pandaBee said:
Homework Statement
In my notes I keep stumbling upon this equation:
Equation 1: E(X-bar^2) = (σ^2)/n + μ
I was wondering why the above equation is true and how it is derived.The Attempt at a Solution
E(X-bar^2)
##Summations are from i/j=1 to n
= E[(Σx_i/n)^2)]
= E[(Σx_i/n)(Σx_j/n)]
=(1/n^2)E[(Σx_i)(Σx_j)]
=(1/n^2)ΣΣE(x_i*x_j)
=E(x_1^2) + E( x_2^2 + ... + E(x_n^2)
+ E(x_1*x_2) + E(x_1*x_3) + ... : Equation 2
I'm stuck at this point. Could someone show me how you get to Equation 1 from Equation 2? Assuming that my derivation to Equation 2 is correct.
If you mean
E \overline{X}^2 = \frac{\sigma^2}{n} + \mu
then that cannot possibly be right! For one thing, the "dimensions" don't match. For example, if ##X## is measured in meters, then ##\sigma^2## is measured in ##\text{meters}^2##, while ##\mu## is in meters.
The easiest approach is to recognize that the mean square and variance of any random variable ##Y## are connected by the equation
\text{Var}(Y) = E(Y^2) - (E Y)^2, \; \text{or} \; E(Y^2) = \text{Var}(Y) + (EY)^2
Apply this to ##Y = \bar{X} = \sum X_i / n##. Assuming the ##X_i## are independent and identically distributed there are easy and well-known formulas for ##EY## and ##\text{Var} (Y)## in terms of ##\mu##, ##\sigma## and ##n##. That will tell you the value of ##E \overline{X}^2##.
On the other hand, if you meant that you want to know
E \overline{X^2} = E \sum X_i^2 / n,
then you have a different set of formulas to use. However, you can approach it again using the relationship between mean square and variance.
