###### Binomial Option Valuation Model

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Heteroskedasticity is the **dependence** of the **variance of the error term** on the **independent variable.** We have been assuming that time series follows the homoskedasticity assumption. Homoskedasticity is the **independence **of the variance of the error term on the independent variable. Violation of this condition leads to heteroskedasticity.

**Heteroskedasticity** in the autoregressive model makes the **standard errors** of the regression coefficients of the model i**nvalid**, leading to misleading interpretations.

Autoregressive conditional heteroskedasticity is a problem associated with the correlation of variances of the error terms. An ARCH(1) model is an AR(1) model with conditional heteroskedasticity.

The error terms in an ARCH(1) model are normally distributed with a mean of 0 and a variance of \(\text{a}_{0}+\text{a}_{1}\epsilon_{\text{t}-1}^{2}\).

$$\epsilon_{\text{t}}∼\text{N}(0, \text{a}_{0}+\text{a}_{1}\epsilon_{\text{t}-1}^{2})$$

**Steps for Testing for ARCH(1) Conditional Heteroskedasticity**

- Regress the squared residuals from each period on the prior period squared residuals.
- Estimate \(\hat{\epsilon}_{\text{t}}^{2}=\alpha_{0}+\alpha_{1}\hat{\epsilon}_{\text{t}-1}^{2}+\mu_{\text{t}}\).
- If the estimated slope coefficient \(\hat{\alpha}_1\) is statistically different from 0, the series shows an ARCH(1) effect, and thus the need to correct for heteroskedasticity.
- We can then predict the variance of the errors for time \(t+1\).

Given an ARCH(1) model, we can use the parameter estimates from our test for conditional heteroskedasticity to predict next period variance as:

$$\hat{\sigma}_{\text{t}+1}^{2}=\hat{\alpha}_{0}+\hat{\alpha}_{1}\hat{\epsilon}_{\text{t}}^{2}$$

The following results have been obtained from regressing the squared residuals of an estimated autoregressive model on a constant and one lag of the squared error terms.

$$\small{\begin{array}{c|c} \text{R-squared} & 0.0430 \\ \hline\text{Standard error} & 24.638 \\ \hline\text{Observations} & 358 \\ \text{Durbin-Watson} & 2.1021\\ \hline\end{array}}$$

$$\small{\begin{array}{c|c|c|c} {}&\textbf{Coefficient}&\textbf{Standard Error}&\textbf{t-Statistic}\\ \hline\text{Intercept}&9.1680&1.4104&6.55002\\ \hline \hat{\epsilon}_{\text{t}-1}^{2}&0.2050&0.0506&4.0514\\ \end{array}}$$

- Test whether the original time series has ARCH(1) errors at the 5% significance level.
- If the squared error in the current period is 0.5%, predicted variance in the next period is
*closest*to:

**Solution**

The test involves the * first lag of the residuals* of the approximated time series model, and thus the number of the observations is less by one. The number of degrees of freedom is \(358-1=357\), and the corresponding critical t-value at the 5% level of significance is 1.97.

The **t-statistic** on \(\hat{\epsilon}_{\text{t}-1}^{2}\) is **greater than **the** critical value.** This implies that \(\hat{\alpha}_{1}\) is statistically significantly different from 0. Therefore, the time series has ARCH(1) errors.

The predicted variance for the next period is determined using the formula:

$$\hat{\sigma}_{\text{t}}^{2}=\hat{\alpha}_{0}+\hat{\alpha}_{1}\hat{\epsilon}_{\text{t}-1}^{2}$$

$$\hat{\sigma}_{\text{t}}^{2}=9.1680+0.2050\hat{\epsilon}_{\text{t}-1}^{2}$$

So that the variance for the next period, in this case, is:

$$\hat{\sigma}_{\text{t}}^{2}=9.1680+0.2050\times0.005=9.169025$$

## Question

Consider the following regression results for an ARCH(1) model:

$$\small{\begin{array}{c|c|c} {}&\textbf{Coefficients} & \textbf{P-value} \\ \hline\text{Constant} & 4.8347 & <0.001 \\ \hline\text{Lag 1} & 0.2678 & <0.001\\ \end{array}}$$

Given that the current period squared error is 0.678, the predicted variance of the error terms in the next period is

closestto:

- 4.835.
- 5.103.
- 5.016.
## Solution

The correct answer is C.The predicted variance for the next period is determined using the formula:

$$\hat{\sigma}_{\text{t}}^{2}=\hat{\alpha}_{0}+\hat{\alpha}_{1}\hat{\epsilon}_{\text{t}-1}^{2}$$

$$\hat{\sigma}_{\text{t}}^{2}=4.8347+0.2678\times0.678=5.016$$

Reading 5: Time Series Analysis

*LOS 5 (m) Explain autoregressive conditional heteroskedasticity (ARCH) and describe how ARCH models can be applied to predict the variance of a time series.*