2 2014-Spring
2.2 Problem 1
Given:
\(x_{i,1}, x_{i,2}\) : Known values For \(1 \leq i \leq n\): $Z_i = 1x{i,1} +_i $ and \(Y_i = \beta_1x_{i,1} +\beta_2x_{i,2} + \epsilon_i\) \(\epsilon_i \sim N(0,1)\) ### (a) Given \({\bf{Z}} =(Z_1,Z_2,\dots,Z_n)\) find MLE estimate of \(\beta_1\):
Since, \(\epsilon_i \sim N(0,1)\) \(\implies\) \(Z_i \sim N(\beta_1x_{i,1}, 1)\) for given \(\beta_1\) The likelihood function is given by:
\(L(\beta_1|\textbf{Z}) = \frac{e^{-\sum_{i=1}^{n}\big(\frac{Z_i-\beta_1x_{i,1}}{2}\big)^2}}{(\sqrt{2\pi})^n}\)
\(LL = log(L(\beta_1|\textbf{Z})) = -\sum_{i=1}^{n}\big(\frac{Z_i-\beta_1x_{i,1}}{2}\big)^2 - \frac{n}{2}log(2\pi)\)
\(\frac{dLL}{d\beta_1} = \sum_{i=1}^n (Z_i-\beta_1x_{i,1})x_{i,1}\)
and hence
\(\hat\beta_1 = \frac{\sum Z_ix_{i,1}}{\sum x_{i,1}^2}\)
\(\frac{d^2LL}{d\beta_1} = -\sum x_{i,1}^2\) and hence \(\hat \beta_1\) is in fact attains the maxima.
Now, \(E\hat\beta_1 = \frac{E(\sum Z_ix_{i,1})}{\sum x_{i,1}^2} = \frac{\sum x_{i,1}E(Z_i)}{\sum x_{i,1}^2} = \beta_1\)
and hence the MLE estimate pf \(\hat \beta_1\) is an unibased estimator of \(\beta_1\)