# Is sample minimum an unbiased estimator for population mean?

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.

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 $$3.5$$. $$E\left[X_{(2)}\right]=\sum_{i=1}^6P(X_{(2)}=i)=\frac{161}{36}=4.472222\ldots$$