Given the following regression model:
$$y_{ij} = \beta_0 + \beta_T x_{ij} + \beta_L(x_{ij} - x_{i1}) + \epsilon_{ij}$$
I need to find the expectation of $\hat \beta$, given by: $\hat \beta = (X'X)^{-1}(X'y)$. So:
$$ E[\hat \beta] = E[(X'X)^{-1}(X'y)]$$ $$ E[\hat \beta] = \sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij}'x_{ij})^{-1} \sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij}'E[y_{ij}]) $$ $$ E[\hat \beta] = \sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij}'x_{ij})^{-1} \sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij}'(\beta_0 + \beta_T x_{ij} + \beta_L(x_{ij} - x_{i1}))) $$
What I did was:
Divide and multiplicate by $\sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij})$: $$ E[\hat \beta] = \frac{\sum_{i = 1}^{N}\sum_{j = 1}^{n}\beta_0 + \beta_T x_{ij} + \beta_L(x_{ij} - x_{i1})}{\sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij})} $$
Remove the sums where I can:
$$ E[\hat \beta] = \frac{(N*n)\beta_0}{\sum_{i = 1}^{N} \sum_{j = 1}^{n}(x_{ij})} + (\beta_T + \beta_L) - \beta_L \sum_{i = 1}^{N} n * x_{i1}$$
And this is where I'm stuck. Also, I don't think everything I did is right. I called the coefficient "biased" because the teacher told me I would verify that.