How to I find the Infinitesimal generator of the following Poisson Process with drift I have a Poisson Process Y with parameter $\lambda$ and I want to find the sequence of infinitesimal generators of the following sequence of processes: 
$$ X_n(t) = \frac{1}{n} \left[ Y(n^2 t)-\lambda n^2 t \right] $$ 
Then the question asks me to identify the limit of the sequence $(A_n)$ as $n \to \infty$ with the corresponding stochastic process. 
My attempt: 
Assuming the poison process starts from a point $x$ then the nth infinitesimal generator is: 
$$A_{n} f(x) = \lim_{t \to 0} \frac{\mathbb{E}^x [f(X_n(t))]-f(x)}{t} \quad (\dagger)$$ 
So my idea was to condition on the number of jumps before time t and show that the conditional expectation of $f(X_n(t))$ given 2 or more jumps is order $O(t^2)$ as $t \to 0$ so that it vanishes in $(\dagger)$ above. That is: 
$$\mathbb{P}(Y(n^2 t) \leq 1) = (1+\lambda n^2 t)\exp(-\lambda n^2 t) =(1+\lambda n^2 t)(1-\lambda n^2 t +O(t^4))=1+O(t^2)$$
$$\therefore \mathbb{P}(Y(n^2 t) \geq 2) = 1-\mathbb{P}(Y(n^2 t) \leq 1) = O(t^2)$$ 
But then when calculating the expectations conditional on 0 or 1 jumps of $Y(n^2 t)$ I keep getting answers which diverge to infinity as n does which contradicts my intuition that the process converges (in  distribution I think) to a time-change Brownian motion $B_{\lambda t}$. I used the central limit theorem (with $Y(n^2 t) = \sum_{k=1}^n Y_k(nt)$ for i.i.d poisson processes) to gain this insight and also sketched what the process $(X_n(t))$ might look like). In which case I expect the limiting infintesimal generator to be $Af(x) = f''(x)/2$
Me and a friend have spent several hours on this so would really appreciate some help here
 A: To simplify calculations, I recall that $\mathbb{P}(Y(n^2t) - Y(0) = 0) = 1 - \lambda n^2t + o(t)$ and $\mathbb{P}(Y(n^2 t) - Y(1) = 1) = \lambda n^2 t + o(t)$. 
Hence, writing $U_{n,t} = Y(n^2t) - Y(0)$, we have that
\begin{align*}
A_nf(x) =& \lim_{t \to 0} \frac{\mathbb{E}^x[f(X_n(t)) - f(x) | U_{n,t} = 0]\mathbb{P}(U_{n,t} = 0) + \mathbb{E}^x[f(X_n(t)) - f(x) |U_{n,t} = 1]\mathbb{P}(U_{n,t} =1)}{t}  
\\ =& \lim_{t \to 0} \frac{[f(x - \lambda nt) - f(x)] (1- \lambda n^2 t) + [f(x + n^{-1} - \lambda nt) - f(x)] \lambda n^2 t}{t}
\\ =& \lim_{t \to 0} \bigg(- \lambda n \frac{ f(x - \lambda n t) - f(x)}{-\lambda n t} + \lambda n^2 [f(x + n^{-1} - \lambda n t) - f(x)]\bigg)
\\=& \lambda n^2 \big ( f(x + n^{-1}) - f(x) - n^{-1} f'(x)\big)
\end{align*}
From here, computing the limiting generator is just one of the usual applications of Taylor's theorem. You can write
$$
\lambda n^2\big(f(x+n^{-1}) - f(x) - n^{-1} f'(x)\big) =  \frac{\lambda}{2} f''(\xi_{x,n})
$$
where $\xi_{x,n} \in [x, x+n^{-1})$ by Taylor's theorem. Hence, by sandwiching, as $n \to \infty$ you get that $$A_nf(x) = \frac{\lambda}{2}f''(\xi_{x,n}) \to \frac{\lambda}{2}f''(x)$$
so that the limiting generator is that of a Brownian motion run at rate $\lambda$.
