Assume that $f,g \in L^2(\mathbb R)$ and define the Fourier transform of $f$ by $$\hat{f}(\xi) = \int_{\mathbb R} \mathrm{e}^{-i\,x\,\xi}\, f(x)\,\mathrm{d}\xi, \quad \xi \in \mathbb R.$$ The well-known Parseval-Plancherel identity links the $2$-norm of $f-g$ and the $2$-norm of $\hat{f} - \hat{g}$ via $$\int_{\mathbb R} |f(x)-g(x)|^2\,\mathrm{d} x = \frac{1}{2\,\pi}\,\int_{\mathbb R} |\hat{f}(\xi)-\hat{g}(\xi)|^2\,\mathrm{d} \xi \label{1}\tag{1}.$$ Now suppose that $f = (f_0,f_1,\ldots) \in \mathcal{P}(\mathbb N)$ and $g = (g_0,g_1,\ldots) \in \mathcal{P}(\mathbb N)$ are probability distributions over $\mathbb N$, and define for each $z \in (0,1)$ the probability generating function of $f \in \mathcal{P}(\mathbb N)$ by $$\hat{f}(z) = \sum_{n\geq 0} z^n\,f_n. $$ I am wondering if there exists a "natural" analog of the Parseval-Plancherel identity \eqref{1} involving the probability generating function. Unfortunately, I did not find any useful reference or related literature about this question.
Edit: Motivated by the answer provided by Andrew, I am wondering if there exists a collection of functions (not necessarily a orthonormal basis) $\{\varphi_n(z)\}_{n \geq 0}$ such that $$f_n = \int_0^1 \hat{f}(z)\,\varphi_n(z)\,\mathrm{d} z \label{2}\tag{2}$$ for each $n \in \mathbb N$. I want to view \eqref{2} as a sort of analog of the inverse Fourier transform (but now it involves the probability generating function instead).
If you allow yourself to extend the domain of $\hat f$ to the closed complex unit circle, you can come up a result. For $|z|\leq 1$, we can see that the sum $\hat f(z)=\sum_{n\geq 0}f_nz^n$ absolutely converges because $$\sum_{n\geq 0}|f_nz^n|=\sum_{n\geq 0}f_n|z|^n\leq \sum_{n\geq 0}f_n =1$$
Consider the function $\bar{f}:\mathbb{Z}\to \mathbb{R}\subset\mathbb{C}$ given by $\bar{f}(n)=f_n$ for $n\geq 0$ and $\bar{f}(n)=0$ for $n<0$. $\bar{f}$ gives the coefficients of the Fourier series of $\hat f(e^{i\theta})=\sum_{n\geq 0}f_ne^{in\theta}$. The same is true for $g$.
We can use Parseval's Theorem for Fourier series.
$$\|\bar g-\bar f\|^2=\|\hat g-\hat f\|^2\implies\sum_{n\geq 0}(f_n-g_n)^2 = \frac{1}{2\pi}\int_0^{2\pi}|\hat{f}(e^{i\theta})-\hat{g}(e^{i\theta})|^2d \theta$$
On the other hand, if we want a identity involving $\int_0^1(\hat{f}(z)-\hat{g}(z))^2dz$, we can use the shifted Legendre polynomials $\tilde P_k$ (defined by $\tilde P_k(z)=P_k(2z-1)$). Let $a_k$ be the following. $$a_k = (2k+1)\int_{0}^1 \tilde P_k(z)\hat{f}(z)dz$$ We can write $a_k$ in terms of the probabilities by using the unform convergence of $\hat f$ as follows. $$a_k = (2k+1)\int_{0}^1\sum_{n\geq 0}f_n \tilde P_k(z)z^ndz=(2k+1)\sum_{n\geq 0}f_n\int_{0}^1 \tilde P_k(z)z^ndz$$ I don't have the exact form of these integrals on hand, but they should be computable in general. You can trivially prove that it is zero if $n<k$. Therefore, there exists some constants $C_{n,k}$ such that $$a_k=\sum_{n\geq k} C_{n,k} f_{n}$$ If we define $b_k$ anaologusly for $g$, we can use the Parseval's theorem for Legendre-Fourier Series. $$\int_0^1(\hat{f}(z)-\hat{g}(z))^2dz=\sum_{k\geq 0}(a_k-b_k)^2$$ This is of course, much less pretty. If you choose other orthogonal polynomials with different weight functions on $[0,1]$ you should be able to get different $C_{n,k}$ but this won't make the final form any nicer.