Show that Exbarmu using the fact that expected value is a li

Solution

Let x1,x2,--,xn  be a random sample of size n from a distribution (population) with mean mu and variance sigma^2.........
Now, x-bar=(x1+x2+....+xn)/n ; E(x-bar) =E( (x1+....+xn) / n ) ;
Then, using the linear operator property of expectation, we get, E(x-bar)= [ E(x1)+E(x2)+..E(xn) ] / n ;

Now, the xi\'s are identically distributed, which means they have the same mean mu.Therefore, replacing E(xi) with the alternative notation mu, we get: E(x-bar)= [mu+mu+.....+mu] /n
Now, because there are n mu\'s in the above formula,we can rewrite the expected value as-

E(x-bar)= n*mu /n = mu (Hence proved)