by this Wiener process. I want to check whether $X_t = cos(W_t)$ and $Y_t = W_t+W_t-2$ is Markovian.
The definition says, if a random process is Markovian, for any Borel-measurable function $f$ there exists another Borel-measurable function $g$ such that $E[f(X_t)|mathscrF_s] = g(X_s)$.
beginalign*
E[f(X_t)|mathscrF_s] &= E[f(cos(W_t))|mathscrF_s]\
&= E[f(cos(W_t-W_s+W_s))|mathscrF_s]\
endalign*
Let $h(w) = E[f(cos(W_t-W_s+w))|mathscrF_s]$, then:
beginalign*
h(w) &= int_-infty^infty f(cos(y+w))frac1sqrt2pi (t-s)e^-fracy^22(t-s) mathrmdy\
&= int_-infty^infty f(cos(y))frac1sqrt2pi (t-s)e^-frac(y-w)^22(t-s) mathrmdy
endalign*
So $g(x)=h(arccos(x))$? It seems strange since $W_s$ is not necessarily between $-1$ and $1$. I intuitively think it is markovian, due to the periodicity of $cos$. How to overcome this problem?
And for $Y_t$, I can only start with
beginalign*
E[f(W_t+W_t-2)|mathscrF_s] &= E[f(W_t+W_t-2-(W_s + W_s-2)+(W_s + W_s-2))|mathscrF_s]\
&= E[f((W_t-W_s)+(W_t-2- W_s-2)+(W_s + W_s-2))|mathscrF_s]
endalign*
I guess it is not Markovian since $W_t-2- W_s-2$ is not necessarily independent from $mathscrF_s$, then I cannot apply the same approach in $X_t$. But I am not sure whether I am right and how to get the result rigorously.
Thank you for any help!
stochastic-processes markov-chains conditional-expectation markov-process
asked Jan 31 at 1:10
Edward Wang
697411
I have difficulty checking the Markov property of some stochastic process.
Let $W(t)$ be a standard Wiener process adapted to $mathscrF_s$ which is a filtration generated by this Wiener process. I want to check whether $X_t = cos(W_t)$ and $Y_t = W_t+W_t-2$ is Markovian.
The definition says, if a random process is Markovian, for any Borel-measurable function $f$ there exists another Borel-measurable function $g$ such that $E[f(X_t)|mathscrF_s] = g(X_s)$.
beginalign*
E[f(X_t)|mathscrF_s] &= E[f(cos(W_t))|mathscrF_s]\
&= E[f(cos(W_t-W_s+W_s))|mathscrF_s]\
endalign*
Let $h(w) = E[f(cos(W_t-W_s+w))|mathscrF_s]$, then:
beginalign*
h(w) &= int_-infty^infty f(cos(y+w))frac1sqrt2pi (t-s)e^-fracy^22(t-s) mathrmdy\
&= int_-infty^infty f(cos(y))frac1sqrt2pi (t-s)e^-frac(y-w)^22(t-s) mathrmdy
endalign*
So $g(x)=h(arccos(x))$? It seems strange since $W_s$ is not necessarily between $-1$ and $1$. I intuitively think it is markovian, due to the periodicity of $cos$. How to overcome this problem?
And for $Y_t$, I can only start with
beginalign*
E[f(W_t+W_t-2)|mathscrF_s] &= E[f(W_t+W_t-2-(W_s + W_s-2)+(W_s + W_s-2))|mathscrF_s]\
&= E[f((W_t-W_s)+(W_t-2- W_s-2)+(W_s + W_s-2))|mathscrF_s]
endalign*
I guess it is not Markovian since $W_t-2- W_s-2$ is not necessarily independent from $mathscrF_s$, then I cannot apply the same approach in $X_t$. But I am not sure whether I am right and how to get the result rigorously.
Thank you for any help!
stochastic-processes markov-chains conditional-expectation markov-process
stochastic-processes markov-chains conditional-expectation markov-process
asked Jan 31 at 1:10
Edward Wang
697411
asked Jan 31 at 1:10
Edward Wang
697411
edited Sep 9 at 2:37
asked Jan 31 at 1:10
Edward Wang
697411
asked Jan 31 at 1:10
Edward Wang
697411
asked Jan 31 at 1:10
Edward Wang
697411
697411
@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54
add a comment |Â
@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54
@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54
add a comment |Â
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@Shalop Should it be that it cannot be written as $g(W_t-1+W_t-3)$? Though it seems to me it suffices to prove it cannot be written as $g(W_t-1)$. Sorry I still cannot see the contradiction. Is it just we cannot write $W_t-2 = h(W_t-1)$?
– Edward Wang
Jan 31 at 4:47
It seems you err similarly in both cases. For example, to show $X$ is Markov, for every $s<t$ and every suitable function $f$, one is supposed to exhibit some function $g$ such that $$E(f(X_t)midmathcal F^X_s)=g(X_s)$$ where $mathcal F^X_s=sigma(X_u;uleqslant s)$, not such that $$E(f(X_t)midmathcal F_s)=g(X_s)$$ where $mathcal F_s=sigma(W_u;uleqslant s)$. Since $mathcal F^X_snemathcal F_s$, these are not equivalent.
– Did
Feb 5 at 19:59
@Did I understand what you mean, yes they are different. I just read from wikipedia, and the definition on that page does not specify what the filtration should be. Only $X_t in mathbbF_t$ is required.
– Edward Wang
Feb 5 at 21:14
No, the definition of a Markov process does not let you the choice of the filtration.
– Did
Mar 7 at 0:54