# Thread: Least Square Derivation - Checking

1. ## Least Square Derivation - Checking

QUESTION

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

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

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

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

By Least-Squares Definition we minimize distance

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

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

If I do this I get

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

This solves to

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

Is this somewhere close?

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

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

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

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

Thanks very much for any help advice

2. I just realised that (2) should actually say:

(2) Find the standard error of the estimate in (a) in terms of $\displaystyle \gamma$.. 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?)??