Large deviation results for a \(U\)-statistical sum with product kernel (Q2716971)

A large deviation result is proved in two forms (under Cramér or Linnik condition) for non-degenerate \(U\)-statistical sum NEWLINE\[NEWLINEU_n={n\choose m}^{-1}\sum_{1\leq i_1<\cdots<i_m\leq n}X_{i_1}\cdots X_{i_m},NEWLINE\]NEWLINE of degree \(m\geq 1\) with kernel \(h(x_1,\dots,x_m)=x_1\cdots x_m,\) where \(X_1,\dots, X_n\) are i.i.d. r.v. with \(EX_1=\mu\neq 0\), \(0<\sigma^2= E(X_1-\mu)^2<\infty.\) The method of proof uses truncation and the contraction technique. The above-mentioned result is an asymptotic relation of the type NEWLINE\[NEWLINE(1-F_n(x))/(1-\Phi(x))= \exp \Biggl\{\frac {x^3}{\sqrt{n}}\lambda_m \biggl(\frac x{\sqrt{n}} \biggr)\Biggr\} \Biggl(1+ O\biggl(\frac {1+x}{\sqrt{n}} \biggr)\Biggr)\tag{1}NEWLINE\]NEWLINE where \(F_n(x)=P(\sqrt{n}(U_n-\mu^m)/(m\sigma|\mu|^{m-1})\leq x)\), \(x\in R\), \(\Phi(x)\) denotes the standard normal distribution function, and \(\lambda_m(u)=\sum_0^{\infty} \lambda_{km}u^k\) represent the so-called Cramér series. It is proved NEWLINENEWLINENEWLINE1) for \(x=o(\sqrt{n})\), \(n\to \infty,\) and under Cramér condition \(E\exp (a|X_1|)<\infty\) for some \(a>0\) and alsoNEWLINENEWLINENEWLINE2) for \(x=o(n^{\alpha})\), \(n\to \infty\), \(0<\alpha<1/2,\) and under Linnik condition \(E\exp (a|X_1|^{4\alpha/(1-2\alpha)})<\infty\) for some \(a>0\); in the last case \(\lambda_m(u)\) must be replaced by truncated Cramér series \(\lambda_m^{[s]}(u)=\lambda_{0m}+\lambda_{1m}u+\cdots+\lambda_{sm}u^s\), with \(s=[4\alpha/(1-2\alpha)]\) \(([b]\) denotes the integer part of \(b).\)NEWLINENEWLINENEWLINEIt must also be mentioned that all existing large deviation results for \(U\)-statistics with \(x\) in this range require the kernels to be bounded and so they do not apply here.