Can I use averages to improve the forecast of a multiple regression? I have a cross-sectional multiple regression that I have estimated and now I would like to apply it to make a simple forecast of the dependent variable.  
Take the data generating process
$$y_i =\alpha+\beta x_i+\Gamma Z_i+\epsilon_i,$$
where $y$ is the dependent variable, $x$ the independent variable and $Z$ some vector of covariates. I have performed the regression for the entire sample and estimated $\alpha$, $\beta$ and $\Gamma$ accordingly.
Now, I would like to make an out-of-sample forecast of $y_j$, say, at indicator $j$. All of $Z_j$ is known to me; however, $x_j$ is not. Now, it itches to take the average or conditional average of $x$, i.e., $E[x|..]$, of the original sample and use the it to predict $y_j$. That is
$$\hat y_j =\hat\alpha+\hat\beta \bar x+\hat \Gamma Z_j.$$
Would using the average or conditional improve the forecast in any way? 
Intuitively it appears to me the average or conditional average does not store any additional information to improve the prediction per se, but, if I have to leave out $x_i$ from the original regression and re-estimate it and then do a forecast without $x_i$, it appears that I am losing out on the information stored in $x_i$ as well.
 A: Let us write the regression problem as:
$$y_i = \alpha + \beta x_i + \Gamma Z_i + e_i$$
We can rewrite $\beta x_i$ as $\beta \bar{x} + \beta (x_i -\bar{x})$:
$$y_i = \alpha + \beta \bar{x} + \beta (x_i-\bar{x}) + \Gamma Z_i + e_i$$
Since $\bar{x}$ is a constant, the term $\beta \bar{x}$ can be absorbed into the constant term:
$$y_i = \alpha' + \beta (x_i-\bar{x}) + \Gamma Z_i + e_i$$
Note that this is mathematically equivalent to the original regression problem and can be estimated as one; you just get out a different parameter estimate for $\alpha$ than in the original problem, but you can recover it via:
$$\hat{\alpha} = \hat{\alpha}' - \hat{\beta} \bar{x}$$
Your estimates for $\beta$ will be the same in the two regressions, but your estimates of the constant terms won't be, unless $\bar{x}=0$, and this is the key to understanding what happens with your problem.
If we don't know $x_j - \bar{x}$ for some future observations indexed by $j$, and we therefore exclude all terms involving $x_j$ and $\bar{x}$, we are setting $x_j-\bar{x} = 0$  in the third equation, so the term $\beta(x_j-\bar{x})$ disappears, but we are also subtracting the term $\hat{\beta}\bar{x}$ from the estimated constant term $\alpha'$, which does not in fact change with changes in the value of $x_j$.  This will cause our prediction of $y_j$ to be biased by an amount equal to $\hat{\beta}\bar{x}$. 
Another way to look at is that if you remove the $x_j$ from the model altogether, you'll get the same answer as if you had substituted $0$ for all the $x_j$.  Naturally, substituting $0$ for all the $x_j$ is not the same as substituting $\bar{x}$, unless $\bar{x}=0$, and, again, you will have effectively removed the term $\hat{\beta}\bar{x}$ from the prediction equation - even though it was part of the estimated equation.
A concrete example follows.  Our true regression equation is $y_i = 1 + x_i + e_i$, with $e_i$ i.i.d. Normal(0,1) and $x_i \sim U(5,15)$, therefore with mean $10$:
> x <- runif(100, 5, 15)
> y <- 1 + x + rnorm(100)
> model <- lm(y~x)
> 
> x_future <- runif(100, 5, 15)
> y_future <- 1 + x_future + rnorm(100)
> 
> # Mean of future observed values of y
> mean(y_future)
[1] 11.05226
> 
> # Mean prediction if we have all the x values available
> mean(predict(model, newdata=data.frame(x_future)))
[1] 10.70136
> 
> # Mean prediction if we substitute the mean x value
> mean(predict(model, 
+              newdata=data.frame(x_future=rep(mean(x_future),100))))
[1] 10.70136
> 
> # Prediction if we remove all the x values (leaving just the intercept)
> coefficients(model)[[1]]
[1] 0.1992937

As we can see, substituting $\bar{x}$ for the $x_j$ preserves the mean of our predictions relative to predictions made with the $x_j$ themselves, but removing $x$ altogether does not.
One caveat, though; your prediction confidence intervals will be too narrow in your situation, because they don't take into account the added variability due to not knowing the true $x_i$ and using $\bar{x}$ instead.  In effect, you have another error term -  $\beta(x_i-\bar{x})$ - that's not accounted for in the estimation procedure.
