Markov property of continuous time stochastic processes
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Suppose $X_t$ is a continuous time stochastic process and $B_t$ is a Brownian motion. Suppose there is a deterministic function $f:mathbb{R}rightarrow mathbb{R}$. What conditions that $f$ needs to make sure that $Y_t=f(X_t)$ is still markov? What conditions that $f$ needs to make sure that $Z_t=f(B_t)$ is a markov process.
Suppose there are a sequence of independent continuous time Markov process $X_{i,t}$ and independent Brownian motions $W_{i,t}$. There is a deterministic function $g:mathbb{R^n}rightarrowmathbb{R}$. Let $Y_t=g(X_{1,t},...,X_{n,t})$ and $Z_t=g(W_{1,t},...,W_{n,t})$. Are $Y_t$ and $Z_t$ still Markov? Is $Z_t$ a martingale.
Few examples of $Y_t$ and $Z_t$ could be
$$
Z_t=aW_{1,t}+bW_{2,t}\
Z_t=W_{1,t}^2+2tW_{1,t}\
Z_t=W_{1,t}^3+W_{2,t}^2\
Z_t=W_{1,t}W_{2,t}.
$$
Usually, we know the definition of a Markov process. But It hard to see problems about how to show a function of stochastic process is Markov in general.
stochastic-processes stochastic-calculus stochastic-analysis
asked Dec 15 '18 at 5:06
YHHYHH
442214
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add a comment |
$begingroup$
Suppose $X_t$ is a continuous time stochastic process and $B_t$ is a Brownian motion. Suppose there is a deterministic function $f:mathbb{R}rightarrow mathbb{R}$. What conditions that $f$ needs to make sure that $Y_t=f(X_t)$ is still markov? What conditions that $f$ needs to make sure that $Z_t=f(B_t)$ is a markov process.
Suppose there are a sequence of independent continuous time Markov process $X_{i,t}$ and independent Brownian motions $W_{i,t}$. There is a deterministic function $g:mathbb{R^n}rightarrowmathbb{R}$. Let $Y_t=g(X_{1,t},...,X_{n,t})$ and $Z_t=g(W_{1,t},...,W_{n,t})$. Are $Y_t$ and $Z_t$ still Markov? Is $Z_t$ a martingale.
Few examples of $Y_t$ and $Z_t$ could be
$$
Z_t=aW_{1,t}+bW_{2,t}\
Z_t=W_{1,t}^2+2tW_{1,t}\
Z_t=W_{1,t}^3+W_{2,t}^2\
Z_t=W_{1,t}W_{2,t}.
$$
Usually, we know the definition of a Markov process. But It hard to see problems about how to show a function of stochastic process is Markov in general.
stochastic-processes stochastic-calculus stochastic-analysis
asked Dec 15 '18 at 5:06
YHHYHH
442214
$endgroup$
1
$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
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– Kavi Rama Murthy
Dec 15 '18 at 5:12
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Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
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– Did
Dec 15 '18 at 6:48
1
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@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58
add a comment |
$begingroup$
Suppose $X_t$ is a continuous time stochastic process and $B_t$ is a Brownian motion. Suppose there is a deterministic function $f:mathbb{R}rightarrow mathbb{R}$. What conditions that $f$ needs to make sure that $Y_t=f(X_t)$ is still markov? What conditions that $f$ needs to make sure that $Z_t=f(B_t)$ is a markov process.
Suppose there are a sequence of independent continuous time Markov process $X_{i,t}$ and independent Brownian motions $W_{i,t}$. There is a deterministic function $g:mathbb{R^n}rightarrowmathbb{R}$. Let $Y_t=g(X_{1,t},...,X_{n,t})$ and $Z_t=g(W_{1,t},...,W_{n,t})$. Are $Y_t$ and $Z_t$ still Markov? Is $Z_t$ a martingale.
Few examples of $Y_t$ and $Z_t$ could be
$$
Z_t=aW_{1,t}+bW_{2,t}\
Z_t=W_{1,t}^2+2tW_{1,t}\
Z_t=W_{1,t}^3+W_{2,t}^2\
Z_t=W_{1,t}W_{2,t}.
$$
Usually, we know the definition of a Markov process. But It hard to see problems about how to show a function of stochastic process is Markov in general.
stochastic-processes stochastic-calculus stochastic-analysis
asked Dec 15 '18 at 5:06
YHHYHH
442214
$endgroup$
Suppose $X_t$ is a continuous time stochastic process and $B_t$ is a Brownian motion. Suppose there is a deterministic function $f:mathbb{R}rightarrow mathbb{R}$. What conditions that $f$ needs to make sure that $Y_t=f(X_t)$ is still markov? What conditions that $f$ needs to make sure that $Z_t=f(B_t)$ is a markov process.
Suppose there are a sequence of independent continuous time Markov process $X_{i,t}$ and independent Brownian motions $W_{i,t}$. There is a deterministic function $g:mathbb{R^n}rightarrowmathbb{R}$. Let $Y_t=g(X_{1,t},...,X_{n,t})$ and $Z_t=g(W_{1,t},...,W_{n,t})$. Are $Y_t$ and $Z_t$ still Markov? Is $Z_t$ a martingale.
Few examples of $Y_t$ and $Z_t$ could be
$$
Z_t=aW_{1,t}+bW_{2,t}\
Z_t=W_{1,t}^2+2tW_{1,t}\
Z_t=W_{1,t}^3+W_{2,t}^2\
Z_t=W_{1,t}W_{2,t}.
$$
Usually, we know the definition of a Markov process. But It hard to see problems about how to show a function of stochastic process is Markov in general.
stochastic-processes stochastic-calculus stochastic-analysis
stochastic-processes stochastic-calculus stochastic-analysis
asked Dec 15 '18 at 5:06
YHHYHH
442214
asked Dec 15 '18 at 5:06
YHHYHH
442214
asked Dec 15 '18 at 5:06
YHHYHH
442214
asked Dec 15 '18 at 5:06
YHHYHH
442214
asked Dec 15 '18 at 5:06
YHHYHH
442214
442214
1
$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 5:12
$begingroup$
Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
$endgroup$
– Did
Dec 15 '18 at 6:48
1
$begingroup$
@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58
add a comment |
1
$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 5:12
$begingroup$
Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
$endgroup$
– Did
Dec 15 '18 at 6:48
1
$begingroup$
@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58
1
1
$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 5:12
$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 5:12
$begingroup$
Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
$endgroup$
– Did
Dec 15 '18 at 6:48
$begingroup$
Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
$endgroup$
– Did
Dec 15 '18 at 6:48
1
1
$begingroup$
@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58
add a comment |
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$begingroup$
Markov property only requires $f$ to be a measurable function. Martingale property is rarely preserved by functions of the process.
$endgroup$
– Kavi Rama Murthy
Dec 15 '18 at 5:12
$begingroup$
Right, these are two exercises copied from your homework.Now, the question is what did you do to try to solve them?
$endgroup$
– Did
Dec 15 '18 at 6:48
1
$begingroup$
@KaviRamaMurthy Thanks. To show martingale, I usually use the definition of Martingale and show that the expectation doesn't change over time. What I'm I confused is how to show the Markov property because the Markov property is in terms of the probability instead of expectation. In discrete time, it is easy. But I don't how to show it for continuous time stochastic processes.
$endgroup$
– YHH
Dec 15 '18 at 15:57
$begingroup$
@KaviRamaMurthy Can you give some references showing the requirement is $f$ be a measurable function.
$endgroup$
– YHH
Dec 15 '18 at 15:58