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Suppose I have the GARCH(1,1) specification: \begin{align} y_t &= \mu + \epsilon_t\\ \sigma_t^2 &= \alpha \epsilon_{t-1}^2 + \beta \sigma_{t-1}^2. \end{align}

In EViews: GARCH(1,1) -> y c

If I then forecast $y_{t+1}$, are the forecasts in any way adjusted for the variance that is modeled by the equation for the variance? Or else, will this simply be the mean of the previous $y_t$'s?

Edit: Would someone be able to complement Richard Hardy's answer by showing an explicit mathematical relation for how $\mu$ is influenced by the variance?

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  • $\begingroup$ I presume that your available data is only a time-series sample on $y$'s? $\endgroup$ – Alecos Papadopoulos Jun 17 '15 at 20:11
  • $\begingroup$ And what is the estimation method? Least-squares? Maximum likelihood? (if the latter, what are the distributional assumptions)? $\endgroup$ – Alecos Papadopoulos Jun 17 '15 at 20:13
  • $\begingroup$ @AlecosPapadopoulos Correct, I have time-series data of $y$. The estimation method is maximum likelihood where $\epsilon_t \sim \mathcal{N} (0,1)$. $\endgroup$ – rbm Jun 17 '15 at 20:14
  • $\begingroup$ A final clarification: $\sigma_t^2$ is meant to represent $E(\epsilon_t^2\mid \epsilon_{t-1}, \epsilon_{t-2}...)$, right? $\endgroup$ – Alecos Papadopoulos Jun 17 '15 at 20:20
  • $\begingroup$ @AlecosPapadopoulos Correct, or put differently $\sigma_t^2 = \mathbb{E}(\epsilon_t^2 \mid \mathcal{I}_{t-1})$, where $\mathcal{I}_{t-1}$ denotes the information set at time $t-1$. $\endgroup$ – rbm Jun 17 '15 at 20:23
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Note that this is a restricted GARCH(1,1) model since the intercept is excluded from the conditional variance equation.

Suppose you start by estimating the model using historical data. You obtain the coefficients $\hat \mu$, $\hat \alpha_1$ and $\hat \beta_1$. Note that $\hat \mu$ will generally be different from the sample mean of $y_t$: $\hat \mu \neq \frac{1}{T} \sum_{t=1}^{T} y_t$, and this will be due to allowing for a time-varying conditional variance.

When you forecast $y_{t+1}$, the point forecast will be $\hat \mu$. The prediction interval (which estimates the uncertainty of the point forecast) will depend on the fitted standard deviation $\hat \sigma_{t+1}^2$ and will vary with $t$ due to the conditional variance being time varying.

In sum, allowing for a time-varying conditional variance affects both the point forecast and the prediction interval.

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  • $\begingroup$ Thanks for your answer. Could you please show me mathematically how $\hat{\mu}$ (i.e. the point forecast) depends on the estimated variance? (I didn't purposely choose the restricted specification btw - the other case would also be fine). $\endgroup$ – rbm May 10 '15 at 10:21
  • $\begingroup$ Sorry, I cannot post an explicit answer right now (and I am not sure how difficult it would be to come up with one). But let me give you some intuition. Suppose you know the true underlying variances of each $\epsilon_t$. Then you would estimate the regression $y_t=\mu+\epsilon_t$ with weighted least squares (WLS) in place of OLS. Now suppose you have estimates of the true underlying variances. Then you can still use those to run a WLS regression hoping that your estimated variances are close enough to the true ones so that WLS would give you a more efficient estimate of $\mu$ than OLS. $\endgroup$ – Richard Hardy May 10 '15 at 12:13
  • $\begingroup$ So when you estimate the two equations (one for the cond. mean and the other for the cond. variance), the cond. mean equation gets treated somewhat like with WLS rather than OLS; the estimated $\sigma_t$ are used for the weighting. Since the estimation is simultaneous, this is not quite as straightforward, but the main idea remains the same. $\endgroup$ – Richard Hardy May 10 '15 at 12:15
  • $\begingroup$ Thank you. It certainly makes sense that you can use the information from the conditional variance equation in a WLS way. Exactly for that reason, my question was actually more whether and how this then would happen in such a specification. Your answer is certainly helpful, so thank you for that (I upvoted it) but I am still hoping that someone could show me a bit more explicitly how this works in this case. $\endgroup$ – rbm May 10 '15 at 12:18
  • $\begingroup$ Very good that you edited the question to make it more specific. That should increase the chance of getting the answer you need. I might try to answer it later myself. $\endgroup$ – Richard Hardy May 10 '15 at 12:28

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