AR(1) to Ornstein-Uhlenbeck for AR(1) process of the form $ln z_{t+1}=rho ln z_t+sigma...
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I have the AR(1) process of the following form:
$$ln z_{t+1}=rho ln z_t+sigma sqrt{(1-rho^2)}epsilon_t$$
And need to find its continous time corresponding Ornstein-Uhlenbeckprocess.
I have the info that an AR(1) can be translated to an Ornstein-Uhlenbeck as follows:
$$dx=theta(bar{x}-x)dt+sigma dW tag{O-U}$$
$$x_{t+1}=thetabar{x}+(1-theta)x_t +sigmaepsilon_ttag{AR(1)}$$
but fail to see how to apply this to this specific case. I have looked at other answers considering the translation between AR(1) and OU, but again do not clearly see it for this specific form.
Any help is highly appreciated :-)
edit:
Using the answer below and a linear approximation of $e^{-theta Delta}approx1-theta Delta$ and setting $Delta=1$ I get
$$dln z_t=-(1-rho) ln z_t +sigma sqrt{(1-rho^2)}dW_t$$
which corresponds to the approximation as given by (AR(1))
stochastic-processes vector-auto-regression
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up vote
1
down vote
favorite
I have the AR(1) process of the following form:
$$ln z_{t+1}=rho ln z_t+sigma sqrt{(1-rho^2)}epsilon_t$$
And need to find its continous time corresponding Ornstein-Uhlenbeckprocess.
I have the info that an AR(1) can be translated to an Ornstein-Uhlenbeck as follows:
$$dx=theta(bar{x}-x)dt+sigma dW tag{O-U}$$
$$x_{t+1}=thetabar{x}+(1-theta)x_t +sigmaepsilon_ttag{AR(1)}$$
but fail to see how to apply this to this specific case. I have looked at other answers considering the translation between AR(1) and OU, but again do not clearly see it for this specific form.
Any help is highly appreciated :-)
edit:
Using the answer below and a linear approximation of $e^{-theta Delta}approx1-theta Delta$ and setting $Delta=1$ I get
$$dln z_t=-(1-rho) ln z_t +sigma sqrt{(1-rho^2)}dW_t$$
which corresponds to the approximation as given by (AR(1))
stochastic-processes vector-auto-regression
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the AR(1) process of the following form:
$$ln z_{t+1}=rho ln z_t+sigma sqrt{(1-rho^2)}epsilon_t$$
And need to find its continous time corresponding Ornstein-Uhlenbeckprocess.
I have the info that an AR(1) can be translated to an Ornstein-Uhlenbeck as follows:
$$dx=theta(bar{x}-x)dt+sigma dW tag{O-U}$$
$$x_{t+1}=thetabar{x}+(1-theta)x_t +sigmaepsilon_ttag{AR(1)}$$
but fail to see how to apply this to this specific case. I have looked at other answers considering the translation between AR(1) and OU, but again do not clearly see it for this specific form.
Any help is highly appreciated :-)
edit:
Using the answer below and a linear approximation of $e^{-theta Delta}approx1-theta Delta$ and setting $Delta=1$ I get
$$dln z_t=-(1-rho) ln z_t +sigma sqrt{(1-rho^2)}dW_t$$
which corresponds to the approximation as given by (AR(1))
stochastic-processes vector-auto-regression
I have the AR(1) process of the following form:
$$ln z_{t+1}=rho ln z_t+sigma sqrt{(1-rho^2)}epsilon_t$$
And need to find its continous time corresponding Ornstein-Uhlenbeckprocess.
I have the info that an AR(1) can be translated to an Ornstein-Uhlenbeck as follows:
$$dx=theta(bar{x}-x)dt+sigma dW tag{O-U}$$
$$x_{t+1}=thetabar{x}+(1-theta)x_t +sigmaepsilon_ttag{AR(1)}$$
but fail to see how to apply this to this specific case. I have looked at other answers considering the translation between AR(1) and OU, but again do not clearly see it for this specific form.
Any help is highly appreciated :-)
edit:
Using the answer below and a linear approximation of $e^{-theta Delta}approx1-theta Delta$ and setting $Delta=1$ I get
$$dln z_t=-(1-rho) ln z_t +sigma sqrt{(1-rho^2)}dW_t$$
which corresponds to the approximation as given by (AR(1))
stochastic-processes vector-auto-regression
stochastic-processes vector-auto-regression
edited Nov 15 at 22:06
asked Nov 14 at 16:26
user469216
567
567
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add a comment |
1 Answer
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We can solve
$$dx=theta(bar{x}-x)ds+sigma dW_s $$
by multiplying by an integrating factor $e^{theta s}$ and integrating over $[t,t+Delta]$ to obtain
$$e^{theta(t+Delta)}x(t + Delta) - e^{theta t}x(t) = bar{x} ( e^{theta(t + Delta)}- e^{theta t}) + sigmaint_t^{t + Delta} e^{theta s}, dW_s$$
Rearranging we get,
$$tag{*}x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + underbrace{sigma e^{-theta Delta}int_t^{t + Delta} e^{theta(s- t)}, dW_s}_{I(t,Delta)}$$
The moments of the stochastic integral on the RHS are
$$mathbb{E}left(I(t, Delta) right) = 0, \ varleft(I(t, Delta) right)= sigma^2 e^{-2theta Delta}int_t^{t + Delta} e^{2theta(s- t)}, ds = frac{sigma^2(1 - e^{-2theta Delta})}{2 theta}$$
Thus, we can write (*) as
$$x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}xi,$$
where $xi sim N(0,1)$ is a standard normal random variable.
Making the association with your AR(1) process we have
$$x(t+Delta) iff ln z_{t+1} quad x(t) iff ln z_t, \ bar{x} = 0, quad rho = e^{-theta Delta}, quad sigmasqrt{1- rho^2} iff sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}$$
So the continuous time process is
$$d ln z_t = frac{ln rho}{Delta}ln z_t ,dt + sigma sqrt{frac{-ln rho}{Delta}} , dW_t$$
where $Delta$ is the time interval between sampled observations.
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
|
show 12 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We can solve
$$dx=theta(bar{x}-x)ds+sigma dW_s $$
by multiplying by an integrating factor $e^{theta s}$ and integrating over $[t,t+Delta]$ to obtain
$$e^{theta(t+Delta)}x(t + Delta) - e^{theta t}x(t) = bar{x} ( e^{theta(t + Delta)}- e^{theta t}) + sigmaint_t^{t + Delta} e^{theta s}, dW_s$$
Rearranging we get,
$$tag{*}x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + underbrace{sigma e^{-theta Delta}int_t^{t + Delta} e^{theta(s- t)}, dW_s}_{I(t,Delta)}$$
The moments of the stochastic integral on the RHS are
$$mathbb{E}left(I(t, Delta) right) = 0, \ varleft(I(t, Delta) right)= sigma^2 e^{-2theta Delta}int_t^{t + Delta} e^{2theta(s- t)}, ds = frac{sigma^2(1 - e^{-2theta Delta})}{2 theta}$$
Thus, we can write (*) as
$$x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}xi,$$
where $xi sim N(0,1)$ is a standard normal random variable.
Making the association with your AR(1) process we have
$$x(t+Delta) iff ln z_{t+1} quad x(t) iff ln z_t, \ bar{x} = 0, quad rho = e^{-theta Delta}, quad sigmasqrt{1- rho^2} iff sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}$$
So the continuous time process is
$$d ln z_t = frac{ln rho}{Delta}ln z_t ,dt + sigma sqrt{frac{-ln rho}{Delta}} , dW_t$$
where $Delta$ is the time interval between sampled observations.
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
|
show 12 more comments
up vote
1
down vote
accepted
We can solve
$$dx=theta(bar{x}-x)ds+sigma dW_s $$
by multiplying by an integrating factor $e^{theta s}$ and integrating over $[t,t+Delta]$ to obtain
$$e^{theta(t+Delta)}x(t + Delta) - e^{theta t}x(t) = bar{x} ( e^{theta(t + Delta)}- e^{theta t}) + sigmaint_t^{t + Delta} e^{theta s}, dW_s$$
Rearranging we get,
$$tag{*}x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + underbrace{sigma e^{-theta Delta}int_t^{t + Delta} e^{theta(s- t)}, dW_s}_{I(t,Delta)}$$
The moments of the stochastic integral on the RHS are
$$mathbb{E}left(I(t, Delta) right) = 0, \ varleft(I(t, Delta) right)= sigma^2 e^{-2theta Delta}int_t^{t + Delta} e^{2theta(s- t)}, ds = frac{sigma^2(1 - e^{-2theta Delta})}{2 theta}$$
Thus, we can write (*) as
$$x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}xi,$$
where $xi sim N(0,1)$ is a standard normal random variable.
Making the association with your AR(1) process we have
$$x(t+Delta) iff ln z_{t+1} quad x(t) iff ln z_t, \ bar{x} = 0, quad rho = e^{-theta Delta}, quad sigmasqrt{1- rho^2} iff sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}$$
So the continuous time process is
$$d ln z_t = frac{ln rho}{Delta}ln z_t ,dt + sigma sqrt{frac{-ln rho}{Delta}} , dW_t$$
where $Delta$ is the time interval between sampled observations.
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
|
show 12 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We can solve
$$dx=theta(bar{x}-x)ds+sigma dW_s $$
by multiplying by an integrating factor $e^{theta s}$ and integrating over $[t,t+Delta]$ to obtain
$$e^{theta(t+Delta)}x(t + Delta) - e^{theta t}x(t) = bar{x} ( e^{theta(t + Delta)}- e^{theta t}) + sigmaint_t^{t + Delta} e^{theta s}, dW_s$$
Rearranging we get,
$$tag{*}x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + underbrace{sigma e^{-theta Delta}int_t^{t + Delta} e^{theta(s- t)}, dW_s}_{I(t,Delta)}$$
The moments of the stochastic integral on the RHS are
$$mathbb{E}left(I(t, Delta) right) = 0, \ varleft(I(t, Delta) right)= sigma^2 e^{-2theta Delta}int_t^{t + Delta} e^{2theta(s- t)}, ds = frac{sigma^2(1 - e^{-2theta Delta})}{2 theta}$$
Thus, we can write (*) as
$$x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}xi,$$
where $xi sim N(0,1)$ is a standard normal random variable.
Making the association with your AR(1) process we have
$$x(t+Delta) iff ln z_{t+1} quad x(t) iff ln z_t, \ bar{x} = 0, quad rho = e^{-theta Delta}, quad sigmasqrt{1- rho^2} iff sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}$$
So the continuous time process is
$$d ln z_t = frac{ln rho}{Delta}ln z_t ,dt + sigma sqrt{frac{-ln rho}{Delta}} , dW_t$$
where $Delta$ is the time interval between sampled observations.
We can solve
$$dx=theta(bar{x}-x)ds+sigma dW_s $$
by multiplying by an integrating factor $e^{theta s}$ and integrating over $[t,t+Delta]$ to obtain
$$e^{theta(t+Delta)}x(t + Delta) - e^{theta t}x(t) = bar{x} ( e^{theta(t + Delta)}- e^{theta t}) + sigmaint_t^{t + Delta} e^{theta s}, dW_s$$
Rearranging we get,
$$tag{*}x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + underbrace{sigma e^{-theta Delta}int_t^{t + Delta} e^{theta(s- t)}, dW_s}_{I(t,Delta)}$$
The moments of the stochastic integral on the RHS are
$$mathbb{E}left(I(t, Delta) right) = 0, \ varleft(I(t, Delta) right)= sigma^2 e^{-2theta Delta}int_t^{t + Delta} e^{2theta(s- t)}, ds = frac{sigma^2(1 - e^{-2theta Delta})}{2 theta}$$
Thus, we can write (*) as
$$x(t + Delta) =bar{x}(1 - e^{-theta Delta}) + e^{-theta Delta}x(t) + sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}xi,$$
where $xi sim N(0,1)$ is a standard normal random variable.
Making the association with your AR(1) process we have
$$x(t+Delta) iff ln z_{t+1} quad x(t) iff ln z_t, \ bar{x} = 0, quad rho = e^{-theta Delta}, quad sigmasqrt{1- rho^2} iff sigmasqrt{frac{1 - e^{-2theta Delta}}{2theta}}$$
So the continuous time process is
$$d ln z_t = frac{ln rho}{Delta}ln z_t ,dt + sigma sqrt{frac{-ln rho}{Delta}} , dW_t$$
where $Delta$ is the time interval between sampled observations.
edited Nov 14 at 22:13
answered Nov 14 at 21:54
RRL
46.7k42366
46.7k42366
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
|
show 12 more comments
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
– user469216
Nov 14 at 22:08
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
– RRL
Nov 14 at 22:11
1
1
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
– RRL
Nov 14 at 22:12
1
1
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
So I left out the $dt$ in the answer by mistake -- I now added it.Clearly you can see how $Delta$ can be absorbed into the time variable. Also the Wiener process is $mathcal{O}(dt)$ which is why the $sqrt{Delta}$ shows up as a coefficient. Short answer -- just set $Delta = 1$.
– RRL
Nov 14 at 22:16
1
1
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
The OU and AR(1) you wrote with the same parameters $theta$ and $sigma$ appearing in both is only correct if the spacing between discrete steps is the same as the time unit for $t$, e.g. 1 year.
– RRL
Nov 14 at 23:25
|
show 12 more comments
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