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Show that \({\bar X_{i \ldots }} - {\bar X_ \ldots }\)=\({\alpha _i}\) so that \({\bar X_{i \ldots }} - {\bar X_ \ldots }\) is an unbiased estimator for \({\alpha _i}\) (in the fixed effects model).

b. With\({\hat \gamma _{ij}} = {\bar X_{ij}} - {\bar X_{i.}} - {\bar X_{jj}} + {\bar X_{..}}\), show that \({\hat \gamma _{ij}}\)is an unbiased estimator for \({\hat \gamma _{ij}}\) (in the fixed effects model).

Short Answer

Expert verified

a. Substitute the averages by the corresponding sums to obtain the expected value;

b. Substitute the averages by the corresponding sums to obtain the unbiased estimator.

Step by step solution

01

Show unbiased estimator

In the fixed effects model, the following is true:

Let

where the are independent normally distributed random variable with mean 0 and variance \({\sigma ^2}\)Denote with \({\bar X_{i \ldots }}\)the average of measurements obtained when factor A is held at level i

with $\bar{X}_{. j .}$, the average of measurements obtained when factor $B$ is held at level $j$

with \({\bar X_{.j.}}\)the average of measurements obtained when factor A is held at level i and factor B is held at level j

and with \({\bar X_ \ldots }\)the grand mean

Observed values are denoted with small x instead of big X. The notations without line over X are just the sums.

(a):

The following holds

Which stands because the sums are equal to zero.

b) The estimate of \({\hat \gamma _{ij}}\) is

\(\begin{aligned}{*{20}{c}}{{{\hat \gamma }_{ij}} = {{\bar X}_{ij.}} - {{\bar X}_{i..}} - {{\bar X}_{.j.}} + {{\bar X}_ \ldots }.}\\{E\left( {{{\hat \gamma }_{ij}}} \right) = {\gamma _{ij}}.}\end{aligned}\)

By substituting every average with corresponding sum, the following holds

\(E\left( {{{\hat \gamma }_{ij}}} \right) = E\left( {{{\bar X}_{ij.}} - {{\bar X}_{i..}} - {{\bar X}_{.j.}} + {{\bar X}_ \ldots }} \right)\)

Where by substituting

The following holds

\(\begin{aligned}{*{20}{c}}{E\left( {{{\hat \gamma }_{ij}}} \right) = \mu + {\alpha _i} + {\beta _j} + {\gamma _{ij}} - \left( {\mu + {\alpha _i}} \right) - \left( {\mu + {\beta _j}} \right) + \mu }\\{ = {\gamma _{ij}}}\end{aligned}\)

Which proves claim

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