# Forecasting MA(1) process

#### lord12

##### New Member
Suppose $$x_{t} = w_{t} + \theta w_{t-1}$$ where $$w_t$$ is white noise with variance $$\sigma_{w}^2$$

Derive the minimum mean square error one-step forecast based on the infinite past and determine the mean square error of this forecast.

Let $$\tilde{x}_n^{n+1}$$ be the truncated one step ahead forecast based on the n previous observations. Show that $$E[(x_{n+1}-\tilde{x}_n^{n+1})^2=\sigma_w^2(1+\theta^{(2+2n)})$$
The solution for the first part states that:
$$x_{n+1} = \sum_{j=1}^{\infty}-\theta^{j}x_{n+1-j} + w_{n+1}$$ and $$\tilde{x}_{n+1}=\sum_{j=1}^{\infty}(-(\theta)^{j}x_{n+1-j})$$ so the $$MSE = E[x_{n+1}-\tilde{x}_{n+1}] = \sigma_{w}^2$$. I am not sure how the expression for $$x_{n+1}$$ or $$\tilde{x}_{n}^{n+1}$$ was arrived at. Can someone explain how this was derived?

For the second part, the solution states $$\tilde{x}_{n}^{n+1}= \sum_{j=1}^{n}-\theta^{j}x_{n+1-j}$$ and $$MSE = E(x_{n+1} - \tilde{x}_{n}^{n+1})^2 =E[\sum_{j=n+1}^{\infty}-\theta^{j}x_{n+1-j} + w_{n+1}]^2$$. I am not sure how this was arrived at.

#### BGM

##### TS Contributor
Introduce the back-shift operator $$B$$ such that $$Bw_t = w_{t-1}$$

Now the MA(1) model can be rewritten as $$x_t = (1 + \theta B)w_t$$

For the time being lets believe that we may "invert" the operator like this

$$w_t = (1 + \theta B)^{-1} x_t$$

And also believe that we can apply the geometric series result:

$$w_t = \sum_{j=0}^{+\infty} (-\theta)^j B^j x_t = \sum_{j=0}^{+\infty} (-\theta)^j x_{t-j}$$

Now combining these facts on $$x_{n+1}$$:

$$x_{n+1} = (1 + \theta B)w_{n+1} = w_{n+1} + \theta B\sum_{j=0}^{+\infty} (-\theta)^j x_{n+1-j}$$

$$= \sum_{j=0}^{+\infty} -(-\theta)^{j+1} x_{n-j} + w_{n+1}$$

$$= \sum_{j=1}^{+\infty} -(-\theta)^j x_{n+1-j} + w_{n+1}$$

The one-step ahead forecast should be the conditional expectation:

$$\tilde{x}_{n+1} = E[x_{n+1}|\mathcal{F}_n] = \sum_{j=1}^{+\infty} -(-\theta)^j x_{n+1-j} + E[w_{n+1}|\mathcal{F}_n]$$

as the former sum is $$\mathcal{F}_n$$-measurable. Note that the latter expectation vanish as it is white noise and independent of the filtration.

For the truncated forecast it is really by definition - truncate the series up to the most recent $$n$$ terms only instead of the infinite past.

For the last part you may try to write $$x_t$$ in terms of those white noises $$w_t$$ again and calculate the MSE. By the zero expectation and independence of white noise, the variance/MSE is not hard to calculate.

#### lord12

##### New Member
I am still not sure how you got $$w_{t} = \sum_{j=0}^{\infty}(-\theta^{j})B^{j}x_{t}$$. What do you mean when you say "believe" we can apply the geometric series result. I am lost at this step.

#### Dason

$$x_t = w_t + \theta w_{t-1}$$ which implies $$w_t = x_t - \theta w_{t-1}$$. (1)
But we also know $$x_{t-1} = w_{t-1} + \theta w_{t-2}$$ so we can rearrange this to solve for $$w_{t-1}$$ and plug that back into (1). If you keep doing these substitutions you might notice a pattern.