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))










share|cite|improve this question




























    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))










    share|cite|improve this question


























      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))










      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago

























      asked yesterday









      user469216

      567




      567






















          1 Answer
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          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.






          share|cite|improve this answer























          • just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
            – user469216
            yesterday










          • It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
            – RRL
            yesterday






          • 1




            Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
            – RRL
            yesterday






          • 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
            yesterday








          • 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
            yesterday











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          active

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          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.






          share|cite|improve this answer























          • just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
            – user469216
            yesterday










          • It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
            – RRL
            yesterday






          • 1




            Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
            – RRL
            yesterday






          • 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
            yesterday








          • 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
            yesterday















          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.






          share|cite|improve this answer























          • just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
            – user469216
            yesterday










          • It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
            – RRL
            yesterday






          • 1




            Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
            – RRL
            yesterday






          • 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
            yesterday








          • 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
            yesterday













          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          RRL

          46.4k42365




          46.4k42365












          • just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
            – user469216
            yesterday










          • It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
            – RRL
            yesterday






          • 1




            Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
            – RRL
            yesterday






          • 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
            yesterday








          • 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
            yesterday


















          • just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
            – user469216
            yesterday










          • It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
            – RRL
            yesterday






          • 1




            Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
            – RRL
            yesterday






          • 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
            yesterday








          • 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
            yesterday
















          just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
          – user469216
          yesterday




          just for notational consistency, how would you get rid of $Delta$ so that I only have "dt" terms?
          – user469216
          yesterday












          It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
          – RRL
          yesterday




          It's just a matter of using a dimensionless time variable $hat{t} = t/Delta$
          – RRL
          yesterday




          1




          1




          Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
          – RRL
          yesterday




          Or I could have integrated the SDE over the interval $[t,t+1]$ and it goes away.
          – RRL
          yesterday




          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
          yesterday






          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
          yesterday






          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
          yesterday




          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
          yesterday


















           

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