How do you show that the expected value of x-bar equals mu?

Question

I understand that the expected value of the mean of a random variable equals the "true" mean of the variable, but how do you show that with rules of expectations? I also need to show the same thing for the variance of x-bar equals (sigma^2 / n).

Answer 1

E[X] = mu = a constant depending on the distribution

x-bar = (x1 + x2 + ... + xn) / n

E[x-bar] = E[ (x1 + x2 + ... + xn) / n ]
= (1/n) (E[x1] + E[x2] + ... + E[xn])
= (1/n) (mu + mu + ... + mu)
= (1/n) (n mu)
= mu

Variance[x-bar] = E[ (x-bar)^2 ] - (E[ x-bar ])^2
= E[ (x-bar)^2 ] - mu^2 (from the above)

E[ (x-bar)^2 ] =
= E[ ((x1 + x2 + ... + xn) / n)^2 ]
= E[ (1/ n^2) (x1^2 + x2^2 + ... + xn^2 + 2 x1 x2 + 2 x1 x3 + ... + 2 x1 xn + ... + 2 x{n-1} xn) ]
= (1 / n^2)(the n E[xi^2] terms + the n(n-1)/2 E[2 xi xj] terms...)
Each E[xi^2] is E[X^2]
Each E[2 xi xj] = 2 E[xi xj] = 2 E[xi] E[xj] = 2 mu mu = 2 mu^2 where we invoke E[y z] = E[y] E[z] for ** INDEPENDENT ** y and z, and we are assuming each of the xi are independent, identically distributed.

So E[ (x-bar)^2 ] = (1 / n^2) (n E[X^2] + n(n-1)/2 (2 mu^2)) - mu^2
= (1/n) (E[X^2] + (n-1) mu^2) - mu^2
= (1/n) (E[X^2] + (n-1) mu^2 - n mu^2)
= (1/n) (E[X^2] - mu^2)
= (1/n) (E[X^2] - (E[X])^2)
= (1/n) Variance[X]

Answer 2

Expectation Of X Squared

Answer 3

E(x) = x_1*f(x_1)+x_2*f(x_2)+.....+x_n*f(x_n)

where E(x) is the expectation of x and f(x) the probability of x, i.e. P(X=x) = f(x)

When all the probabilities are equal and f(x)=1, we have the arithmetic mean (for discrete random variables):
f(x)*[x_1+x_2+...x_n]/n

The second part of your question is unclear.