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pset 6.1 Prove that the OLS estimator of $\boldsymbol{\beta}$ in the linear regression model with individualspecific intercepts, that is$$y_{i t}=\boldsymbol{x}_{i t}^{\prime} \boldsymbol{\beta}+\alpha_{i}+u_{i t},$$where $\alpha_{i}$ is a parameter to be estimated, for $i=1, \ldots, n$ is equal to $\hat{\boldsymbol{\beta}}_{W G}$.

Question

pset 6.1 Prove that the OLS estimator of β\boldsymbol{\beta} in the linear regression model with individualspecific intercepts, that isy_{i t}=\boldsymbol{x}{i t}^{\prime} \boldsymbol{\beta}+\alpha{i}+u_{i t},where αi\alpha_{i} is a parameter to be estimated, for i=1,,ni=1, \ldots, n is equal to β^WG\hat{\boldsymbol{\beta}}_{W G}.

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Solution

To prove that the OLS estimator of β in the linear regression model with individual-specific intercepts is equal to β̂_WG, we can follow these steps:

Step 1: Start with the given linear regression model: y_it = x_it'β + α_i + u_it

Step 2: Rewrite the model by subtracting the individual-specific intercept α_i from both sides: y_it - α_i = x_it'β + u_it

Step 3: Rearrange the equation to isolate the dependent variable on the left-hand side: y_it - α_i = x_it'β + u_it y_it - x_it'β = α_i + u_it

Step 4: Now, let's take the average of both sides of the equation over time for each individual i: (1/T) ∑(t=1)^(T) (y_it - x_it'β) = (1/T) ∑(t=1)^(T) (α_i + u_it)

Step 5: Since the individual-specific intercept α_i is constant over time, we can rewrite the left-hand side of the equation as: (1/T) ∑_(t=1)^(T) (y_it - x_it'β) = y_i - x_i'β

Step 6: Similarly, the right-hand side of the equation becomes: (1/T) ∑(t=1)^(T) (α_i + u_it) = α_i + (1/T) ∑(t=1)^(T) u_it

Step 7: Now, let's substitute the expressions from Step 5 and Step 6 back into the equation: y_i - x_i'β = α_i + (1/T) ∑_(t=1)^(T) u_it

Step 8: Since the OLS estimator minimizes the sum of squared residuals, we can differentiate the sum of squared residuals with respect to β and set it equal to zero: ∂/∂β [∑(i=1)^(n) ∑(t=1)^(T) (y_it - x_it'β - α_i)^2] = 0

Step 9: Simplify the expression by expanding the squared term: ∑(i=1)^(n) ∑(t=1)^(T) 2(y_it - x_it'β - α_i)(-x_it) = 0

Step 10: Rearrange the terms and divide by -2: ∑(i=1)^(n) ∑(t=1)^(T) (x_it'β + α_i - y_it)x_it = 0

Step 11: Distribute the x_it term: ∑(i=1)^(n) ∑(t=1)^(T) x_it'βx_it + ∑(i=1)^(n) ∑(t=1)^(T) α_ix_it - ∑(i=1)^(n) ∑(t=1)^(T) y_itx_it = 0

Step 12: Recognize that ∑(t=1)^(T) x_it'βx_it is a constant term that does not depend on β, so we can ignore it for now: ∑(i=1)^(n) ∑(t=1)^(T) α_ix_it - ∑(i=1)^(n) ∑_(t=1)^(T) y_itx_it = 0

Step 13: Rearrange the terms to isolate β: ∑(i=1)^(n) ∑(t=1)^(T) y_itx_it = ∑(i=1)^(n) ∑(t=1)^(T) α_ix_it

Step 14: Divide both sides of the equation by nT: (1/nT) ∑(i=1)^(n) ∑(t=1)^(T) y_itx_it = (1/nT) ∑(i=1)^(n) ∑(t=1)^(T) α_ix_it

Step 15: Recognize that (1/nT) ∑(i=1)^(n) ∑(t=1)^(T) y_itx_it is the OLS estimator of β, denoted as β̂_OLS, and (1/nT) ∑(i=1)^(n) ∑(t=1)^(T) α_ix_it is the within-group estimator of β, denoted as β̂_WG: β̂_OLS = β̂_WG

Therefore, we have proven that the OLS estimator of β in the linear regression model with individual-specific intercepts is equal to β̂_WG.

This problem has been solved

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