equation on expectation I am reading a note on the probability theory and they are verifying that the sample mean $\bar y=\frac 1n \sum_{i=1}^n y_i$ is an unbiased estimator of the population mean $\bar Y$ by taking expectation
$$E(\bar y)=\frac 1n \frac{1}{{N \choose n}}\sum_{i=1}^{N \choose n}\sum_{i=1}^ny_i.
$$
I do not understand how they have gotten the following relation:
$$
\sum_{i=1}^{N \choose n}\sum_{i=1}^ny_i={N-1 \choose n-1}\sum_{i=1}^N y_i.
$$
 A: The sum
$$
E(\bar y)=\frac 1n \frac{1}{\binom Nn}\sum_{i=1}^{\binom Nn}\sum_{i=1}^ny_i.
$$
appears to be using the indices of the sums in a non-standard way with an intended meaning that cannot be deduced from the literal formula alone, even if you change one of the index variables from $i$ to $j.$
What I think they have in mind is that if the population has $N$ members,
there are $\binom Nn$ subsets of the population that could be selected as a sample of size $n.$
The summation $\sum_{i=1}^{\binom Nn}$ is meant to iterate through that list of subsets, visiting each subset exactly once.
Moreover, each subset is equally likely to be selected, so it occurs with probability $1/\binom Nn.$
Now within each subset of $n$ members of the population, we can arbitrarily label the members of the subset as "member number $1$," "member number $2$,"
up to "member number $n$," and then we can write the observed $y$-values of those members as $y_1, y_2,\ldots y_n$ respectively.
Then the sample mean for that particular subset is 
$\frac 1n \sum_{i=1}^ny_i.$
Notice that this means $y_1$ in one of the $\binom Nn$ samples is not necessarily the same as $y_1$ in one of the other samples. In fact, if $N > n$ and each member of the population has a different $y$-value, it is impossible for $y_1$ to have the same value in each subset.
So I'm confident we are absolutely not meant to take the formula for this sum literally.
A more accurate mathematical notation might be to number each of the members of the population from $1$ to $N$ and assign $y_i$ to the $y$-value of member number $i.$
Number each subset of $n$ members with a number from $1$ to $\binom Nn$
and let $\kappa_j$ be the mapping of the set $\{1,\ldots,n\}$ to the set of indices that tells which members of the population belong to subset number $j.$
For example, if the $j$th subset contains members numbered $k_1,$ $k_2,\ldots, k_n$
with $y$-values $y_{k_1},$ $y_{k_2},\ldots, y_{k_n}$ respectively,
then $\kappa_j(i) = k_i.$
Then the first sum can be written
$$
E(\bar y)=\frac 1n \frac{1}{\binom Nn}
  \sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)}. \tag1
$$
But now let's consider the original "member numbers" of the members of the population, according to which there is a $y$-value $y_m$ for each integer $m$ such that $1\leq m\leq N.$
How many times does that particular subscript of $y$ occur in the summation
in Equation $(1),$ summed over all possible values of $j$ and all possible values of $i$?
Given any particular index $m$ whose occurrences we want to count,
it should be easy to see that for any $j$ there can be at most one $i$ such that $\kappa_j(i) = m.$
Moreover, there will be one such $i$ only if $m$ is the member number of one of the members of the population that belongs to subset number $j.$
That is, $y_m$ will occur in the sum precisely once for each $n$-member subset of the population that includes member number $m.$
The number of occurrences of $y_m$ is the number of $n$-member subsets that contain member $m$ of the population.
How many $n$-member subsets contain the $m$th member of the population?
To form such a subset, we must choose exactly $n-1$ of the remaining $N-1$ members of the population. We can do this in $\binom{N-1}{n-1}$ ways.
So that's how many subsets there are. Summing $y_m$ across all its occurences,
it contributes $\binom{N-1}{n-1} y_m$ to the total sum.
If we consider this for each $m$ from $1$ to $N,$ we find that the double summation in Equation $(1)$ can be rewritten:
$$
\sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)} = \sum_{m=1}^N \binom{N-1}{n-1} y_m.
$$
Since $i$ does not occur on the right-hand side, we can change the index variable there from $m$ to $i$:
$$
\sum_{j=1}^{\binom Nn}\sum_{i=1}^n y_{\kappa_j(i)} = \sum_{i=1}^N \binom{N-1}{n-1} y_i.
$$
And there you have the desired result.
A: To prove this
$$
\sum_{i=1}^{N \choose n}\sum_{i=1}^ny_i={N-1 \choose n-1}\sum_{i=1}^N y_i.
$$
notice that
$$\sum_{i=1}^ny_i$$
Is a constant.
So it would be enough to show that
$$
1+2+\cdots+{N \choose n}={N-1 \choose n-1}.
$$
However, this is not true. As an example, for $N=2$ and $n=1$, ${1 \choose 0}=1$,  ${2 \choose 1}=2$,
and $1+2=3.$
