Problem
Suppose a random walk in 2D with step sizes determined by $f(m) = (\frac{1}{2} + m)$, where the final position is $X = \sum_{m=0}^{N-1} x_m f(m)$ and $x_m \in \mathbb{C}$ (for example $x_m = e^{- i \phi_m}$ where $\phi_m$ is uniformly distributed over [0, $2\pi$]). What is the expectation value $\mathbb{E}[e^{-y^2|X|^2}]$, if $y\sim \mathcal{N}(0, \sigma^2)$ is normally distributed and independent of $X$?
Attempted solution
To first verify the behaviour of the exponentiated random walk, we assume that $y$ is constant and we set $y = \sigma$. With this, the expectation value becomes $\sum_{k=0}^{\infty} \frac{(-\sigma^2)^k}{k!}\mathbb{E}[(|X|^2)^k] \approx \sum_{k=0}^{\infty} (- \sigma^2)^k (\frac{N^3}{3} + \mathcal{O}(N))^k = \frac{1}{1 + \sigma^2 N^3/3}$. This solution has been verified with numerical simulations, by random sampling the result of the random walk and calculating the mean of the exponentiated distance.
To consider the original problem, we must calculate the mean $\mathbb{E}[ \sum_{k=0}^{\infty} \frac{(y^2)^k}{k!}(-|X|^2)^k] = \sum_{k=0}^{\infty} \frac{\mathbb{E}[(y^2)^k]}{k!}\mathbb{E}[(-|X|^2)^k]$. Here, we can use the fact that $\mathbb{E}[y^{2k}] = \sigma^{2k} (2k-1)!!$. With this, and with the approximation $\mathbb{E}[(-|X|^2)^k] = k! (N^3/3)^k$, the mean becomes $\sum_{k=0}^{\infty} (2k-1)!! (\sigma^2 N^3/3)^k$. This final expression however does not converge. From numerical simulations, we verify that the mean does converge.