Vrftsjtryk

In a example about U-statistics, $h(x_1,x_2)=frac 12(x_1-x_2)^2$, then
$$U_n=frac{2}{n(n-1)}sum_{i<j}frac{(X_i-X_j)^2}{2}=frac{1}{n-1}sum_{i=1}^{n}(X_i-bar{X})^2$$
I don't know how to prove it completely.

Hint 1 :
$sum_{i<j}{(X_i-X_j)^2} = frac{1}{2}sum_{i}sum_{j}(X_i-X_j)^2$

Hint 2 : Add and subtract $bar{X} $ to simplify sum of squares.

You will arrive at your result.

We know that (I found it here)
begin{equation}
left( sum_{n=1}^N a_n right)^2 = sum_{n=1}^N a_n^2 + 2 sum_{j=1}^{N}sum_{i=1}^{j-1} a_i a_j
end{equation}
So using the above identity

begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
sum_{i=1}^{n}(X_i-frac{1}{n}sum_{j=1}^nX_j)^2\
&=
sum_{i=1}^{n}(X_i^2-frac{2}{n}X_isum_{j=1}^nX_j + frac{1}{n^2}(sum_{j=1}^nX_j)^2 )\
&=
sum_{i=1}^{n}(X_i^2-frac{2}{n}X_isum_{j=1}^nX_j + frac{1}{n^2}(sum_{j=1}^nX_j^2 + 2sum_{j=1}^nsum_{k=1}^{j-1}X_jX_k) )
end{align}
The last term above is independent of $i$ so it sums up $n$ times as
begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
sum_{i=1}^{n}(X_i^2-frac{2}{n}X_isum_{j=1}^nX_j) + frac{n}{n^2}(sum_{j=1}^nX_j^2 + 2sum_{j=1}^nsum_{k=1}^{j-1}X_jX_k)
end{align}
which is also
begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
sum_{i=1}^{n}(X_i^2-frac{2}{n}X_isum_{j=1}^nX_j) + frac{1}{n}(sum_{j=1}^nX_j^2 + 2sum_{j=1}^nsum_{k=1}^{j-1}X_jX_k)
end{align}
which could also be written as
begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
(1 + frac{1}{n})
sum_{i=1}^{n}X_i^2-frac{2}{n}sum_{i=1}^{n}X_isum_{j=1}^nX_j) + frac{1}{n}( 2sum_{j=1}^nsum_{k=1}^{j-1}X_jX_k)
end{align}
Rewriting differently we have
begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
(1 + frac{1}{n})
sum_{i=1}^{n}X_i^2-frac{2}{n}sum_{i,j}X_iX_j + frac{2}{n}sum_{i<j}X_iX_j
end{align}
The last two terms above are the same terms with missing terms. Notice that $sum_{i,j}X_iX_j$ spans all $i = 1 ldots n$ and $j = 1 ldots n$ but the other one spans an upper triangular version of it. This means that their difference will span the lower triangular version of it as
begin{align}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
(1 + frac{1}{n})
sum_{i=1}^{n}X_i^2 - frac{2}{n}sum_{igeq j}X_iX_j
end{align}
Factor $n$ on the right hand side, then divide by $n-1$ on both sides, then Multiply/divide by $2$ on the right hand side
begin{align}
frac{1}{n-1}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
frac{2}{n(n-1)}
Big(
frac{(n + 1)
sum_{i=1}^{n}X_i^2 - 2sum_{igeq j}X_iX_j}{2}
Big)
end{align}
Notice that $i geq j$ could be split to two summations
begin{align}
frac{1}{n-1}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
frac{2}{n(n-1)}
Big(
frac{(n + 1)
sum_{i=1}^{n}X_i^2 - 2sum_{i = j}X_iX_j - 2sum_{i > j}X_iX_j}{2}
Big)
end{align}
but when $i = j$, it is the same as a single summation, hence
begin{align}
frac{1}{n-1}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
frac{2}{n(n-1)}
Big(
frac{(n + 1)
sum_{i=1}^{n}X_i^2 - 2sum_{i=1}^n X_i^2 - 2sum_{i > j}X_iX_j}{2}
Big)
end{align}
which gives
begin{align}
frac{1}{n-1}
sum_{i=1}^{n}(X_i-bar{X})^2
&=
frac{2}{n(n-1)}
Big(
frac{(n -1)
sum_{i=1}^{n}X_i^2- 2sum_{i > j}X_iX_j}{2}
Big)
end{align}
The numerator above is nothing other than $sum_{i<j} (X_i - X_j)^2 = sum_{i<j} X_i^2 - 2 sum_{i<j} X_iX_j + sum_{i<j} X_j^2$. It is easy to see the cross terms, however it is not as straightforward to see that we have $n-1$ terms of the form $X_i^2$. This should conclude
begin{align}
frac{1}{n-1}sum_{i=1}^{n}(X_i-bar{X})^2
=
frac{2}{n(n-1)}sum_{i<j}frac{(X_i-X_j)^2}{2}
end{align}

A one-line proof summary:$$sum_{i<j}(X_i-X_j)^2=frac{1}{2}sum_{ij}(X_i-X_j)^2=nsum_iX_i^2-sum_{ij}X_iX_j=nsum_i X_i(X_i-overline{X})=nsum_i(X_i-overline{X})^2.$$The first $=$ uses the fact that $(X_i-X_j)^2$ is $ileftrightarrow j$-symmetric and $0$ if $i=j$. The second $=$ expands the square and separates squares from cross terms. The third $=$ is a trivial rearrangement. The last $=$ uses $$X_i(X_i-overline{X})-(X_i-overline{X})^2=overline{X}(X_i-overline{X}),$$which becomes $0$ under $sum_i$.