Find a conditional expectation of a difference of two independent Poisson process

Question

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...

Answer 1

First here, $E(f(X(t)|X(0)=x)$ is, not a conditional probability, but a conditional expectation. Second, the event $X(0)=x$ is certain to occur; therefore, conditioning on it does not affect the expectations or probabilities. So, $$E(f(X(t)|X(0)=x)=Ef(X(t).$$ Finally, $$Ef(X(t))=\sum_{k=0}^\infty\sum_{l=0}^\infty f\Big(x+\frac{k-l}{\sqrt n}\Big)\frac{(t\lambda)^{k+l}}{k!\,l!}e^{-2t\lambda}.$$