Given $\mu$ as the population mean and $X_{(1)}$ as the lowest value of a sample extracted from this population, I want to know if $X_{(1)}$ is an unbiased estimator for $\mu$, i.e., if $E(X_{(1)}) = \mu$
I understand that the sample mean $\overline X$ is an unbiased estimator for the population mean. However, I couldn't find anything about the lowest sample value, except this quote from Wikipedia:
Due to their sensitivity to outliers, the sample extrema cannot reliably be used as estimators unless data is clean
What I tried: Since $E(X_i) = \mu$, for every $X_i$ value of a sample, then $X_{(1)}$, as a value of the sample, should be an unbiased estimator for population mean.
Best Answer
The expected value of the minimum Will be lower than the mean. One way of seeing it, if you take a sample of a population, what are the chances that the smallest value une the sample is bigger than the mean of the population? This might happen, but it won't happen very often, and most certainly not half of the time.
Here is an example to show you. Let's $X$ be the value on a fair 6-sides dice., Then $E[X]=3.5$. Now lets throw the dice twice and have $X_{(1)}$ be the smallest of the two values. First we compute the probability for $X_{(1)}$. $$P(X_{(1)}=1)=\frac{11}{36}$$ $$P(X_{(1)}=2)=\frac{9}{36}$$ $$P(X_{(1)}=3)=\frac{7}{36}$$ $$P(X_{(1)}=4)=\frac{5}{36}$$ $$P(X_{(1)}=5)=\frac{3}{36}$$ $$P(X_{(1)}=6)=\frac{1}{36}$$ The highest value has a very low probability and the lowest one is high. We now the expected value will be less than $3.5$. $$E\left[X_{(1)}\right]=\sum_{i=1}^6P(X_{(1)}=i)=\frac{91}{36}=2.527777\ldots$$
The same is true for the highest value. With the same example, but we keep the highest value, named $X_{(2)}$. $$P(X_{(2)}=1)=\frac{1}{36}$$ $$P(X_{(2)}=2)=\frac{3}{36}$$ $$P(X_{(2)}=3)=\frac{5}{36}$$ $$P(X_{(2)}=4)=\frac{7}{36}$$ $$P(X_{(2)}=5)=\frac{9}{36}$$ $$P(X_{(2)}=6)=\frac{11}{36}$$ The highest value has a very high probability and the lowest one is low. We now the expected value will be more than $3.5$. $$E\left[X_{(2)}\right]=\sum_{i=1}^6P(X_{(2)}=i)=\frac{161}{36}=4.472222\ldots$$