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Let $\pi_K$ be Poisson random measure with intensity $\mu = K\lambda$, where $\lambda$ is Lebesgue measure. I have a task to do and the first part was to show that for $f \in L^1({\mathbb{R}^d}) \cap L^3(\mathbb{R}^d$) (shouldn't $3$ be $2$ actually?) $$\frac{1}{\sqrt{K}}\int_{\mathbb{R}^d}f(x)\bar{\pi}_K(dx)$$ converges in distribution as $K \to \infty$, where $\bar{\pi}_K = \pi_K - \mu$.

I've looked at Laplace transform and got that the limit is gaussian random variable with mean $0$ and variance $\sigma^2 = \int f^2(x)dx$ (if it is correct...).

Now the second part is to look at $X_K = \frac{1}{\sqrt{K}}\left(\pi_K\left([0,t]\right)-Kt\right)$ and find the limit of finite dimentional distributions, so the vector $\left(X_K\left(t_1\right),\ldots,X_K\left(t_n\right)\right)$.

I'm not really sure how to start. I see that $X_K$ is my previous random variable with $f = \chi_{[0,t]}$, so I guess the limit should be some gaussian vector, but I cannot show it.

Barabara
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It seems that we need $f$ integrable in order to make the integral $\int_{\mathbb{R}^d}f(x)\bar{\pi}_K(dx)$ well defined.

For the convergence of the finite-dimensional distribution, we could establish it for $\left(X_K(t_1),X_K(t_2)-X_K(t_1),\dots,X_K(t_n)-X_K(t_{n-1})\right)$ by using the Cramer-Wold device: we are reduced to investigate the convergence of $$ \frac 1{\sqrt K}\sum_{i=1}^nc_j\left(\pi_K\left([t_{i-1},t_i]-K\left(t_{i}-t_{i-1}\right)\right)\right)$$ for all $c_1,\dots,c_n\in\mathbb R$.

Davide Giraudo
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  • Isn't it actually $\pi_K\left((t_{i-1}, t_i]\right)$ under this sum? Then I think that we can again look at laplace transform and use the fact those this random variables are independent, because sets measured by Poisson random measure are disjoint. Am I right? – Barabara Oct 24 '19 at 20:37
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    @Barabara Exactly, this is the idea. Is actually $X_K = \frac{1}{\sqrt{K}}\left(\pi_K\left((0,t]\right)-Kt\right)$? – Davide Giraudo Oct 24 '19 at 20:40
  • no, its [0,t], but maybe its the mistake – Barabara Oct 25 '19 at 11:58
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    Or the contribution of $\pi_K({t})$ is negligible. – Davide Giraudo Oct 25 '19 at 12:24