Consider $N,M$ two independent Poisson process with rate $\lambda$, and define $$X(t):=x+\dfrac{1}{\sqrt{n}}[N(t)-M(t)].$$ From this formula we know that $X(0)=x$. Now, I want to computed the conditional probability $$\mathbb{E}[f(X(t))|X(0)=x].\ \ \ (1)$$ I know that for a single Poisson process, it's transition probability is $$\mathbb{P}(i,t,\{j\})=e^{-\lambda t}\dfrac{(\lambda t)^{j-i}}{(j-i)!},\ j\in\mathbb{Z}, j\in\{i,i+1,\cdots\}.$$ Thus, I know that $\mathbb{E}f(N(t)|N(0)=y)=\mathbb{E}f(y+N(t))=\sum_{k=0}^{\infty}f(x+k)e^{-t\lambda}\dfrac{(t\lambda)^{k}}{k!},$ but I don't know how to compute equation $(1)$ from here...

Consider two independent Poisson processes $N,M$ with rate $\lambda$, and define $$X(t):=x+\dfrac{1}{\sqrt{n}}[N(t)-M(t)].$$ From this formula we know that $X(0)=x$. Now, I want to compute the conditional probability $$\mathbb{E}[f(X(t))|X(0)=x].\ \ \ (1)$$ I know that for a single Poisson process, its transition probability is $$\mathbb{P}(i,t,\{j\})=e^{-\lambda t}\dfrac{(\lambda t)^{j-i}}{(j-i)!},\ j\in\mathbb{Z}, j\in\{i,i+1,\cdots\}.$$ Thus, I know that $\mathbb{E}f(N(t)|N(0)=y)=\mathbb{E}f(y+N(t))=\sum_{k=0}^{\infty}f(x+k)e^{-t\lambda}\dfrac{(t\lambda)^{k}}{k!},$ but I don't know how to compute equation $(1)$ from here...