Consistent of the two definitions of L-moments [closed]

Question

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

Answer 1

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: \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.)

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}