# Unbiased Estimator of $\sigma^2$ using Least square estimates

Suppose I have $$Y_i \sim N(\beta_i ,\sigma^2)$$ for $$i=1,2,3$$.

It is given that $$\beta_1 + \beta_2= \beta_3$$ and $$\hat{\beta}_1 = \frac{2Y_1 - Y_2 +Y_3}{3} , \hat{\beta}_2 = \frac{2Y_2-Y_1+Y_3}{3},\hat{\beta}_3= \frac{Y_1+Y_2+2Y_3}{3}$$ where $$\hat{\beta}_i$$'s are the least square estimates of $$\beta_i$$'s. I am trying to find an unbiased estimator for $$\sigma^2$$ using the Least squares estimates, but I can't guess how should I combine the $$\hat{\beta}_i$$'s?

Let $$S=\frac{1}{3}\sum\limits_{i=1}^3(y_i-\hat{\beta_i})^2=\frac{1}{3}\sum\limits_{i=1}^3(y_i-\beta_i)^2+(\hat{\beta_i}-\beta_i)^2$$, then we have

$$E[S]=\sigma^2+E[(\hat{\beta_i}-\beta_i)^2]=\sigma^2+\frac{\sigma^2}{\sum\limits_{i=1}^3 y_i^2}=\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)\sigma^2$$,

since $$E[(\hat{\beta_i}-\beta_i)^2]=\sigma^2(Y^TY)^{-1}$$, for least-squares estimates.

$$\implies \hat{\sigma}^2_{unbiased}=\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)^{-1}S$$

$$\quad \quad =\frac{1}{3}\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)^{-1}\sum\limits_{i=1}^3(y_i-\hat{\beta_i})^2$$

Also, we have

$$$$\quad y_1=2\hat{\beta_3}-\hat{\beta_1}\\ \quad y_2=2\hat{\beta_3}-\hat{\beta_2}\\ \quad y_3=\hat{\beta_1}+\hat{\beta_2}-\hat{\beta_3}$$$$

$$\implies \hat{\sigma}^2_{unbiased}=\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)^{-1}\frac{1}{3}\sum\limits_i(y_i-\hat{\beta_i})^2$$

$$\quad \quad \quad \quad \quad \quad=\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)^{-1}(\hat{\beta_1}+\hat{\beta_2}-2\hat{\beta_3})^2$$

$$\quad \quad \quad \quad \quad \quad =\left(1+\frac{1}{\sum\limits_{i=1}^3 y_i^2}\right)^{-1}\hat{\beta_3^2}$$, since $$\hat{\beta_1}+\hat{\beta_2}=\hat{\beta_3}$$