Fourier transform of a simple random walk

Question

Consider the usual simple random walk on $\mathbb{Z}$, taking steps of +1 or -1 with equal probability. Of course, each trajectory corresponds uniquely to an element of $\{-1,1\}^\infty$. Now, there is an obvious way to identify this with $\mathbb{Z}_2^\infty$, and this is in fact a nice compact abelian group.

So, the functional of a simple random walk which, say, gives the hitting time of some $k$, or the indicator function of the event that we hit $n$ before $-n$, can be seen as functions from $\mathbb{Z}_2^\infty$ into $\mathbb{R}$. Thus, these functions have Fourier transforms.

Unless I am completely mistaken, the Fourier transform of these functions should look like (taking our omegas to be -1 or 1, for convenience) $$f(\omega_1,\omega_2,\ldots) = \sum_{S\subset\mathbb{N},\ |S|<\infty} \hat{f}(S)\prod_{i\in S}\omega_i$$ with $\hat{f}(S)$ real-valued Fourier coefficients, analogously to the Fourier analysis of functions of finitely many $\omega$.

Has this been studied before? Do the Fourier coefficients of these functions encode anything interesting about the simple random walk?

I tried to analyse the second example a bit, trying to compute the level one Fourier coefficients (i.e. the ones corresponding to singleton subsets of $\mathbb{N}$). Letting $\tau_{x,y}$ be the first time the random walk hits $y$ after it has hit $x$, and let $\tau_x$ be the hitting time of $x$ from $0$, and further letting $\tau^i_x$ be the first time after time $i$ at which the random walk hits $x$, I get the following formula: $$\hat{f}(\{i\}) = \mathbb{P}\left(\omega_i = 1, i \leq \tau_n, \tau^i_{-(n-1)} < \tau_{n+1}-i\ \middle|\ \tau_n < \tau_{-n}\right)$$ which I don't really immediately see how to compute.

Answer 1

I doubt that this is of any use but the 1-element modes are not hard to compute. Let $X_i:=\sum^i_{k=1}\omega_k$ be the random walk and $\tau=\min\{i:X_i=a\text{ or }X_i=-b\}$. We have $$ \mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i)=\mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i\mathbb{1}_{\tau<i})+\mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i\mathbb{1}_{\tau>i-1})=\mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i\mathbb{1}_{\tau>i-1}), $$ by strong Markov property. Now, you can write $$ \mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i\mathbb{1}_{\tau>i-1})=\mathbb{E}(\mathbb{1}_{\tau>i-1}\mathbb{E}(\mathbb{1}_{X_\tau=a}\omega_i |\omega_1,\dots,\omega_{i-1})). $$ On the event $\tau>i-1$, the conditional expectation of $\mathbb{1}_{X_\tau=a}$ given $\omega_1,\dots,\omega_i$ is just $(X_i+b)/(a+b)$. Therefore, we have on that event $$ \mathbb{E}(\mathbb{1}_{X_\tau=a}\omega_i |\omega_1,\dots,\omega_{i-1})=\frac{1}{2}\frac{X_{i-1}+1+b}{a+b}-\frac{1}{2}\frac{X_{i-1}-1+b}{a+b}=\frac{1}{a+b}, $$ so that $$ \mathbb{E }(\mathbb{1}_{X_\tau=a} \omega_i)=\frac{\mathbb{P}(\tau>i-1)}{a+b}. $$ The latter probability is not hard to compute using discrete Fourier transfrom, or find in any book.

The last simple observation is that if you take $a=-b$, then all the modes for even-size sets vanish, since $$ \mathbb{E}(\mathbb{1}_{\tau_a<\tau_-a}(\omega)\omega_S)=\mathbb{E}(\mathbb{1}_{\tau_a<\tau_{-a}}(-\omega)\omega_S)=\mathbb{E}((1-\mathbb{1}_{\tau_a<\tau_{-a}}(\omega))\omega_S). $$