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Uniform Distribution: Expectation $E(\bar{X})$

Assuming that the samples are independently and identically distributed (iid):

First i was asked to find $E(X_k)$

$E(X_k)= $$\int_{0}^{θ} dn P_θ(n) n = θ/2 $

And apparently you're supposed to get

And using this $E(\bar{X})= n/n E(x_1) = θ/2 $

Why is this?

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Knowing that $\overline{X}=\sum_{i=1}^{n} \frac{X_i}{n}$, when we take the expected value of $\overline {X}$ we get the following:

$E[\sum_{i=1}^{n} \frac{X_i}{n}]=\frac{1}{n}E[\sum_{i=1}^{n} X_i]=\frac{n}{n}E[X_1]$

The last line is true because I am assuming the samples are iid and therefore, the first sample has the same expected value as all the samples. So if take the expected value of the sum of all the samples this is equivalent to taking the expected value of one sample and multiplying it by n.

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