A question on a compound Poisson process: $P(|\xi(t)|\leqslant0.3)\underset{t\rightarrow\infty}{\rightarrow}0$? Let $\xi(t)$ be a compound Poisson point process such that the number of summands is a Poisson point process with parameter $\lambda$ and the summands are random variables: $P(\xi_k=\pm1)=0.5.$ Is it true that $P(|\xi(t)|\leqslant0.3)\underset{t\rightarrow\infty}{\rightarrow}0$?
I know a central limit theorem:

Theorem. Let $\{\mu_k\}_k$ be a sequence of independently equally distributed random variambles, $\mu_k>0$, $N(t)=\max\{n|\sum\limits_{k=1}^n\leqslant t\}$, $a=E\mu_1$, $\sigma^2=D\mu_1$ ($D$ denotes variance) then $P\Bigl(\cfrac{N(t)-\frac{t}{a}}{\sigma a^{-\frac{3}{2}}\sqrt{t}}<x\Bigl)\underset{t\rightarrow\infty}{\rightarrow}\Phi(x)$ where $\Phi(x)$ is the distribution function of $N(0,1).$

I suppose it must be the key but I am not good at probability theory at all. I don't really know how to apply this theorem (and, furthermore, if this is the right way). Can you please help me?
 A: We can write $\xi(t)$ as
$$\xi(t)=\sum_{k=1}^{N(t)} \xi_k$$
where $\{N(t)\}_{t\in\mathbb{R}}$ is a counting Poisson process with rate $\lambda$. It is not specified in your question but I assume that $\lambda>0$ otherwise we would have $N(t)=0$ for all $t$ and thus $\xi(t)=0$ which contradicts what we are trying to prove. I am also going to assume that the $\xi_k$ random variables are independent from each other and from $N(t)$ because I'll need this assumption later and I couldn't see how to solve the problem without them.
The main idea is that, since $\xi_k\in\{-1,+1\}$ for all $k$ then $\xi(t)$ is always an integer hence $\mathbb{P}(|\xi(t)|\leq 0.3)=\mathbb{P}(\xi(t)=0)$ and the only way for $\xi(t)$ to be equal to $0$ is if there is the same number of $\xi_k$ in the sum that is equal to $1$ than there is equal to $-1$. Since there is $N(t)$ random variables in the sum then this is equivalent to $N(t)$ being even and that exactly $N(t)/2$ of the $\xi_k$ for $k\in\{1,\cdots,N(t)\}$ are equal to $1$.
Now, by the law of total probability,
$$\mathbb{P}(\xi(t)=0)=\sum_{n=0}^{\infty}\mathbb{P}(\xi(t)=0|N(t)=n)\mathbb{P}(N(t)=n).$$
As a consequence of $\{N(t)\}_{t\in\mathbb{R}}$ being a counting Poisson process, then for a given $t\in\mathbb{R}$ we have that $N(t)$ follows a Poisson distribution with parameter $\lambda t$ so
$$\mathbb{P}(N(t)=n)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}.$$
We already saw that $\xi(t)\neq 0$ if $N(t)$ is not even hence
$$\forall p\in\mathbb{N},~\mathbb{P}(\xi(t)=0|N(t)=2p+1)=0$$
If $n$ is even, let's say $n=2p$ with $p\in\mathbb{N}$, then the number of $\xi_k$ for $k\in\{1,\cdots,2p\}$ equal to $1$ follows a binomial distribution with parameters $2p$ and $1/2$ hence the probability that exactly $p$ of them are equal to $1$ is $\binom{2p}{p}\left(\frac{1}{2}\right)^{2p}$. Therefore,
$$\forall p\in\mathbb{N},~\mathbb{P}(\xi(t)=0|N(t)=2p)=\binom{2p}{p}\left(\frac{1}{2}\right)^{2p}.$$
In the end, we obtain
$$\mathbb{P}(\xi(t)=0)=\sum_{p=0}^{\infty}\binom{2p}{p}\left(\frac{1}{2}\right)^{2p}\frac{(\lambda t)^{2p} e^{-\lambda t}}{(2p)!}=e^{-\lambda t}\sum_{p=0}^{\infty}\frac{(\lambda t)^{2p}}{2^{2p}(p!)^2}=e^{-\lambda t}I_0(\lambda t),$$
where $I_0$ is the modified Bessel function of the first kind. Using the asymptotic
$$I_0(\lambda t)\substack{\sim \\ t\rightarrow \infty}\frac{e^{\lambda t}}{\sqrt{2\pi\lambda t}}$$
(see the wikipedia page for Bessel functions for example https://en.wikipedia.org/wiki/Bessel_function) we can conclude that $\mathbb{P}(\xi(t)=0)\substack{\longrightarrow \\ t\rightarrow \infty}0$ and thus $\mathbb{P}(|\xi(t)|\leq 0.3)\substack{\longrightarrow \\ t\rightarrow \infty}0$
