Gauss Markov assumptions 
I think these two methods would differ, and the second method would be more accurate. But I am wondering that we know nothing about these parameters, maybe they could be the same, so from what angle should we approach this problem?
But I also think that the second method would be preferred under the Gauss Markov assumptions since "Distinct error terms are uncorrelated"is preferred. Thanks.
 A: Simple intuition:
Imagine we're estimating the probability that a thumbtack lands heads.
Experiment 1 flips a thumbtack 5 times.
Experiment 2 flips a thumbtack 100 times.
Let's say from experiment 1, $\hat{b}_1 = .8$ and from experiment 2 $\hat{b}_2 = .6$ Which estimate is more precise? For an overall estimate based upon both experiments, should we calculate $\frac{1}{2} .8 + \frac{1}{2} .6$ (i.e. take the simple arithmetic mean) or is that in some sense sub-optimal? What weighting would correspond with a single experiment that flipped the thumbtack 105 times?
Linear algebra based argument:
Let's split the sample in two.
$$ X = \left[ \begin{array}{c} X_1 \\ X_2 \end{array} \right]\quad \quad \mathbf{y} = \left[ \begin{array}{c} \mathbf{y}_1 \\ \mathbf{y}_2 \end{array} \right] $$
Now let's write the ordinary least squares (OLS) estimate $\hat{\mathbf{b}}$ for the full sample as a weighted sum of the OLS estimates for the subsamples.
$$ 
\begin{align*} \hat{\mathbf{b}} &= (X'X)^{-1}X'y \\
&= \left[\begin{array}{c} X_1'X_1 + X_2'X_2 \end{array}  \right]^{-1}\left[X_1'\mathbf{y}_1 + X_2'\mathbf{y}_2 \right]  \\
&= \left[\begin{array}{c} X_1'X_1 + X_2'X_2 \end{array}  \right]^{-1}\left[X_1'X_1(X_1'X_1)^{-1} X_1'\mathbf{y}_1 + X_2'X_2(X_2'X_2)^{-1} X_2'\mathbf{y}_2 \right] \\
&= \left[\begin{array}{c} X_1'X_1 + X_2'X_2 \end{array}  \right]^{-1}\left[X_1'X_1\hat{\mathbf{b}}_1 + X_2'X_2\hat{\mathbf{b}}_2 \right] \\
&= \left( \left[\begin{array}{c} X_1'X_1 + X_2'X_2 \end{array}  \right]^{-1}X_1'X_1\right) \hat{\mathbf{b}}_1 + \left( \left[\begin{array}{c} X_1'X_1 + X_2'X_2 \end{array} \right]^{-1} X_2'X_2\right)\hat{\mathbf{b}}_2 
\end{align*} $$
Now define $\hat{M}_{1} = \frac{X_1'X_1}{n_1}  \quad \hat{M}_{2} = \frac{X_2'X_2}{n_2} \quad \quad \hat{M} = \frac{X'X}{n} = \frac{n_1}{n_1 + n_2} M_{1} + \frac{n_2}{n_1 + n_2} M_{2}$ where $n_1$ is size of subsample $X_1$ etc... And the idea is that $M = \mathrm{E}\left[\mathbf{x}\mathbf{x}' \right] $.
Then:
$$\hat{\mathbf{b}} = \left( \frac{n_1}{n_1 + n_2}\right)\left(\hat{M}^{-1}\hat{M}_1 \right) \hat{\mathbf{b}}_1 + \left( \frac{n_2}{n_1 + n_2}\right)\left(\hat{M}^{-1}\hat{M}_2 \right) \hat{\mathbf{b}}_2 $$
$\hat{\mathbf{b}}$ from OLS is in some sense a sophisticated weighted-average of the subsample estimates. How does this compare to the simple arithmetic average of $\hat{\mathbf{b}}_1$ and $\hat{\mathbf{b}}_2$? Which way of combining  $\hat{\mathbf{b}}_1$ and $\hat{\mathbf{b}}_2$ is optimal in a best linear unbiased estimator sense?
Note also that $\left( \frac{n_1}{n_1 + n_2}\right)\left(\hat{M}^{-1}\hat{M}_1 \right)$ is a matrix, not a scalar. And for $\hat{M} \approx \hat{M}_1$ you'll probably have something in some sense close to $\frac{n_1}{n_1+n_2}$ as the weight on $\hat{\mathbf{b}}_1$.
