explaination of a passage in proof of consistency of OLS-estimator

Question

I am proving that an OLS-estimator is consistent. In the solution, they show that $$ \hat{\beta_3} = \frac{\sum_{i=1}^nT_i(\beta_3 \cdot T_i + u_{i2} - u_{i1})}{\sum_{i=1}^n T_i} = \beta_3 + \frac{\sum_{i=1}^n T_i(u_{i2} - u_{i1})}{\sum_{i=1}^n T_i}$$

I am not sure how they have gotten $\beta_3$ out of the bracket and reduced the sum in the numerator. Any ideas?

This is probably not relevant, but will still mention anyway just incase: i=1,2, T is time period, a dummy variable, thus $T_i = \{0,1\} $, $\beta_3$ is a OLS estimator.

Answer 1

distributing $T_i$ and splitting the sum in two parts you get that

$$ \hat{\beta_3} = \frac{\sum_{i=1}^nT_i(\beta_3 \cdot T_i + u_{i2} - u_{i1})}{\sum_{i=1}^n T_i} = \frac{\sum_{i=1}^n\beta_3 \cdot T_i^2 + \sum_{i=1}^n T_i(u_{i2} - u_{i1})}{\sum_{i=1}^n T_i}$$

so

$$\hat{\beta_3} = \beta_3 \frac{\sum_{i=1}^n T_i^2}{\sum_{i=1}^n T_i} + \frac{\sum_{i=1}^n T_i(u_{i2} - u_{i1})}{\sum_{i=1}^n T_i}$$

but since it is true that $T_i \in \{0,1\}$ then $\frac{\sum_{i=1}^n T_i^2}{\sum_{i=1}^n T_i} = 1$. Therefore you obtain the desired identity.