# Maximum Likelihood estimator for $\alpha$.

I'm trying to work this question and I need an explaination ( The starting point ). How do I find the MLE for $\alpha$ and MSE for the MLE for $\alpha$ if we have the following: $X_{1}, ..., X_{n}$ are constants and $u_1, ...u_n$ are iid $N(0, \sigma^2 )$ and $\sigma^2$ is assumed to be known. Suppose $Z_1, ..., Z_n$ are random sample that satisfy $$Z_j = \alpha X_j + u_j$$

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Somebody should probably mention that to denote real numbers by capital letters and random variables by lower cases is the opposite of the conventions of the field. – Did Mar 11 '13 at 7:56
@Did: Yes, I should have -- but we've found before that this particular convention isn't sufficiently engrained in me :-) – joriki Mar 11 '13 at 12:45
@joriki You simply followed the conventions of the text of the question, which is a reasonable choice too. – Did Mar 11 '13 at 16:37

## 1 Answer

The density for obtaining $Z_j$, given $\alpha$, is that of $u_j=Z_j-\alpha X_j$, so the likelihood is proportional to

$$\prod_{j=1}^n\exp\left(-\frac1{2\sigma^2}\left(Z_j-\alpha X_j\right)^2\right)=\exp\left(-\frac1{2\sigma^2}\sum_{j=1}^n\left(Z_j-\alpha X_j\right)^2\right)\;.$$

Thus the maximum likelihood is attained at the minimum of the quadratic function of $\alpha$ in the exponent. Since you asked for a starting point, I'll leave it at that for now.

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@ Joriki, I thought $1/sqrt(2\pi)\sigma$ was suppose to be in front of the exponential? or? – Eugene Mettle Mar 11 '13 at 12:41
@Eugene: It says "proportional to", not "equal to". For finding the maximum likelihood, only the dependence on $\alpha$ is relevant; a factor independent of $\alpha$ has no bearing on the value of $\alpha$ that maximizes the expression. – joriki Mar 11 '13 at 16:43