Least Square Derivation - Checking

**ninmaster** · March 21st 2008, 04:51 PM

QUESTION

I had a 10cm pole which broke in half. The two peices were measured and this resulted in values of $Z_1$ and $Z_2$ respectively.

If the true 'correct' length of the first peice is given as $\gamma$ and the standard deviation of the two measurements is $\sigma$

1) Then find the least-square estimate of $\gamma$
2) Find the standard error of the estimate in (1) in terms of $\sigma$

My Answer so Far

$Z_1 = \gamma + \epsilon_1$
$Z_2 = 10 - \gamma + \epsilon_2$

By Least-Squares Definition we minimize distance

$\epsilon_1 = (Z_1 - \gamma)^2$
$\epsilon_2 = (Z_2 - 10 + \gamma)^2$

Take derivative and set to zero? This is where I am unsure? *Thanks*

If I do this I get

$-2(Z_1 - \gamma) + 2(Z_2 - 10 + \gamma) = 0$

This solves to

$\hat\gamma\ = \frac{Z_1 - Z_2 + 10}{2}$

Is this somewhere close?

2) $Var(\hat\gamma) = \frac{1}{2^2} Var(Z_1 -Z_2 + 10)$

Now Im also unsure of this variance property.. but I solve to get

$Var(\hat\gamma) = \frac{\sigma^2 + \sigma^2}{4}$
$Var(\hat\gamma) =\frac{\sigma^2}{2}$

Is it right that the variance of the 10 is zero.. so it just leaves the equation?

Thanks very much for any help advice

**ninmaster** · March 21st 2008, 11:15 PM

I just realised that (2) should actually say:

(2) Find the standard error of the estimate in (a) in terms of $\gamma$ .. so i think I shouldn't be finding variance.. but standard error?

Would that just be the sum of errors squared divided by the degrees of freedom (which would be 1?)??

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