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An engineer wishes to estimate the mean yield of chemical process based on the yield measurements x1, x2, x3 from three independent runs of an experiment with variance σ2.

Consider the following two estimators of the mean yield θ:

(x1+x2+x3)/3 or (x1+2(x2)+x3)/4

Which is a better estimator? Justify with the appropriate statistical procedures.

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1 Answer 1

We prefer to use caps for random variables. Note that $$E\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{3}\left(E(X_1)+E(X_2)+E(X_3)\right)=\mu,$$ where $\mu$ is the common mean of the $X_i$.

Similarly, $$E\left(\frac{X_1+2X_2+X_3}{4}\right)=\frac{1}{4}\left(E(X_1)+2E(X_2)+E(X_3)\right)=\mu.$$

So our two estimators are both unbiased estimators of $\mu$.

Since the $X_i$ are independent, we have $$\text{Var}\left(\frac{X_1+X_2+X_3}{3}\right)=\frac{1}{9}\left(\text{Var}(X_1)+\text{Var}(X_2)+\text{Var}(X_3)\right)=\frac{\sigma^2}{3}.$$

Also, $$\text{Var}\left(\frac{X_1+2X_2+X_3}{4}\right)=\frac{1}{16}\left(\text{Var}(X_1)+2^2\text{Var}(X_2)+\text{Var}(X_3)\right)=\frac{6\sigma^2}{16}.$$

The variance of the second estimator is greater than the variance of the first. Thus the first estimator is "better" than the second.

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