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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

$\begin{aligned} &y_{T+1}=E(y_{T+1}|I_T)=E(\epsilon_{T+1}-\frac 1 2 \epsilon_T|I_T)\\ &=E(\epsilon_{T+1}|I_T)-\frac 1 2E(\epsilon_T|I_T)\\ &=0-\frac 1 2\epsilon_T=-\frac 1 2 \epsilon_T\\ &\epsilon_T=y_T+\frac 1 2y_{T-1}+\frac 1 4 y_{T-2}+......+{1\over 2^{t-1}}y_1+0\\ &y_{T+1}=-\frac 12 \epsilon_T=-\frac 12y_T-\frac 1 4y_{T-1}-\frac 1 8 y_{T-2}-......-{1\over 2^{t}}y_1 \end{aligned}$

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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.

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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.

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

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