# Sample distribution and expected value.

1. Dec 10, 2015

### kidsasd987

Consider a scenario where samples are randomly selected with replacement. Suppose that the population has a probability distribution with mean µ and variance σ 2 . Each sample Xi , i = 1, 2, . . . , n will then have the same probability distribution with mean µ and variance σ 2 . Now, let us calculate the mean and variance of X_bar: E(X_bar) = 1/n*(E(X1) + E(X2) + · · · + E(Xn)) = 1/n (µ + µ + · · · + µ ) = µ

*X_i is independent random variable.

Hello. I wonder why the expected values of Xi are the same as population average µ.

Last edited: Dec 10, 2015
2. Dec 10, 2015

### BvU

Hi,

Not sure what you mean with the probability distribution of a single sample. What's that ?

3. Dec 10, 2015

### kidsasd987

I guess it means that random variable has the same probability for P(X=x), like Bernoulli random variable.

Last edited: Dec 10, 2015
4. Dec 10, 2015

### BvU

It's probably more like a short form of saying that the set of all possible individual xi has the same probability distribution as ... (because it's the same population).

Well, that is because the expression in the definition of $\mu$ and the expression for the expectation value are identical.

5. Dec 10, 2015

### kidsasd987

I am sorry. Maybe I am too dumb to understand at once. Can you help me to figure out the questions below?
(*they are not hw questions but I wrote them in statement form because It'd be easier to answer.)

1. Xi are the samples with n size.
Does that mean X1 can have n number of data within it? For example, lets say our population has a data set {1,2,3,4,5,6,7,8,9,10}
and X1 has a size of 2, then {1,2},{1,4},... on can be the sample X1.

2. (if 1 is correct) I understand why E(X)=μ, but how their samples E(X1),E(X2).. and on equal to μ.
E(X)=sigma(P(X=xi)*xi)
E(X1)=sigma(P(X1=xj)*xj) but the sum will be significantly smaller than E(X)?

Thanks.

6. Dec 11, 2015

### Amith2006

$X_i$ is not a sample. It is a random variable. We find the expectation value of that random variable defined as,
$E(X_i) = \Sigma{x_iP(x_i)} = \mu$
Hope this helps!