4 added 29 characters in body edited May 19 '17 at 9:05 Richard Hardy 29.1k66 gold badges4848 silver badges139139 bronze badges It seems that you did not get the equations (nor the idea) quite right. The formula for the AR($$p$$) model should be $$y_t=\beta_0+\beta_1 y_{t-1}+\dotsb+\beta_p y_{t-p}+\varepsilon_t;$$ the formula for the ARCH($$s$$) model should be $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}.$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) We can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a linear regression (although there are $$p$$ missing observations for lagged series), where conditional heteroskedasticity leads to inefficiency but does not prevent consistency. Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Then the errors $$e_t$$ can be used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH$$(s)$$ model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 e^2_{t-1}+\dotsb+\alpha_s e^2_{t-s}$$ Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit for the conditional variance. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS cannot work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate both models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (nor the idea) quite right. The formula for the AR($$p$$) model should be $$y_t=\beta_0+\beta_1 y_{t-1}+\dotsb+\beta_p y_{t-p}+\varepsilon_t;$$ the formula for the ARCH($$s$$) model should be $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}.$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) We can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a linear regression (although there are $$p$$ missing observations for lagged series), where conditional heteroskedasticity leads to inefficiency but does not prevent consistency. Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Then the errors $$e_t$$ can be used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH$$(s)$$ model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 e^2_{t-1}+\dotsb+\alpha_s e^2_{t-s}$$ Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS cannot work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate both models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (nor the idea) quite right. The formula for the AR($$p$$) model should be $$y_t=\beta_0+\beta_1 y_{t-1}+\dotsb+\beta_p y_{t-p}+\varepsilon_t;$$ the formula for the ARCH($$s$$) model should be $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}.$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) We can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a linear regression (although there are $$p$$ missing observations for lagged series), where conditional heteroskedasticity leads to inefficiency but does not prevent consistency. Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Then the errors $$e_t$$ can be used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH$$(s)$$ model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 e^2_{t-1}+\dotsb+\alpha_s e^2_{t-s}$$ Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit for the conditional variance. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS cannot work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate both models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. 3 added 27 characters in body edited Jun 9 '15 at 14:12 Richard Hardy 29.1k66 gold badges4848 silver badges139139 bronze badges It seems that you did not get the equations (andnor the idea) quite right. First, weThe formula for the AR($$p$$) model should be $$y_t=\beta_0+\beta_1 y_{t-1}+\dotsb+\beta_p y_{t-p}+\varepsilon_t;$$ the formula for the ARCH($$s$$) model should be $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}.$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) We can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a simplelinear regression (although there are $$p$$ missing observations for lagged series), where conditional heteroskedasticity leads to inefficiency but does not prevent consistency. Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Second, Then the model errors $$e_t$$ arecan be used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH($$s$$)$$(s)$$ model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.)$$\sigma^2_t=\omega+\alpha_1 e^2_{t-1}+\dotsb+\alpha_s e^2_{t-s}$$ Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS can notcannot work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate theboth models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (and the idea) quite right. First, we can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a simple regression (although there are $$p$$ missing observations for lagged series). Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Second, the model errors $$e_t$$ are used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH($$s$$) model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS can not work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate the models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (nor the idea) quite right. The formula for the AR($$p$$) model should be $$y_t=\beta_0+\beta_1 y_{t-1}+\dotsb+\beta_p y_{t-p}+\varepsilon_t;$$ the formula for the ARCH($$s$$) model should be $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}.$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) We can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a linear regression (although there are $$p$$ missing observations for lagged series), where conditional heteroskedasticity leads to inefficiency but does not prevent consistency. Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Then the errors $$e_t$$ can be used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH$$(s)$$ model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 e^2_{t-1}+\dotsb+\alpha_s e^2_{t-s}$$ Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS cannot work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate both models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. 2 added 27 characters in body edited Jun 9 '15 at 14:05 Richard Hardy 29.1k66 gold badges4848 silver badges139139 bronze badges It seems that you did not get the equations (and the idea) quite right. First, we can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a simple regression (although there are $$p$$ missing observations for lagged series). Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Second, the model errors $$e_t$$ are used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH($$s$$) model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS can not work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. SomeGiven that, some of your further questions then become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) modelsmodel and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate the models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (and the idea) quite right. First, we can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error because AR($$p$$) can be expressed as a simple regression (although there are $$p$$ missing observations for lagged series). Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Second, the model errors $$e_t$$ are used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH($$s$$) model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic. This is in contrast to stochastic volatility models. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Some of your further questions then become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) models and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate the models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. It seems that you did not get the equations (and the idea) quite right. First, we can estimate an AR($$p$$) model consistently even if we neglect the ARCH structure in the model error -- because AR($$p$$) can be expressed as a simple regression (although there are $$p$$ missing observations for lagged series). Errors $$e_t$$ obtained from such a model will be consistent estimates of the true underlying disturbances $$\varepsilon_t$$ as long as the AR($$p$$) model is the correct model for the conditional mean of the process. Second, the model errors $$e_t$$ are used in place of the true underlying disturbances $$\varepsilon_t$$ to estimate the ARCH($$s$$) model (usually via maximum likelihood): $$\sigma^2_t=\omega+\alpha_1 \varepsilon^2_{t-1}+\dotsb+\alpha_s \varepsilon^2_{t-s}$$ (I used $$\omega$$ where you used $$\alpha_0$$ but this makes no difference.) Note that there is no error term on the right hand side (since $$\varepsilon$$ is treated as a variable, not an error in the ARCH model). The ARCH model assumes a perfect fit. That is, the conditional variance is assumed to be deterministic -- not stochastic. This is in contrast to stochastic volatility models. Now, as with any VMA model, estimating this model using OLS/ML is impossible, because the error term is not observable. This is not really true. Although OLS can not work directly with latent variables, ML can. VMA can be estimated via ML. Where does the equivalence of $$\sigma^2$$ and $$e^2_t$$ come from? There is no equivalence between $$\sigma^2_t$$ and $$\varepsilon^2_t$$. Given that, some of your further questions become irrelevant. Do they use same amount of lags for both equations? The lag order of the AR($$p$$) model and the ARCH($$s$$) model need not be the same. Selecting the best lag orders may be tough in practice because the models are interdependent. (The AR($$p$$) model may be estimated first and ARCH($$s$$) model estimated subsequently, but the efficient way is to estimate the models simultaneously.) One option is to use a grid of $$p$$ and $$s$$ values, estimate all models over the grid and pick the model with the lowest AIC or BIC value. However, this approach is subject to overfitting, especially if the grid is fine. 1 answered Jun 9 '15 at 13:59 Richard Hardy 29.1k66 gold badges4848 silver badges139139 bronze badges