# Standard deviation of LS-estimator

I am trying to solve this problem:

Three objects have weights $m_1$, $m_2$ and $m_3$. We measure the weight of the pairs twice and get the following result:

$m_1 + m_2$ have the collective weight $44.60$ in trial $1$ and $44.70$ in trial $2$.

$m_1 + m_3$ have the collective weight $50.53$ in trial $1$ and $49.49$ in trial $2$.

$m_2 + m_3$ have the collective weight $54.32$ in trial $1$ and $55.42$ in trial $2$.

All observations have a normally distributed random error equal to $\sigma$.

First, I was supposed to estimate $m_1$, $m_2$ and $m_3$ using the least square method. This was no problem, I reached the results $m_1 = 19.895$, $m_2 = 24.755$ and $m_3 = 30.115$.

But then I am supposed to determine the standard deviation of the estimators, written in terms of $\sigma$. And this I haven't been able to figure out.

I tried using that, for the sum of two independent random variables, the standard deviation is given by:

$\sigma = \sqrt{\sigma_1^2 + \sigma_2^2}$

I arranged it as an equation system and reached the result that $\mathrm{sd}( \sigma_1 ) = \mathrm{sd}(\sigma_2) = \mathrm{sd}(\sigma_3) = \frac{ \sigma }{ \sqrt{2} }$. That was not the right answer.

I have tried what feels like a trillion more things that have taken me no where. Any advice?

• The sd should be $\sqrt{3/8}\sigma$. – Hans Engler Nov 25 '15 at 13:08
• Right. That I know, but I don't know why it is that way. :) – m.bing Nov 25 '15 at 13:15

The problem of estimating the true weights $m_i$ is best studied using a matrix formulation. Consider the "weighing matrix" $$A = \begin{pmatrix} 1&1&0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \, .$$ Let $m$ be the vector of true weights. You have two observations per combination of objects, namely $b_1 = Am + \xi_1$ and $b_2 = A m + \xi_2$ where the error vectors $\xi_i$ all have independent $N(0,\sigma^2)$ distributed components. The least squares estimator for the vector $m$ then is $$\hat m = (A^TA)^{-1}(A^Tb_1 + A^Tb_2)/2$$ leading to your weight estimates. The estimation error $\hat m - m$ is $$\zeta = (A^TA)^{-1}(A^T\xi_1 + A^T\xi_2)/2 \, .$$ Now work out the covariance $$\mathcal{E ( \zeta \zeta^T)} = \frac{\sigma^2}{2}(A^TA)^{-1} = \frac{\sigma^2}{2} \cdot \begin{pmatrix}\frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} \\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{pmatrix}$$
and you see that $var \, \hat m_i = \frac{3 \sigma^2}{8}$ for all $i$ and $cov(\hat m_i, \hat m_j) = - \frac{\sigma^2}{8}$ for all $i \ne j$.