bias of a point estimator of 2 samples

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Kevin Sly

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Jan 21, 2016
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If I have a point estimator rule of (1/2)(Xsub1 + Xsub2) and I want to find the bias of an exponential distribution, I use

E(theta hat) - theta
E((1/2)(Xsub1 + Xsub2)) - theta = 0
So bias = 0.

Since Xsub1 and Xsub2 are 2 separate samples, why can I equate both to theta? Can't my point estimator change resulting in 2 different thetas?


...sorry for my lack of latex, that comes later this semester ;)
 
Kevin Sly said:
If I have a point estimator rule of (1/2)(Xsub1 + Xsub2) and I want to find the bias of an exponential distribution, I use

E(theta hat) - theta
E((1/2)(Xsub1 + Xsub2)) - theta = 0
So bias = 0.

Since Xsub1 and Xsub2 are 2 separate samples, why can I equate both to theta? Can't my point estimator change resulting in 2 different thetas?


...sorry for my lack of latex, that comes later this semester ;)
Click to expand...
A point estimate is just that, one single point to estimate something about the population you are measuring. In this particular case you have two samples; x1 and x2. The sample average, \(\displaystyle \frac{1}{2}(x_1\, +\, x_2)\) is an unbiased point estimator of the population mean [which is the expected value for the population].
 
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