If I have a [compound Poisson process](http://en.wikipedia.org/wiki/Compound_Poisson_process) 
$$Y(t) = \sum\_{i=1}{N(t)}D\_{i}$$ 
where $ \{\,N(t) : t \geq 0\,\}$ is a Poisson process with rate $\lambda$, and $ \{\,D\_i : i \geq 1\,\}$ are i.i.d random variables with distribution function $G$, which are also independent of $ \{\,N(t) : t \geq 0\,\}.\,$


Then Wikipedia informs me that it has the following properties 
$$E[Y(t)]=\lambda t E[D]$$
$$Var[Y(t)]=\lambda t E[D^{2}]$$


which can be shown using Wald's equation. As an aside does anyone have a better source that they would refer me to read about this?


Then if I have a compound Poisson process occurring but I chop it up into $N$ segments of length $\bar{t}$ then I think I have a sequence $\{Y\_{1}(\bar{t}), Y\_{2}(\bar{t}), Y\_{3}(\bar{t}), \dots\}$ where each is it's own compound Poisson process.


Then I think $Y\_{i}$ are i.i.d, but does this mean I can use the central limit theorem to say that: provided $\lambda$ is sufficiently large the distribution of $Y\_{i}$ should be normally distributed. Any comments appreciated.



Compound Poisson process and central limit theorem
I think the following shows what you're after. Similar problems/exercises can be found in many texts on probability theory (I guess it's more common in textbooks to consider the limit as time grows to infinity).


By the stationary and independent increments of the Poisson process, together with the independence of the $D\_i$'s (amongst themselves as well as with $N$), each $Y\_i$ is an independent copy of the random variable
$$Z = \sum \_{i=1} ^{N(\bar t)} D\_i,$$
where $\bar t >0$ is fixed. Wald's equation (rather, the relations in the original post that follow from Wald's eq.) suggests that for a CLT we want to consider
$$ \tilde Z = \frac{Z - \lambda \bar t E[D\_1] }{\sqrt{\lambda \bar t E[D\_1 ^2]}}. $$
In order to show convergence to the normal distribution we can use MGF's. Note that the MGF of $Z$ is
$$ \Psi \_{Z} (\theta) = E [e^{\theta Z}] = e^{\lambda \bar{t} (\Psi \_{D}(\theta) -1)}.$$
It follows that
$$ \Psi \_{\tilde Z} = \textrm{exp}\left\{ \lambda \bar t (\Psi \_D(\theta / \sqrt{\lambda \bar t E[D\_1 ^2]}) - 1) - (\lambda \bar t \theta)/ \sqrt{\lambda \bar t E[D\_1 ^2]} \right\}.$$
Under the assumption that the distribution $G$ has a nice enough MGF, Taylor expansion as $\lambda \to \infty$ gives (with some simplifications)
$$ \Psi \_D(\theta / \sqrt{\lambda \bar t E[D\_1 ^2]}) = 1 + \frac{\theta}{\sqrt{\lambda \bar t E[D\_1 ^2]}}E[D\_1] + \frac{\theta ^2}{2 \lambda \bar t} + o(1/\lambda), \ \textrm{as } \lambda \to \infty. $$ 
Inserting this into the expression for the MGF for $\tilde Z$,
$$ \Psi \_{\tilde Z} (\theta) = \textrm{exp}\left\{ \frac{\theta ^2}{2} + o(1) \right\}, $$
where the $o$-term is with respect to $\lambda \to \infty$. Hence, 
$$ \Psi \_{\tilde Z} (\theta) \to \textrm{exp}\{\theta^2 / 2\}, \ \lambda \to \infty,$$
and it follows that $\tilde Z$ converges in distribution to a $N(0,1)$-distributed random variable. 


Note that this does not use the independence of the $Y\_i$'s. Rather, we just consider the compound Poisson process at a fixed time, i.e., you might as well consider $Y(t)$ for a fixed $t$ and send $\lambda$ to infinity. Therefore, I'm not quite sure if this is really what you are after. Still, hopefully it's helpful in some way.


