<p>Your equation <span class="math-container">$y_{T+1} = E(y_{T+1}|I_T)$</span> clearly does not make sense.</p>

<p>In any case,
<span class="math-container">$$
E(y_{T+1}|I_T)
$$</span> is the <em>best one step ahead forecast conditional on <span class="math-container">$Y_1, \cdots, Y_T$</span></em>, whereas you're asked to find the <em>best linear one step ahead forecast given <span class="math-container">$Y_T$</span></em>. They are in general not the same.
The best linear forecast for <span class="math-container">$Y_{T+1}$</span> given <span class="math-container">$Y_T$</span> is
<span class="math-container">$$
\mu + \frac{\gamma(1)}{\gamma(0)} (Y_T - \mu),
$$</span>
where <span class="math-container">$\mu$</span> is the unconditional mean of the series and <span class="math-container">$\gamma$</span> is the autocovariance function. In your case, <span class="math-container">$\mu = 0$</span> and <span class="math-container">$\gamma$</span> is that of a MA(1) series (<span class="math-container">$\frac{\gamma(1)}{\gamma(0)} = \frac{\theta}{1+\theta^2}, \; \theta = \frac12$</span>). </p>

<p>In general, the best linear forecast for covariance-stationary time series can be obtained by solving the Yule-Walker equations.</p>

<p>For best linear forecast, the assumption that innovations are i.i.d. is extraneous.</p>

CC BY-SA 4.0
2020-02-15T03:34:09.330Z
2020-02-15T04:40:45.993Z
2020-02-15T04:40:45.993Z
Michael
<p>EDIT: I had put this answer earlier and then I realized that it is the same as the OP's result so I deleted it. Now, I see that other people are claiming that the OP's result is incorrect and I disagree ( I believe it's correct ), so I'm undeleting my answer which is below. The result is the same as what the OP obtained but just obtained in a slightly different way.</p>

<p>Given the MA process for <span class="math-container">$y_{t}$</span>, we can replace with <span class="math-container">$t$</span> with <span class="math-container">$T+1$</span> and write <span class="math-container">$y_{T+1} = \epsilon_{T+1} - \frac{1}{2} \epsilon_{T}$</span></p>

<p>Also, since <span class="math-container">$y_{t} = \epsilon_{t} - \frac{1}{2} \times \epsilon_{t-1} 
= \epsilon_t(1 - \frac{1}{2} L) $</span></p>

<p>this implies that, <span class="math-container">$\epsilon_t = \frac{y_{t}}{(1 - \frac{1}{2}L)}$</span>.</p>

<p>Then we can use the formula for a geometric series which results in:</p>

<p><span class="math-container">$\epsilon_t = y_{t} + 1/2 \times y_{t-1} + (1/2)^2 \times y_{t-2} +  \ldots (1/2)^n \times y_{t-n} + \ldots  $</span></p>

<p>Now, going back to the first equation at the top ( but replacing <span class="math-container">$t$</span> with <span class="math-container">$T$</span> ) we have <span class="math-container">$E(\epsilon_{T+1}|T) = 0$</span> and we are left with <span class="math-container">$-1/2 \times \epsilon_T$</span>. But it was just shown that <span class="math-container">$\epsilon_T$</span> can be written in terms of the infinite series of previous <span class="math-container">$y_{t}$</span>. So, we put that for <span class="math-container">$\epsilon_T$</span>. Then, multiplying that by <span class="math-container">$-\frac{1}{2}$</span>, results in the same expression as that of the OP.</p>

<p>ANOTHER EDIT:</p>

<p>Also, to write the forecast as was requested,  we have that </p>

<p><span class="math-container">$y_{T+1} = -\frac{1}{2} \sum_{i=0}^{\infty}(\frac{1}{2} L)^{k} y_{T}) =  -\frac{1}{2} \left(\frac{y_T}{1 - \frac{1}{2} L }\right)  = - \frac{y_T}{2 -L}$</span></p>

<p>So, <span class="math-container">$C = - \frac{1}{2 - L}$</span></p>

CC BY-SA 4.0
2020-02-14T04:58:41.897Z
2020-03-15T13:24:09.987Z
2020-03-15T13:24:09.987Z
mark leeds
<p>If we have a model MA(1): <span class="math-container">$y_t=\epsilon_t-\frac{1}{2}\epsilon_{t-1}$</span> where <span class="math-container">$\epsilon_t\sim (0,\sigma^2)$</span> and <span class="math-container">$\forall_{t}E(\epsilon_t)=0$</span> then <span class="math-container">$\forall_{l\geq 1}E(y_{T+l})=0$</span>. The best choice of C is 0.</p>

CC BY-SA 4.0
2020-03-14T23:16:36.800Z
2020-03-14T23:16:36.800Z
Sławomir Jarek
<p>Suppose that we have an MA(1) model where <span class="math-container">$y_t=\epsilon_t-\frac 1 2 \epsilon_{t-1}$</span> and <span class="math-container">$\epsilon_t$</span> are i.i.d.
I want to find an optimal forecast of <span class="math-container">$y_{T+1}$</span> in the form of <span class="math-container">$Cy_T$</span> where C is a constant. What is the best choice of C?
I have computed that </p>

<p><span class="math-container">$\begin{aligned}
&amp;y_{T+1}=E(y_{T+1}|I_T)=E(\epsilon_{T+1}-\frac 1 2 \epsilon_T|I_T)\\
&amp;=E(\epsilon_{T+1}|I_T)-\frac 1 2E(\epsilon_T|I_T)\\
&amp;=0-\frac 1 2\epsilon_T=-\frac 1 2 \epsilon_T\\
&amp;\epsilon_T=y_T+\frac 1 2y_{T-1}+\frac 1 4 y_{T-2}+......+{1\over 2^{t-1}}y_1+0\\
&amp;y_{T+1}=-\frac 12 \epsilon_T=-\frac 12y_T-\frac 1 4y_{T-1}-\frac 1 8 y_{T-2}-......-{1\over 2^{t}}y_1
\end{aligned}$</span></p>

<econometrics><applied-econometrics>
What is the optimal forecast for the following moving average model?
Alex Wang
# What is the optimal forecast for the following moving average model?

(Asked by: username_0 on 2020-02-13)

Suppose that we have an MA(1) model where $y\_t=\epsilon\_t-\frac 1 2 \epsilon\_{t-1}$ and $\epsilon\_t$ are i.i.d.
I want to find an optimal forecast of $y\_{T+1}$ in the form of $Cy\_T$ where C is a constant. What is the best choice of C?
I have computed that

$\begin{aligned}
&y\_{T+1}=E(y\_{T+1}|I\_T)=E(\epsilon\_{T+1}-\frac 1 2 \epsilon\_T|I\_T)\\
&=E(\epsilon\_{T+1}|I\_T)-\frac 1 2E(\epsilon\_T|I\_T)\\
&=0-\frac 1 2\epsilon\_T=-\frac 1 2 \epsilon\_T\\
&\epsilon\_T=y\_T+\frac 1 2y\_{T-1}+\frac 1 4 y\_{T-2}+......+{1\over 2^{t-1}}y\_1+0\\
&y\_{T+1}=-\frac 12 \epsilon\_T=-\frac 12y\_T-\frac 1 4y\_{T-1}-\frac 1 8 y\_{T-2}-......-{1\over 2^{t}}y\_1
\end{aligned}$

**Tags:** econometrics, applied-econometrics

---

## 3 answers

### Answer by username_1 on 2020-02-15. Score: 1

Your equation $y\_{T+1} = E(y\_{T+1}|I\_T)$ clearly does not make sense.

In any case,
$$
E(y\_{T+1}|I\_T)
$$ is the *best one step ahead forecast conditional on $Y\_1, \cdots, Y\_T$*, whereas you're asked to find the *best linear one step ahead forecast given $Y\_T$*. They are in general not the same.
The best linear forecast for $Y\_{T+1}$ given $Y\_T$ is
$$
\mu + \frac{\gamma(1)}{\gamma(0)} (Y\_T - \mu),
$$
where $\mu$ is the unconditional mean of the series and $\gamma$ is the autocovariance function. In your case, $\mu = 0$ and $\gamma$ is that of a MA(1) series ($\frac{\gamma(1)}{\gamma(0)} = \frac{\theta}{1+\theta^2}, \; \theta = \frac12$).

In general, the best linear forecast for covariance-stationary time series can be obtained by solving the Yule-Walker equations.

For best linear forecast, the assumption that innovations are i.i.d. is extraneous.


---

### Answer by username_2 on 2020-02-14. Score: 0

EDIT: I had put this answer earlier and then I realized that it is the same as the OP's result so I deleted it. Now, I see that other people are claiming that the OP's result is incorrect and I disagree ( I believe it's correct ), so I'm undeleting my answer which is below. The result is the same as what the OP obtained but just obtained in a slightly different way.

Given the MA process for $y\_{t}$, we can replace with $t$ with $T+1$ and write $y\_{T+1} = \epsilon\_{T+1} - \frac{1}{2} \epsilon\_{T}$

Also, since $y\_{t} = \epsilon\_{t} - \frac{1}{2} \times \epsilon\_{t-1}
= \epsilon\_t(1 - \frac{1}{2} L) $

this implies that, $\epsilon\_t = \frac{y\_{t}}{(1 - \frac{1}{2}L)}$.

Then we can use the formula for a geometric series which results in:

$\epsilon\_t = y\_{t} + 1/2 \times y\_{t-1} + (1/2)^2 \times y\_{t-2} + \ldots (1/2)^n \times y\_{t-n} + \ldots $

Now, going back to the first equation at the top ( but replacing $t$ with $T$ ) we have $E(\epsilon\_{T+1}|T) = 0$ and we are left with $-1/2 \times \epsilon\_T$. But it was just shown that $\epsilon\_T$ can be written in terms of the infinite series of previous $y\_{t}$. So, we put that for $\epsilon\_T$. Then, multiplying that by $-\frac{1}{2}$, results in the same expression as that of the OP.

ANOTHER EDIT:

Also, to write the forecast as was requested, we have that

$y\_{T+1} = -\frac{1}{2} \sum\_{i=0}^{\infty}(\frac{1}{2} L)^{k} y\_{T}) = -\frac{1}{2} \left(\frac{y\_T}{1 - \frac{1}{2} L }\right) = - \frac{y\_T}{2 -L}$

So, $C = - \frac{1}{2 - L}$


---

### Answer by username_3 on 2020-03-14. Score: 0

If we have a model MA(1): $y\_t=\epsilon\_t-\frac{1}{2}\epsilon\_{t-1}$ where $\epsilon\_t\sim (0,\sigma^2)$ and $\forall\_{t}E(\epsilon\_t)=0$ then $\forall\_{l\geq 1}E(y\_{T+l})=0$. The best choice of C is 0.


---

