Discussion
Let \(X_{1}\), \(X_{2}\), .......... \(X_{n}\) be iid r.v.s satisfying the following regression equation :
\(X_{i} = \alpha z_{i} + e_{i}\); \(i = 1, 2, ....... n\)
where \(z_{1}\), \(z_{2}\), ....... \(z_{n}\) are fixed and \(e_{i}\)'s \((i = 1, 2, ..... n)\) are iid r.v.s with \(N(0, \sigma^{2})\), \(\sigma^{2}\) is unknown MLE of \(\alpha\) is given as :
Answer: D
The MLE is \(\hat{\alpha} = \sum x_{i}z_{i}/\sum z_{i}^{2}\).
With normal errors, maximising the likelihood in \(\alpha\) is the same as minimising \(\sum(x_{i} - \alpha z_{i})^{2}\), whatever \(\sigma^{2}\) is. Setting the derivative to zero:
\[ \begin{aligned} -2\sum z_{i}(x_{i} - \alpha z_{i}) &= 0 \\ \hat{\alpha} &= \frac{\sum x_{i}z_{i}}{\sum z_{i}^{2}} \end{aligned} \]
This is least squares through the origin. Option A divides by \(\sum x_{i}^{2}\), which would regress \(z\) on \(x\); option B divides by \(n\), which is correct only if every \(z_{i}^{2} = 1\); and option C does not involve the data at all.
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