# Help with proof that expected value of x_i is the populaton mean X bar

#### sid9221

##### New Member
I'm having a little trouble with the proof that the expected value of $$x_i$$ is $$\bar{X}$$.

What I have is

$$E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j)$$

Then

$$Pr(x_i=X_j) = 1/N$$

This is the bit I can't understand, how does that probability evaluate to that value.

I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.

#### BGM

##### TS Contributor

Usually in probability notation, upper case letter $$X$$ is reserved to random variable while lower case letter $$x$$ is for constant.

Please try to state the assumptions and model behind the question. What is given and what is asking for.

#### sid9221

##### New Member
$$\bar{x}$$ is the sample mean and $$\bar{X}$$ is the population mean.

I'm trying to work through the proof that $$E[\bar{x}]=\bar{X}$$

Hence, I need to evaluate $$E[x_i]$$

Which is where I get the expression I can't understand above.

#### BGM

##### TS Contributor
Do you mean

$$X_1, X_2, \ldots, X_n$$ are random variables with a common population mean $$\mu$$

You are asked to show the expected value of sample mean $$\bar{X}$$, which is defined to be

$$\bar{X} \triangleq \frac {1} {n} \sum_{i=1}^n X_i$$

is equal to the population mean $$\mu$$?

i.e.

$$E[\bar{X}] = \mu$$

BTW what is the source of that question? The notation / wordings are asked quite badly.

#### sid9221

##### New Member
Yes, the only difference is that your $$X_i$$ is my $$x_i$$ and $$\bar{X} = \mu$$

#### ANDS!

##### New Member
In these situations it's always best to write out exactly what $$x_{i}$$ actually is. In this case $$x_{i} =\frac{ \sum_{j=1}^{n} X_{ij}}{n}$$ (assuming the i is meant to indicate which cluster we're in). From there is simply a matter of distributing the expectation over the summation.