# The expected value of the square of the sample mean?

1. Aug 23, 2015

### pandaBee

1. The problem statement, all variables and given/known data
In my notes I keep stumbling upon this equation:

Equation 1: E(X-bar^2) = (σ^2)/n + μ^2

I was wondering why the above equation is true and how it is derived.

3. 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. Last edited: Aug 23, 2015 2. Aug 23, 2015 ### Ray Vickson 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.

Last edited: Aug 23, 2015
3. Aug 23, 2015

### SteamKing

Staff Emeritus
I think your Equation 1 may contain a slight error, as Ray pointed out

I think the correct equation is E(x-bar)2 = (σ2 / n) + μ2

4. Aug 23, 2015

### pandaBee

Yes, you are both correct, the correct equation was actually

Equation 1: E(X-bar^2) = (σ^2)/n + μ^2
The mu had to be squared.

In fact, I can see that I have made a huge blunder in how I was looking at this problem. Thank you both for the insight!