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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.

Any advice ?

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  • $\begingroup$ $x_i$ is the random variable and the $X_j$'s are the possible values that it can take? $\endgroup$ Apr 5, 2014 at 18:17
  • $\begingroup$ $x_i$ is a unit in a random sample of size n, from a population $X_j$ where j=1..N $\endgroup$ Apr 5, 2014 at 18:23
  • $\begingroup$ Then, the RHS of your first expression is a weighted sum of random variables and hence a random variable itself -it cannot express an unconditional expected value (the LHS), which is a number. $\endgroup$ Apr 5, 2014 at 18:36
  • $\begingroup$ @AlecosPapadopoulos I think the $X_i$ represent potentially values from a finite population - they are not random - and $x_i$ are realizations from a simple random sample of that population. Hence $\bar X$ is a number. $\endgroup$
    – guy
    Apr 5, 2014 at 20:42
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    $\begingroup$ @guy Yes, that could be - and then $\bar X$ would be the expected value. But then, using the sample-mean notation to denote $E(X)$ is just confusing practice. $\endgroup$ Apr 5, 2014 at 21:08

2 Answers 2

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Every $X_j$ is equally likely to be selected (this isn't true for every population, but it is a common situation [for example, a die, where 1, 2, 3, 4, 5, 6 are all equally likely to be thrown]). So since there are $N$ possible values, the probability of each is ${1 \over N}.$

So you have \begin{align}\mathrm E[x_i]&=\sum_{j=1}^N X_j\,\mathrm P[x_i=X_j]\\ &={1 \over N} \sum_{j=1}^N X_j\\ &={1 \over N} (N \bar{X})\\&= \bar{X}\;.\end{align}

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If, as some of the comments have suggested, that $x_i$ is a single (discrete) random variable, and the $X_j$ are all the potential values that $x_i$ can take, then basically the definition of expectation of $x_i$ is $\mathbb{E}(x_i) := \sum_{j=1}^n X_j \mathbb{P}(x_i = X_j)$. However, in no way does this imply anything about the values that $\mathbb{P}(x_i = X_j)$ can take (other than that they are non-negative and must sum to unity over $j$)!! The probabilities $\mathbb{P}(x_i = X_j)$ define the distribution of the discrete random variable $x_i$, and since the expectation of $x_i$ is just a quantity associated with its distribution, then $\mathbb{E}(x_i)$ changes as the probabilities $\mathbb{P}(x_i = X_j)$ change.

Perhaps you're missing something from your question - maybe you've been given a question where you know that $\mathbb{E}(x_i)$ is fixed to be equal to a given value, and you're also given the values $X_j$, then it may be that you're asked to solve for all possible values of $\mathbb{P}(x_i = X_j)$?

Several comments:

1) Your use of capitals to represents values and lower-case to represent random variables is very confusing - mathematicians use the opposite convention.

2) Your use of the index $i$ is very confusing - it is not used anywhere (usefully) in your question. Come to think of it, your question mentions some combinatorial statement - perhaps you have written your question badly, in which case I urge you to rephrase your question. If this question was prompted by some other question that you're trying to solve, I would ask that original question.

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EDIT in response to comment:

Ah, if you're trying to prove that the sample mean is unbiased estimator of the population mean, then let's start with only two random variables $x_1$ and $x_2$ (using your notation). We need to prove that the expectation operator is linear i.e. $\mathbb{E}(x_1 + x_2) = \mathbb{E}(x_1) + \mathbb{E}(x_2)$ (note that if this is the case, then it's trivial to extend this to the case of there being $n$ random variables - use induction). So to prove linearity (and letting the $X_1$ and $X_2$ are the possibles values that $x_1$ and $x_2$ can take, respectively):

$\begin{align} \mathbb{E}(x_1 + x_2) &= \sum_{X_1, X_2} (X_1+X_2) \mathbb{P}(x_1 = X_1, x_2 = X_2) \\ &= \sum_{X_1} \sum_{X_2} \left[X_1 \mathbb{P}(x_1 = X_1, x_2 = X_2) + X_2 \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] \\ &= \sum_{X_1} \sum_{X_2} X_1 \mathbb{P}(x_1 = X_1, x_2 = X_2) + \sum_{X_2} \sum_{X_1} X_2 \mathbb{P}(x_1 = X_1, x_2 = X_2) \\ &= \sum_{X_1} \left[ X_1 \sum_{X_2} \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] + \sum_{X_2} \left[ X_2\sum_{X_1} \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] \\ &= \sum_{X_1}X_1\mathbb{P}(x_1 = X_1) + \sum_{X_2}X_2\mathbb{P}(x_2 = X_2)\\ &= \mathbb{E}(x_1) + \mathbb{E}(x_2) \,\,\,\,\,\text{QED} \end{align}$

From this, it should be easy to see that $\mathbb{E}(\bar{x})$ is equal to the population mean.

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  • $\begingroup$ I appreciate all the comments on notation, but it is taken exactly from Sampling Techniques, Cochran(1977) which is the 'bible' for sampling theory. Also, all I am trying to do is work through the proof that the sample mean $\bar{x}$ is an unbiased estimator of the population mean $\bar{X}$. $\endgroup$ Apr 6, 2014 at 16:35
  • $\begingroup$ Please see my edit. $\endgroup$
    – queenbee
    Apr 6, 2014 at 16:59

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