Let \(P=\{P(k)\), \(k=0,1,...\}\) be a probability distribution on \({\mathbb{Z}}_+=\{0,1,...\}\), and let \(B_{n,p}(k)=\left( \begin{matrix} n\\ k\end{matrix} \right)p^ k(1-p)^{n-k},\) \(0\leq k\leq n\), \(0\leq p\leq 1\). Write \[ P\circ B_ p(k)=\sum^{\infty}_{n=k}P(n)B_{n,k}(k),\quad k=0,1,.... \] A distance between \(P\circ B_ p(k)\) and a Poisson distribution \(\Pi_{\lambda}=\{\Pi_{\lambda}(k)=(\lambda^ k/k!)e^{-\lambda},\quad k=0,1,...\}\) is estimated.
For a function F: \({\mathbb{Z}}=\{0,\pm 1,\pm 2,...\}\to {\mathbb{R}}\) and probability distributions \(P_ 1\) and \(P_ 2\) on \({\mathbb{Z}}\) we define \[ D_ s(P_ 1,P_ 2)=\sup_{f\in {\mathcal F}_ s}| \sum^{\infty}_{k=0}f(k)(P_ 1(k)-P_ 2(k))|, \] where \({\mathcal F}_ s=\{f:| f'(k+i)-f'(k)| <i^{s-1}\), \(i\in {\mathbb{Z}}_+\), \(k\in {\mathbb{Z}}\}\), \(1<s\leq 2\), with \(f'(k)=f(k+1)-f(k)\), and \[ d_ s(P_ 1,P_ 2)=\sum^{\infty}_{k=0}k^ s| P_ 1(k)-P_ 2(k)|,\quad 1\leq s\leq 2,\quad d^*_ 2(P_ 1,P_ 2)=\sum^{\infty}_{k=0}k(k-1)| P_ 1(k)-P_ 2(k)|. \] Theorem. Let \(\sum^{\infty}_{k=0}kP(k)=\lambda\). Then \[ D_ s(P\circ B_ p,\Pi_{\lambda p})\leq p^ sd_ s(P,\Pi_{\lambda}),\quad 1<s<2;\quad D_ 2(P\circ B_ p,\Pi_{\lambda p})\leq (p^ 2/2)d^*_ 2(P,\Pi_{\lambda}). \]