I have the following model:





$\varepsilon_i\thicksim \text{iid } \mathcal{N}(0,\sigma^2), \forall i$

I would like to obtain the best (unbiased and with minimum variance) estimator of $\beta$. That is, I would like to know $\hat{\beta}=f(Y_1,Y_2,Y_3,Y_4)$. How should I obtain it?

I will really appreciate your help.

  • 4
    $\begingroup$ Cross-posted. $\endgroup$ – Did Jul 18 '12 at 13:14
  • $\begingroup$ Thank you very much to all those who answered the question! $\endgroup$ – Nikita Samoylov Jul 19 '12 at 6:55

A general result which should be in your lecture notes says that $\hat\beta$ is an unbiased affine transform of the vector $(Y_k)_k$. Unbiasedness for every $\beta$ further imposes that $\hat\beta$ is linear and the coefficients of $\beta$ in $(Y_k)_k$ impose that $\hat\beta=\sum\limits_{k=1}^4x_kY_k$ for some $(x_k)_k$ such that $\sum\limits_{k=1}^4x_k=1$.

At this point, Lagrange multiplier's method readily yields $(x_k)_k$ but, in the present case, symmetry considerations offer a nice alternative proof.

To see this, note that the symmetry $\varepsilon_1\leftrightarrow\varepsilon_3$, $\varepsilon_2\leftrightarrow\varepsilon_4$, exchanges $Y_1$ and $Y_2$ and exchanges $Y_3$ and $Y_4$. Since the distribution of $(Y_k)_k$ is invariant by this operation, this yields $x_1=x_2$ and $x_3=x_4$. Hence $x_1=x_2=\frac12(1-x)$ and $x_3=x_4=\frac12x$ for some $x$.

The variance of $\hat\beta$ is $\sigma^2$ times $(x_1+x_3)^2+(x_1+x_4)^2+(x_2+x_4)^2+(x_2+x_3)^2+(x_3+x_4)^2$ and, when $(x_k)_k$ is as above, this sum is $\frac14+\frac14+\frac14+\frac14+x^2$, which is minimum for $x=0$.

Finally, $\hat\beta=\frac12(Y_1+Y_2)$.


I gave an answer to this cross-posted on CrossValidation. Just compute the likelihood and get the maximum likelihood estimate of beta. The nice theory about mles still apply to this example.

Another option is the best linear unbiased estimator. b=aY_1 +b Y_2 +c Y_3 + d Y_4 where a, b, c and d are chosen so that the variance is minimized.


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