I don't know how to prove that the definition


\begin{equation}
 \lambda\_r = \frac{1}{r} \sum\_{j=0}^{r-1} (-1)^j {r - 1 \choose j} E[X\_{r-j:r}]
\end{equation}


where


\begin{equation}
 E[X\_{r:n}] = \frac{n!}{(r - 1)! \: (n - r)!} \int\_{0}^{1} x(u) \: u^{r-1} \:(1-u)^{n-r} \: du
\end{equation}


is consistent with


\begin{equation}
 \lambda\_r = \int\_{0}^{1} x(u) P^\*\_{r-1}(u) du
\end{equation}


where


\begin{equation}
 P^\*\_r(u) = \sum\_{k=0}^{r-1} p^\*\_{r-1,k} u^k,
\end{equation}


and


\begin{equation}
 p^\*\_{r,k} = (-1)^{r-k} {r \choose k} {r + k \choose k} = \frac{(-1)^{r-k} (r + k)!}{(k!)^2 (r - k)!}
\end{equation}


The indication that gives me is "Substituting the definition of the expectation of an order statistic in the first formula, expanding the binomials in $u$ and summing the coefficients of each power of $u$". The indication is from Hosking (1990) - <https://doi.org/10.1111/j.2517-6161.1990.tb01775.x>


Thanks



Consistent of the two definitions of L-moments
We have
\begin{equation}
 EX\_{r-j:r}=\frac{r!}{(r-j-1)!\,j!} \int\_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j \\ 
 =r\binom{r-1}j \int\_0^1 du\,x(u)\,u^{r-j-1}\,(1-u)^j
\end{equation}
and
\begin{equation}
 (1-u)^j=\sum\_{i=0}^j(-1)^i \binom ji u^i. 
\end{equation}
So,
\begin{equation}
 \lambda\_r=\int\_0^1 du\,x(u)\,p\_r(u),
\end{equation}
where
\begin{equation}
 p\_r(u):=\sum\_{j=0}^{r-1} (-1)^j \binom{r-1}j^2 u^{r-j-1}\sum\_{i=0}^j(-1)^i \binom ji u^i
 =\sum\_{k=0}^{r-1} p\_{r,k}u^k,
\end{equation}
\begin{equation}
 p\_{r,k}:=(-1)^{r-1-k}\sum\_{j=r-1-k}^{r-1} \binom{r-1}j^2 \binom j{r-k-1}
 =\frac{(-1)^{k+r-1} (k+r-1)!}{(k!)^2 (r-k-1)!}, 
\end{equation}
as desired.
The latter equality (which I obtained with Mathematica's help) can be obtained by using [Vol. 1 (in Russian), Formula 4.2.9.5](https://rads.stackoverflow.com/amzn/click/com/B07282HHH2):
\begin{equation}
 \sum\_{k=0}^n\binom nk^2 \binom k{n-m}=\binom nm \binom{n+m}m \tag{1}
\end{equation}
with $r-1,k$ in place of $n,m$, respectively. (This book is available [online](https://ru.b-ok2.org/book/2889120/e35933).)


Actually, it is easy to see why (1) is true. Indeed,
\begin{equation}
 \binom nk^2 \binom k{n-m}=\binom nm \binom nk \binom m{n-k}
\end{equation}
and hence
\begin{align}
 \sum\_{k=0}^n\binom nk^2 \binom k{n-m}
 & =\binom nm \sum\_{k=0}^n\binom nk \binom m{n-k}\\
 &=\binom nm \binom{n+m}n 
 =\binom nm \binom{n+m}m. 
\end{align}


