I would like to show that
$$ \bigg(e^{itc} – 1 – \frac{it c}{1+ c^2} \bigg) \frac{1+ c^2}{c^2} G(c) – G(c-\delta)$$
is the characteristic function of a poisson distribution.
This follows from the claim in Komogorov's book [Limit distributions for sum of independent random variables]
The characteristic function of a poisson $X \sim$ Poi($\lambda$)
$$\phi_{X}(t) = \Bbb{E}[e^{it X}] = \exp\{\lambda(e^{it} – 1)\}$$
If we multiply $X$ by $\alpha$ we obtain:
$$\phi_{\alpha X}(t) = \Bbb{E}[e^{it\alpha X}] = \exp\{\lambda(e^{i\alpha t} – 1)\}$$
So I guess that $\alpha = c $. I would also guess that $\lambda = \frac{1+ c^2}{c^2} G(c) – G(c-\delta)$ but that seems wrong since it misses the term $\frac{it c}{1+ c^2}$ .
Best Answer
If $X$ is a Poisson type random variable i.e. for $k\in \mathbb{N}_0$ and $\lambda>0$
$$P\{X=x_0+k\cdot h\}=\frac{\lambda^k e^{-\lambda}}{k!}$$
its characteristic function is given by
$$\varphi_X(t)=\sum_{k=0}^{\infty}e^{it(x_0+kh)}\frac{\lambda^k e^{-\lambda}}{k!}=e^{itx_0}\sum_{k=0}^{\infty}e^{it(kh)}\frac{\lambda^k e^{-\lambda}}{k!}=e^{ix_0t}e^{\lambda(e^{ith}-1)}$$