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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?

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1 Answer 1

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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

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

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

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

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

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