Let the disaggregrate model be:
$Y_{ia} = X_a \beta + \epsilon_i$
where
$\epsilon_i \sim N(0,\sigma^2)$
Your aggregate model is given by:
$Y_a = \frac{\sum_i(Y_{ia})}{n_a}$
where,
$n_a$ is the number of observations you have corresponding to the $a$ index.
Therefore, it follows that:
$Y_a = X_a \beta + \epsilon_a$
where
$\epsilon_a \sim N(0, \frac{\sigma^2}{n_a} )$ and
$a=1, 2, ... A$
Therefore, the OLS estimate would be given by minimizing:
$\sum_a(Y_a - X_a \beta)^2$
Which yields the usual solution. So, I do not think there is any difference as far as the estimate for the slope parameters are concerned.
Edit 1
Here is a small simulation in R which illustrates the above idea (apologies for the flaky code as I am using questions such as the above to learn R).
set.seed(1);
n <- c(4,2,8);
x <- c(1,2,3);
data <- matrix(0,14,2)
mean_data <- matrix(0,3,2)
index <- 1;
for (i in 1 : 3)
{
for(obs in 1:n[i])
{
data[index,1] <- x[i];
data[index,2] <- x[i]*8 + 1.5*rnorm(1);
mean_data[i,1] = x[i];
mean_data[i,2] = mean_data[i,2] + data[index,2];
index = index + 1;
}
mean_data[i,2] = mean_data[i,2] / n[i];
}
beta <- lm(mean_data[,2] ~ mean_data[,1]);
The above code yields the output when you type beta
:
Call:
lm(formula = mean_data[, 2] ~ mean_data[, 1])
Coefficients:
(Intercept) mean_data[, 1]
-0.03455 7.99326
Edit 2
However, OLS is not efficient as error variances are not equal. Thus, using MLE ideas, we need to minimize:
$\sum_a{n_a (Y_a - X_a \beta)^2}$
In other words, we want to minimize:
$\sum_a{(\sqrt{n_a} Y_a - \sqrt{n_a} X_a \beta)^2}$
Thus, the MLE can be written as follows:
Let $W$ be a diagonal matrix with the $\sqrt{n_a}$ along the diagonal. Thus, the MLE estimate can be written as:
$(X' X)^{-1} X' Y$
where,
$Y = W [Y_1,Y_2,...Y_A]'$ and
$X = W [X_1,X_2,...X_A]'$
Another way to think about this is:
Consider the variance of $Y$. The transformation given above for $Y$ ensures that the variance of the individual values of $Y$ are identical thus satisfying the conditions of the Gauss-Markov theorem that OLS is BLUE.