16
Jun 21

## Solution to Question 1 from UoL exam 2020

The assessment was an open-book take-home online assessment with a 24-hour window. No attempt was made to prevent cheating, except a warning, which was pretty realistic. Before an exam it's a good idea to see my checklist.

Question 1. Consider the following ARMA(1,1) process:

(1) $z_{t}=\gamma +\alpha z_{t-1}+\varepsilon _{t}+\theta \varepsilon _{t-1}$

where $\varepsilon _{t}$ is a zero-mean white noise process with variance $\sigma ^{2}$, and assume $|\alpha |,|\theta |<1$ and $\alpha+\theta \neq 0$, which together make sure $z_{t}$ is covariance stationary.

(a) [20 marks] Calculate the conditional and unconditional means of $z_{t}$, that is, $E_{t-1}[z_{t}]$ and $E[z_{t}].$

(b) [20 marks] Set $\alpha =0$. Derive the autocovariance and autocorrelation function of this process for all lags as functions of the parameters $\theta$ and $\sigma$.

(c) [30 marks] Assume now $\alpha \neq 0$. Calculate the conditional and unconditional variances of $z_{t},$ that is, $Var_{t-1}[z_{t}]$ and $Var[z_{t}].$

Hint: for the unconditional variance, you might want to start by deriving the unconditional covariance between the variable and the innovation term, i.e., $Cov[z_{t},\varepsilon _{t}].$

(d) [30 marks] Derive the autocovariance and autocorrelation for lags of 1 and 2 as functions of the parameters of the model.

Hint: use the hint of part (c).

## Solution

### Part (a)

Reminder: The definition of a zero-mean white noise process is

(2) $E\varepsilon _{t}=0,$ $Var(\varepsilon _{t})=E\varepsilon_{t}^{2}=\sigma ^{2}$ for all $t$ and $Cov(\varepsilon _{j},\varepsilon_{i})=E\varepsilon _{j}\varepsilon _{i}=0$ for all $i\neq j.$

A variable indexed $t-1$ is known at moment $t-1$ and at all later moments and behaves like a constant for conditioning at such moments.

Moment $t$ is future relative to $t-1.$  The future is unpredictable and the best guess about the future error is zero.

The recurrent relationship in (1) shows that

(3) $z_{t-1}=\gamma +\alpha z_{t-2}+...$ does not depend on the information that arrives at time $t$ and later.

Hence, using also linearity of conditional means,

(4) $E_{t-1}z_{t}=E_{t-1}\gamma +\alpha E_{t-1}z_{t-1}+E_{t-1}\varepsilon _{t}+\theta E_{t-1}\varepsilon _{t-1}=\gamma +\alpha z_{t-1}+\theta\varepsilon _{t-1}.$

The law of iterated expectations (LIE): application of $E_{t-1},$ based on information available at time $t-1,$ and subsequent application of $E,$ based on no information, gives the same result as application of $E.$

$Ez_{t}=E[E_{t-1}z_{t}]=E\gamma +\alpha Ez_{t-1}+\theta E\varepsilon _{t-1}=\gamma +\alpha Ez_{t-1}.$

Since $z_{t}$ is covariance stationary, its means across times are the same, so $Ez_{t}=\gamma +\alpha Ez_{t}$ and $Ez_{t}=\frac{\gamma }{1-\alpha }.$

### Part (b)

With $\alpha =0$ we get $z_{t}=\gamma +\varepsilon _{t}+\theta\varepsilon _{t-1}$ and from part (a) $Ez_{t}=\gamma .$ Using (2), we find variance

$Var(z_{t})=E(z_{t}-Ez_{t})^{2}=E(\varepsilon _{t}^{2}+2\theta \varepsilon_{t}\varepsilon _{t-1}+\theta ^{2}\varepsilon _{t-2}^{2})=(1+\theta^{2})\sigma ^{2}$

and first autocovariance

(5) $\gamma_{1}=Cov(z_{t},z_{t-1})=E(z_{t}-Ez_{t})(z_{t-1}-Ez_{t-1})=E(\varepsilon_{t}+\theta \varepsilon _{t-1})(\varepsilon _{t-1}+\theta \varepsilon_{t-2})=\theta E\varepsilon _{t-1}^{2}=\theta \sigma ^{2}.$

Second and higher autocovariances are zero because the subscripts of epsilons don't overlap.

Autocorrelation function: $\rho _{0}=\frac{Cov(z_{t},z_{t})}{\sqrt{Var(z_{t})Var(z_{t})}}=1$ (this is always true),

$\rho _{1}=\frac{Cov(z_{t},z_{t-1})}{\sqrt{Var(z_{t})Var(z_{t-1})}}=\frac{\theta \sigma ^{2}}{(1+\theta ^{2})\sigma ^{2}}=\frac{\theta }{1+\theta ^{2}},$ $\rho _{j}=0$ for $j>1.$

This is characteristic of MA processes: their autocorrelations are zero starting from some point.

### Part (c)

If we replace all expectations in the definition of variance, we obtain the definition of conditional variance. From (1) and (4)

$Var_{t-1}(z_{t})=E_{t-1}(z_{t}-E_{t-1}z_{t})^{2}=E_{t-1}\varepsilon_{t}^{2}=\sigma ^{2}.$

By the law of total variance

(6) $Var(z_{t})=EVar_{t-1}(z_{t})+Var(E_{t-1}z_{t})=\sigma ^{2}+Var(\gamma+\alpha z_{t-1}+\theta \varepsilon _{t-1})=$

(an additive constant does not affect variance)

$=\sigma ^{2}+Var(\alpha z_{t-1}+\theta \varepsilon _{t-1})=\sigma^{2}+\alpha ^{2}Var(z_{t})+2\alpha \theta Cov(z_{t-1},\varepsilon_{t-1})+\theta ^{2}Var(\varepsilon _{t-1}).$

By the LIE and (3)

$Cov(z_{t-1},\varepsilon _{t-1})=Cov(\gamma +\alpha z_{t-2}+\varepsilon _{t-1}+\theta \varepsilon _{t-2},\varepsilon _{t-1})=\alpha Cov(z_{t-2},\varepsilon _{t-1})+E\varepsilon _{t-1}^{2}+\theta EE_{t-2}\varepsilon _{t-2}\varepsilon _{t-1}=\sigma ^{2}+\theta E(\varepsilon _{t-2}E_{t-2}\varepsilon _{t-1}).$

Here $E_{t-2}\varepsilon _{t-1}=0,$ so

(7) $Cov(z_{t-1},\varepsilon _{t-1})=\sigma ^{2}.$

$Var(z_{t})=Var(\gamma +\alpha z_{t-1}+\varepsilon _{t}+\theta \varepsilon _{t-1})=\alpha ^{2}Var(z_{t-1})+Var(\varepsilon _{t})+\theta ^{2}Var(\varepsilon _{t-1})+$

$+2\alpha Cov(z_{t-1},\varepsilon _{t})+2\alpha \theta Cov(z_{t-1},\varepsilon _{t-1})+2\theta Cov(\varepsilon _{t},\varepsilon _{t-1})=\alpha ^{2}Var(z_{t})+\sigma ^{2}+\theta ^{2}\sigma ^{2}+2\alpha \theta \sigma ^{2}$

and, finally,

(8) $Var(z_{t})=\frac{(1+2\alpha \theta +\theta ^{2})\sigma ^{2}}{1-\alpha ^{2}}.$

### Part (d)

From (7)

(9) $Cov(z_{t-1},\varepsilon _{t-2})=Cov(\gamma +\alpha z_{t-2}+\varepsilon _{t-1}+\theta \varepsilon _{t-2},\varepsilon _{t-2})=\alpha Cov(z_{t-2},\varepsilon _{t-2})+\theta Var(\varepsilon _{t-2})=(\alpha +\theta )\sigma ^{2}.$

It follows that

$Cov(z_{t},z_{t-1})=Cov(\gamma +\alpha z_{t-1}+\varepsilon _{t}+\theta \varepsilon _{t-1},\gamma +\alpha z_{t-2}+\varepsilon _{t-1}+\theta \varepsilon _{t-2})=$

(a constant is not correlated with anything)

$=\alpha ^{2}Cov(z_{t-1},z_{t-2})+\alpha Cov(z_{t-1},\varepsilon _{t-1})+\alpha \theta Cov(z_{t-1},\varepsilon _{t-2})+$

$+\alpha Cov(\varepsilon _{t},z_{t-2})+Cov(\varepsilon _{t},\varepsilon _{t-1})+\theta Cov(\varepsilon _{t},\varepsilon _{t-2})+$

$+\theta \alpha Cov(\varepsilon _{t-1},z_{t-2})+\theta Var(\varepsilon _{t-1})+\theta ^{2}Cov(\varepsilon _{t-1},\varepsilon _{t-2}).$

From (7) $Cov(z_{t-2},\varepsilon _{t-2})=\sigma ^{2}$ and from (9) $Cov(z_{t-1},\varepsilon _{t-2})=(\alpha +\theta )\sigma ^{2}.$

From (3) $Cov(\varepsilon _{t},z_{t-2})=Cov(\varepsilon _{t-1},z_{t-2})=0.$

Using also the white noise properties and stationarity of $z_{t}$

$Cov(z_{t},z_{t-1})=Cov(z_{t-1},z_{t-2})=\gamma _{1},$

we are left with

$\gamma _{1}=\alpha ^{2}\gamma _{1}+\alpha \sigma ^{2}+\alpha \theta (\alpha +\theta )\sigma ^{2}+\theta \sigma ^{2}=\alpha ^{2}\gamma _{1}+(1+\alpha \theta )(\alpha +\theta )\sigma ^{2}.$

Hence,

$\gamma _{1}=\frac{(1+\alpha \theta )(\alpha +\theta )\sigma ^{2}}{1-\alpha ^{2}}$

and using (8)

$\rho _{0}=1,$ $\rho _{1}=\frac{(1+\alpha \theta )(\alpha +\theta )}{ 1+2\alpha \theta +\theta ^{2}}.$

The finish is close.

$Cov(z_{t},z_{t-2})=Cov(\gamma +\alpha z_{t-1}+\varepsilon _{t}+\theta \varepsilon _{t-1},\gamma +\alpha z_{t-3}+\varepsilon _{t-2}+\theta \varepsilon _{t-3})=$

$=\alpha ^{2}Cov(z_{t-1},z_{t-3})+\alpha Cov(z_{t-1},\varepsilon _{t-2})+\alpha \theta Cov(z_{t-1},\varepsilon _{t-3})+$

$+\alpha Cov(\varepsilon _{t},z_{t-3})+Cov(\varepsilon _{t},\varepsilon _{t-2})+\theta Cov(\varepsilon _{t},\varepsilon _{t-3})+$

$+\theta \alpha Cov(\varepsilon _{t-1},z_{t-3})+\theta Cov(\varepsilon _{t-1},\varepsilon _{t-2})+\theta ^{2}Cov(\varepsilon _{t-1},\varepsilon _{t-3}).$

This simplifies to

(10) $Cov(z_{t},z_{t-2})=\alpha ^{2}Cov(z_{t-1},z_{t-3})+\alpha (\alpha +\theta )\sigma ^{2}+\alpha \theta Cov(z_{t-1},\varepsilon _{t-3}).$

By (7)

$Cov(z_{t-1},\varepsilon _{t-3})=Cov(\gamma +\alpha z_{t-2}+\varepsilon _{t-1}+\theta \varepsilon _{t-2},\varepsilon _{t-3})=\alpha Cov(z_{t-2},\varepsilon _{t-3})=$

$=\alpha Cov(\gamma +\alpha z_{t-3}+\varepsilon _{t-2}+\theta \varepsilon _{t-3},\varepsilon _{t-3})=\alpha \sigma ^{2}+\alpha \theta \sigma ^{2}=\alpha (1+\theta )\sigma ^{2}.$

Finally, using (10)

$\gamma _{2}=\alpha ^{2}\gamma _{2}+\alpha (\alpha +\theta )\sigma ^{2}+\alpha^2 \theta (1 +\theta )\sigma ^{2}=\alpha ^{2}\gamma _{2}+\alpha\sigma^2 (\alpha +\theta +\alpha\theta +\alpha\theta^2)\sigma ^{2},$

$\gamma _{2}=\frac{\alpha\sigma^2 (\alpha +\theta +\alpha\theta +\alpha\theta^2)\sigma ^{2}}{1-\alpha ^{2}},$

$\rho _{2}=\frac{\alpha\sigma^2 (\alpha +\theta +\alpha\theta +\alpha\theta^2)}{1+2\alpha \theta +\theta ^{2}}.$

A couple of errors have been corrected on June 22, 2021. Hope this is final.