Campbell's theorem variance From Wikipedia, 

For a Poisson point process $N$ and a measurable function  $f:  \textbf{R}^d\rightarrow \textbf{R}$, the random sum
   $$\Sigma=\sum_{x\in {N}}f(x)$$
  [...for complex value $\theta$,]
  $$
E(e^{\theta\Sigma})=\textrm{exp} \left(\int_{\textbf{R}^d} [e^{\theta f(x)}-1]\Lambda (dx)\right)
$$
  [...]
  $$
E(\Sigma)=\int_{\textbf{R}^d} f(x)\Lambda (dx)
$$
  $$
\text{Var}(\Sigma)=\int_{\textbf{R}^d} f(x)^2\Lambda (dx)
$$

How does the variance of the random sum follow from the second equation? 
Expanding the left side of the equation into a power series of $\theta$, we have
\begin{align*}
E[e^{\theta\Sigma}] &=  E[1 + \Sigma\theta + \frac{1}{2}\Sigma^2\theta^2+\ldots] \\
 &= 1 + E[\Sigma]\theta + \frac{1}{2}E[\Sigma^2]\theta^2+\ldots \\
\end{align*}
Expanding the right side into a power series of $\theta$, we have
\begin{align*}
\exp\left(\int_{\mathbf{R}^d}[e^{\theta f(x)}-1]\Lambda(dx)\right) &= 
\exp\left(\int_{\mathbf{R}^d}\left[\left(1+f(x)\theta+\frac{1}{2}f(x)^2\theta^2+\ldots\right)-1\right]\Lambda(dx)\right) \\
 &= \exp\left(\int_{\mathbf{R}^d}\left[f(x)\theta+\frac{1}{2}f(x)^2\theta^2+\ldots\right]\Lambda(dx)\right) \\
 &= \exp\left(\theta\int_{\mathbf{R}^d}f(x)\Lambda(dx)+\frac{\theta^2}{2}\int_{\mathbf{R}^d}f(x)^2\Lambda(dx)+\ldots\right) \\
 &= 1 + \int_{\mathbf{R}^d}f(x)\Lambda(dx)\theta+\int_{\mathbf{R}^d}f(x)^2\Lambda(dx)\theta^2 + \ldots\\
\end{align*}
Equating the expansions,
$$
1 + E[\Sigma]\theta + \frac{1}{2}E[\Sigma^2]\theta^2+\ldots =
1 + \int_{\mathbf{R}^d}f(x)\Lambda(dx)\theta+\int_{\mathbf{R}^d}f(x)^2\Lambda(dx)\theta^2 + \ldots
$$
and setting the coefficients of the powers of $\theta$ equal to each other, we have
$$
E[\Sigma]=\int_{\mathbf{R}^d}f(x)\Lambda(dx)
$$
which is consistent with the Wikipedia article. However,
$$
E[\Sigma^2]=2\int_{\mathbf{R}^d}f(x)^2\Lambda(dx)
$$
so 
\begin{align*}
\text{Var}(\Sigma) &= E[\Sigma^2] - E[\Sigma]^2 \\
 &= 2\int_{\mathbf{R}^d}f(x)^2\Lambda(dx) - \left(\int_{\mathbf{R}^d}f(x)\Lambda(dx)\right)^2 \\
\end{align*}
which is inconsistent with the Wikipedia article.
 A: Let $\phi_{\theta}(\Sigma) = \mathbb E[e^{i\theta\Sigma}]$ be the characteristic function of the random variable $\Sigma$. Then, from the properties of characteristic functions,
$$
\begin{align}
\mathbb E[\Sigma] = -i\frac{d}{d\theta}\phi_{\theta}(\Sigma)\Big\vert_{\theta=0} &= \left(\int_{\mathbb R^d}f(x)e^{i\theta f(x)}\Lambda(dx)\right)e^{\int_{\mathbb R^d}(e^{i\theta f}-1)\Lambda(dx)}\Big\vert_{\theta=0} \\ &=\int_{\mathbb R^d}f(x)\Lambda(dx)\tag{1}
\end{align}
$$
And,
$$
\begin{eqnarray*}
\mathbb E[\Sigma^2] & = &-\frac{d^2}{d\theta^2}\phi_{\theta}(\Sigma)\Big\vert_{\theta=0} \\ & = &\left(\int_{\mathbb R^d}f^2(x)e^{i\theta f(x)}\Lambda(dx)\right)e^{\int_{\mathbb R^d}(e^{i\theta f}-1)\Lambda(dx)}\Big\vert_{\theta=0} \hspace{0.5cm}+ \left(\int_{\mathbb R^d}f(x)e^{i\theta f(x)}\Lambda(dx)\right)^2e^{\int_{\mathbb R^d}(e^{i\theta f}-1)\Lambda(dx)}\Big\vert_{\theta=0} \\ & = & \int_{\mathbb R^d}f^2(x)\Lambda(dx) + \left(\int_{\mathbb R^d}f^2(x)\Lambda(dx)\right)^2\tag{2}
\end{eqnarray*}
$$
Consequently,

$$ {\bf V}\text{ar}(\Sigma) = \mathbb E[\Sigma^2] - (\mathbb E[\Sigma])^2 = \int_{\mathbb R^d}f^2(x)\Lambda(dx)$$

