Moving Average (MA) process: numerical intuition

Question

This forum is full of questions regarding MA processes; for instance: Confusion about Moving Average(MA) Process.

There seem to be a lot of confusion wrt MA processes. I think having a numerical example would help.

Let us say I want to model the following observations:

t  1  2  3  4  5  6
Y  5  6 -4  8 10 -2

I find that I can model it using the following MA(2) process:

$\hat{Y}_t=\frac{1}{2}\epsilon_t+\frac{1}{2}\epsilon_{t-1}$

The average is zero, so I guess the errors are equal to the observations, so that:

t   1   2   3  4  5  6
Y   5   6  -4  8 10 -2
e   5   6  -4  8 10 -2
e-1 NA  5   6 -4  8 10
Ŷ   NA  5.5 1  2  9  4

Is this how Y is forecast in a MA model?

Answer 1

Essentially, I agree with @IrishStat, but I would like to "rephrase" the answer a little.

If you assume that $Y_t$ follows an MA(2) process, then you have

$$Y_t = \varepsilon_t + \theta_1 \varepsilon_{t-1} + \theta_2 \varepsilon_{t-2}$$

(I assume no intercept for simplicity.) Note that this is not what you have in your equation.

Now if you are going to forecast $Y_t$ using the information available up to time $t-1$, $I_{t-1}$, the point forecast of $Y_t$ will be

$$ \begin{multline} \begin{split} \operatorname{E}(Y_t|I_{t-1}) &= \operatorname{E}(\varepsilon_t + \theta_1 \varepsilon_{t-1} + \theta_2 \varepsilon_{t-2}|I_{t-1}) \\ &= \operatorname{E}(\varepsilon_t|I_{t-1}) + \theta_1 \operatorname{E}(\varepsilon_{t-1}|I_{t-1}) + \theta_2 \operatorname{E}(\varepsilon_{t-2}|I_{t-1}) \\ &= 0 + \theta_1 \varepsilon_{t-1} + \theta_2 \varepsilon_{t-2} \\ &= \theta_1 \varepsilon_{t-1} + \theta_2 \varepsilon_{t-2} \end{split} \end{multline} $$

Example:
if
$\varepsilon_1 = 5$,
$\varepsilon_2 = 6$,
$\theta_1 = 0.5$,
$\theta_2 = -0.25$,
then the point forecast of $Y_3$ given the information at time 2 is

$$ \begin{multline} \begin{split} \operatorname{E}(Y_3|I_2) &= \theta_1 \varepsilon_{3-1} + \theta_2 \varepsilon_{3-2} \\ &= \theta_1 \varepsilon_2 + \theta_2 \varepsilon_1 \\ &= 0.5 \cdot 6 - 0.25 \cdot 5 \\ &= 1.75 \end{split} \end{multline} $$

Answer 2

The current error $e_t$ is never known until after the $Y_t$ is observed thus it is set to 0.0 . The MA(2) process is $Y_t= + .5 * e_{t-1}+ .5* e_{t-2} + e_t$ where $e_t= 0.0$. No forecast is possible until period 3.