# Finding the coefficients of an MA(1) process given the expectation and variance.

The following is preparation of an exam I have coming up, any help would be appreciated.

An MA(1) process is selected to model a stationary time series $\{ X_t \}$.

We are given the lag one correlation of $\{ X_t \}$ is $-0.5$, the mean of $\{ X_t \}$ is $10$ and the variance of $\{ X_t \}$ is $4$.

So, my working:

We may model $\{ X_t \}$ as $X_t = \mu + Z_t + \beta Z_{t-1}$. Where $Z$ is a Normal$(0, \sigma^2_Z)$ random variable and all $Z$s are independent from one another.

So, $\mathbb{E}(X_t) = 10$ which implies $\mu = 10$.

$\text{Var}(X_t) = \mathbb{E}(X_t^2) - 100 = 4$, so,

$$\mathbb{E}((\mu + Z_t + \beta Z_{t-1})^2) - 100 = 4.$$

From here, I get $$\sigma^2_Z + \beta^2 \sigma^2_Z = 4.$$

And finally, $\gamma_X(1) = \mathbb{E}((X_t - 10)(X_{t + 1} - 10)) = -0.5.$

This yields $$\beta \sigma^2_Z = -0.5.$$

From here, using the fact that $\sigma^2_Z \geq 0$ I can get a solution, but I'm not entirely convinced I've done this correctly and would like some confirmation. Sadly I have no solutions available.

Much thanks.

Edit: Spotted a mistake. The lag one correlation is -0.5, not covariance.

This leads to $$\beta \sigma^2_Z = -2,$$ which gives a much nicer answer.

Looks ok to me. I would have done the following (essentially equivalent, but a little neater IMO).

First substract the mean, define: $Y_t = X_t - \mu_X=X_t - 10$ . Then, the mean of $Y_t$ is zero (a little easier to work) and the second moments (variance, correlations, etc) are not changed.

Now, as $Y_t$ is MA(1), $Y_t = a_0 N_t + a_1 N_{t-1}$ where $N_t$ is white noise with zero mean and unit variance.

Then $$E(Y_t^2) = a_0^2 +a_1^2 \tag{1}$$ $$E(Y_t Y_{t-1}) = a_0 a_1 \tag{2}$$

(I've used the facts that $E(N_t)=0$, $E(N_t^2)=1$, $E(N_t N_{t-k})=0$ for $k\ne0$)

But because $E(Y_t^2) = Var(Y_t)=Var(X_t)$, and $E(Y_t Y_{t-1})/Var(Y_t)=\rho_1$ :

$$a_0^2 +a_1^2 =4\tag{3}$$ $$a_0 a_1 = -2\tag{4}$$

Solving this, we get $a_0 = \sqrt{2}$, $a_1 = -\sqrt{2}$

Notice also that the system of equations is not linear, hence the existence (and unicity) of solutions is not guaranteed. (this is typical of MA models - in AR models the equations are linear). And indeed, here we really have another solution, but it's trivial (change signs).