Statement concerning Poisson random variables that appeared in a proof I am studying So I am studying a proof and the following line appeared. Assume $X_1,...,X_n,X_1'...X_n'$ i.i.d. Poisson random variables of mean $\frac{\mu}{n}$ each, and $f$ a real valued function such that $|f| \leq k$. Then the statement is 
\begin{equation} 
\frac{1}{2}\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2\right] = e^{\frac{-2\mu}{n}}\frac{\mu}{n}\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right] + O(\frac{k^2}{n^2})
\end{equation}
where $X^{(i)}=\sum_{j\neq i}X_j$. I am trying to understand how this result is derived. Any help would be appreciated. Thanks guys ! 
 A: Start out by using the law of iterated expectations:
$$\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2\right] = \mathbb{E}\left[\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2|X_i,X_i'\right]\right].$$
When calculating the outer expectation, split it up into the following cases: $(X_i,X_i')=(0,0)$, $(X_i,X_i')=(1,0)$, $(X_i,X_i')=(0,1)$, and $X_i,X_i'\geq 1$, so
$$\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2\right] = \mathbb{E}\left[\left(f(0+X^{(i)})-f(0+X^{(i)})\right)^2\right]P(X_i=0,X_i'=0)+\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right]P(X_i=1,X_i'=0)+\mathbb{E}\left[\left(f(0+X^{(i)})-f(1+X^{(i)})\right)^2\right]P(X_i=0,X_i'=1)+\sum_{r,s\geq 1}\mathbb{E}\left[\left(f(r+X^{(i)})-f(s+X^{(i)})\right)^2\right]P(X_i=r,X_i'=s).$$
Notice that the first expectation on the right hand side vanishes, and notice that the second and third terms on the right hand side are equal (because of the square inside the expectations). The equality thus reduces to
$$\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2\right]\\=2\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right]P(X_i=1,X_i'=0) + \sum_{r,s\geq 1}\mathbb{E}\left[\left(f(r+X^{(i)})-f(s+X^{(i)})\right)^2\right]P(X_i=r,X_i'=s).$$
Since $P(X_i=1,X_i'=0)=P(X_i=1,X_i'=0)=e^{-\frac{\mu}{n}}\frac{\left(\frac{\mu}{n}\right)^1}{1!}e^{-\frac{\mu}{n}}\frac{\left(\frac{\mu}{n}\right)^0}{0!}=e^{-\frac{2\mu}{n}}\frac{\mu}{n}$, the first term of the right hand side is
$$2\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right]P(X_i=1,X_i'=0)=2e^{-\frac{2\mu}{n}}\frac{\mu}{n}\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right].$$
As for the second term, recall that $|f|\leq k$, so
$$\left|\mathbb{E}\left[\left(f(r+X^{(i)})-f(s+X^{(i)})\right)^2\right]\right|\leq \mathbb{E}\left[\left|f(r+X^{(i)})-f(s+X^{(i)})\right|^2\right] \leq \mathbb{E}\left[(2k)^2\right]=4k^2,$$
and also $P(X_i=r,X_i'=s) = e^{-2\frac{\mu}{n}}\frac{\left(\frac{\mu}{n}\right)^{r+s}}{r!s!}=e^{-2\frac{\mu}{n}}\frac{\mu^{r+s}}{r!s!}\frac{1}{n^{r+s}}$.
We thus get the bound
$$\left|\sum_{r,s\geq 1}\mathbb{E}\left[\left(f(r+X^{(i)})-f(s+X^{(i)})\right)^2\right]P(X_i=r,X_i'=s)\right|\\\leq \sum_{r,s\geq 1}e^{-2\frac{\mu}{n}}\frac{\mu^{r+s}}{r!s!}\frac{1}{n^{r+s}}4k^2\\\leq \frac{k^2}{n^2}4e^{-2\frac{\mu}{n}}\sum_{r,s\geq 1}\frac{\mu^{r+s}}{r!s!},$$
where the last inequality follows from $n^{-(r+s)}\leq n^{-2}$, since $r,s\geq 1$. The double sum on the right hand side is finite. Indeed, $$\sum_{r,s\geq 1}\frac{\mu^{r+s}}{r!s!}=\sum_{r\geq 1}\frac{\mu^r}{r!}\sum_{s\geq 1}\frac{\mu^s}{s!},$$ and you recognize the Taylor series of the exponential. We conclude that
$$\sum_{r,s\geq 1}\mathbb{E}\left[\left(f(r+X^{(i)})-f(s+X^{(i)})\right)^2\right]P(X_i=r,X_i'=s)=O\left(\frac{k^2}{n^2}\right),$$
and combining this with our calculation for $(X_i,X_i')=(1,0)$ from above, we get
$$\frac{1}{2}\mathbb{E}\left[\left(f(X_i+X^{(i)})-f(X_i'+X^{(i)})\right)^2\right]\\=e^{-\frac{2\mu}{n}}\frac{\mu}{n}\mathbb{E}\left[\left(f(1+X^{(i)})-f(0+X^{(i)})\right)^2\right]+O\left(\frac{k^2}{n^2}\right).$$
