I consider $N_t$ an inhomogeneous Poisson process, especially $N_t$ follows a Poisson law with parameter $\int_{0}^{t}\lambda(s)ds$.
We assume that $\frac{1}{t}\int_{0}^{t}\lambda(s)ds\to \sigma^{2}$ when t goes to infinity.
The aim is to prove that
$$ \sqrt{t}\left(\frac{N_t}{t} - \frac{1}{t}\int_{0}^{t}\lambda(s)ds\right)\to N(0,\sigma^2) $$
Where the convergence take place in law.
I am stuck when I am trying to find the limit of the characteristic function.
Here is my attempt in order to make things clear.
Consider
$$X_t = \sqrt{t}\left(\frac{N_t}{t} - \frac{1}{t}\int_{0}^{t}\lambda(s)ds\right) =\frac{1}{\sqrt{t}}\left(N_t -\int_{0}^{t}\lambda(s)ds\right) $$
The law of $X_t$ is a Poisson with parameter $\lambda(t) = \int_{0}^{t}\lambda(s)ds$ (I don't put the calculations but I could provide them if you think they are the cause of the problem).
Then we have
$$ \varphi_{X_t}(x) = e^{\lambda(t)(e^{ix} - 1)} = e^{\lambda(t)( ix - \frac{x^{2}}{2} + o(x^2) )} $$
But from there I cannot continue and I have the feeling there is something wrong in this characteristic function.
If you have already encounter this theorem or have an idea on what is/are my mistake do not hesitate please.
Thank you
The inhomogeneous Poisson process is s.t. $N_t\sim \textrm{Poisson}(\Lambda_t)$ where $\Lambda_t:=\int_0^t\lambda_udu$ is the (deterministic) intensity process. We define $Z_t:=t^{-1/2}(N_t-\Lambda_t)$. We get: $$\begin{aligned}E[e^{i\xi Z_t}]&=e^{-i(t^{-1/2}\xi)\Lambda_t}E[e^{i(t^{-1/2}\xi)N_t}]\\ &=e^{-i(t^{-1/2}\xi)\Lambda_t}e^{\Lambda_t(e^{i(t^{-1/2}\xi)}-1)}\\ &=e^{(\Lambda_t/t)U(\xi,t)} \end{aligned}$$ where $$U(\xi,t)=t(e^{i\xi/t^{1/2}}-1-i\xi/t^{1/2})\stackrel{t\to \infty}{\to}-\frac{\xi^2}{2}$$ so we conclude with usual rules of limits: $E[e^{i\xi Z_t}]\to e^{-\sigma^2\xi^2/2}$.