[proofplan] Well-definedness is the easy part: since $|\phi(k)| = |\mathbb{E}[\pi^k]| \leq \mathbb{E}[|\pi|^k] = \psi(k)$ by Jensen's inequality and $\psi$ satisfies the summability condition by hypothesis, $|\phi|$ does too, hence the $\phi$-signature kernel series converges absolutely. For the representation formula we expand $k_\phi^{x,y}$ as a series, swap the expectation $\mathbb{E}_\pi$ with the $\sum_i$ using Fubini-Tonelli (justified by the same summability bound), and use the homogeneity of iterated integrals $S(\pi x)^i_{a,s} = \pi^i\, S(x)^i_{a,s}$ to identify the inner sum as the constant-weight signature kernel $k^{\pi x, y}(s,t)$. The symmetric formula with $\pi y$ follows by applying the same argument to the second slot. [/proofplan]

[step:Verify well-definedness via the dominating sequence $\psi$]By Jensen's inequality applied to the convex function $z \mapsto |z|$, \begin{align*} |\phi(k)| = |\mathbb{E}[\pi^k]| \leq \mathbb{E}[|\pi^k|] = \mathbb{E}[|\pi|^k] = \psi(k) \qquad \text{for all } k \geq 0. \end{align*} Therefore $|\phi|$ is dominated termwise by $\psi$. Since $\psi$ satisfies the summability condition of *Sufficient Condition for Signature Membership* by hypothesis, comparison gives that $|\phi|$ does too. Hence the $\phi$-signature kernel \begin{align*} k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \phi(i)\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} \end{align*} converges absolutely on $\widetilde{\mathcal{C}_p} \times \widetilde{\mathcal{C}_p}$ and is well-defined as a function $k_\phi^{x,y} : [a,b]^2 \to \mathbb{R}$.[/step]

[guided]We must check two things before manipulating $k_\phi^{x,y}$: first, that the defining series converges, and second, that we may apply Fubini below to swap a sum and an expectation. Both reduce to controlling $|\phi|$. The bound $|\phi(k)| \leq \psi(k)$ follows from a one-line application of Jensen's inequality. Jensen for the convex map $z \mapsto |z|$ states $|\mathbb{E}[X]| \leq \mathbb{E}[|X|]$ whenever the expectations exist. With $X = \pi^k$, \begin{align*} |\phi(k)| = |\mathbb{E}[\pi^k]| \leq \mathbb{E}[|\pi^k|] = \mathbb{E}[|\pi|^k] = \psi(k). \end{align*} The hypothesis that $\pi$ has finite moments of all orders is consumed here: it ensures all these expectations exist. Why is termwise domination $|\phi(k)| \leq \psi(k)$ enough to conclude that $|\phi|$ satisfies the same summability condition as $\psi$? The summability condition of the *Sufficient Condition for Signature Membership* requires \begin{align*} \sum_{i=0}^\infty |\phi(i)|\, \frac{R^i}{(i!)^2} < \infty \end{align*} for every $R > 0$ (or for sufficiently large $R$, depending on the precise statement). If $\psi$ satisfies this for $R = \|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]}$, then so does $|\phi|$ by direct comparison. With $|\phi|$ summable in this sense, the kernel series $k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \phi(i)\langle S(x)^i_{a,s}, S(y)^i_{a,t}\rangle_{V^{\otimes i}}$ converges absolutely (by Cauchy-Schwarz on $V^{\otimes i}$ and the factorial decay $\|S(x)^i_{a,s}\| \leq \|x\|_{1,[a,s]}^i / i!$, the absolute terms are bounded by $|\phi(i)|\,\|x\|_{1,[a,s]}^i\|y\|_{1,[a,t]}^i / (i!)^2$). Hence $k_\phi^{x,y}(s,t)$ is a well-defined real number.[/guided]

[step:Expand $k_\phi^{x,y}$ as a series and pull $\mathbb{E}_\pi$ inside via Fubini-Tonelli]By the definition of the $\phi$-signature kernel and $\phi(i) = \mathbb{E}[\pi^i]$, \begin{align*} k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \mathbb{E}[\pi^i]\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} We apply Fubini's theorem to interchange the sum and the expectation. Fubini for series-expectation interchange requires \begin{align*} \sum_{i=0}^\infty \mathbb{E}\!\left[\,\left|\pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right|\,\right] < \infty. \end{align*} We verify this. By Cauchy-Schwarz on $V^{\otimes i}$ and the factorial decay estimate $\|S(x)^i_{a,s}\|_{V^{\otimes i}} \leq \|x\|_{1,[a,s]}^i / i!$, \begin{align*} \left|\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right| \leq \frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2}. \end{align*} Therefore \begin{align*} \sum_{i=0}^\infty \mathbb{E}\!\left[|\pi|^i\right]\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} = \sum_{i=0}^\infty \psi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} < \infty, \end{align*} where the last inequality is the assumed summability of $\psi$. Fubini-Tonelli applies, so \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}\!\left[\sum_{i=0}^\infty \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right]. \end{align*}[/step]

[guided]By definition of $\phi(i) = \mathbb{E}[\pi^i]$, \begin{align*} k_\phi^{x,y}(s,t) = \sum_{i=0}^\infty \mathbb{E}[\pi^i]\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} We want to move $\mathbb{E}$ outside the sum so we can recognise the inner content as a single signature kernel. Why is this not free? Interchanging $\sum$ and $\mathbb{E}$ is exactly the kind of move that requires Fubini-Tonelli: the sum is a sum over $i \in \mathbb{N}$ (counting measure on $\mathbb{N}$), the expectation is an integral on the probability space, and we are interchanging two integrals. Fubini's theorem (the general version for $\sigma$-finite measures, with counting measure on $\mathbb{N}$ being directly $\sigma$-finite, and probability measure being finite hence $\sigma$-finite) requires absolute integrability of the joint integrand: \begin{align*} \sum_{i=0}^\infty \mathbb{E}\!\left[\,\left|\pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right|\,\right] < \infty. \end{align*} We must verify this. Bound the inner product first: Cauchy-Schwarz on $V^{\otimes i}$ gives $|\langle S(x)^i_{a,s}, S(y)^i_{a,t}\rangle_{V^{\otimes i}}| \leq \|S(x)^i_{a,s}\|\,\|S(y)^i_{a,t}\|$, and the factorial decay estimate (which uses $x, y \in C_1$ — the bounded variation hypothesis) gives $\|S(x)^i_{a,s}\| \leq \|x\|_{1,[a,s]}^i / i!$ and similarly for $y$. So \begin{align*} \sum_{i=0}^\infty \mathbb{E}[|\pi|^i]\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2} = \sum_{i=0}^\infty \psi(i)\,\frac{(\|x\|_{1,[a,s]}\,\|y\|_{1,[a,t]})^i}{(i!)^2}. \end{align*} This is exactly the summability of $\psi$ that we have assumed, so finiteness holds. Fubini applies and we may pull $\mathbb{E}$ inside: \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}\!\left[\sum_{i=0}^\infty \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right]. \end{align*} Notice that the hypothesis on $\psi$, not $\phi$, is what licenses the swap. We need a bound on $\mathbb{E}[|\pi|^i]$, not on $|\mathbb{E}[\pi^i]|$, because absolute values must be inside the expectation in the Fubini check. This is why the theorem is stated with $\psi$ in the hypothesis even though $\phi$ defines the kernel.[/guided]

[step:Use homogeneity $S(\pi x)^i = \pi^i S(x)^i$ to identify the inner series]The signature is homogeneous in its driving path: for any scalar $c \in \mathbb{R}$ and $i \geq 0$, \begin{align*} S(c\,x)^i_{a,s} = c^i\,S(x)^i_{a,s}, \end{align*} since the level-$i$ signature is an $i$-fold iterated integral and rescaling the integrator by $c$ produces a factor of $c^i$. Multilinearity of the inner product on $V^{\otimes i}$ then yields \begin{align*} \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle \pi^i\,S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(\pi x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} Substituting into the expression from the previous step, \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}\!\left[\sum_{i=0}^\infty \langle S(\pi x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right]. \end{align*} The inner series is, by definition, the constant-weight (i.e. $\phi \equiv 1$) signature kernel of the rescaled path $\pi x$ against $y$: \begin{align*} \sum_{i=0}^\infty \langle S(\pi x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = k^{\pi x, y}(s,t). \end{align*} Therefore \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}_\pi\!\left[ k^{\pi x, y}(s,t) \right]. \end{align*}[/step]

[guided]We need to identify the random series inside the expectation. The first ingredient is the homogeneity property of the signature: the level-$i$ component is an $i$-fold iterated integral \begin{align*} S(x)^i_{a,s} = \int_{a < t_1 < \cdots < t_i < s} dx_{t_1} \otimes \cdots \otimes dx_{t_i}, \end{align*} and rescaling the path by a scalar $c$ rescales each $dx_{t_j}$ by $c$, producing $i$ factors of $c$ in the integrand: \begin{align*} S(c\,x)^i_{a,s} = c^i\,S(x)^i_{a,s}. \end{align*} Setting $c = \pi$ (a fixed real number on each sample path of the random variable, so this makes pathwise sense) gives $S(\pi x)^i_{a,s} = \pi^i\,S(x)^i_{a,s}$. Second, the inner product $\langle \cdot, \cdot \rangle_{V^{\otimes i}}$ is bilinear, so the scalar $\pi^i$ moves into the first slot: \begin{align*} \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle \pi^i\,S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(\pi x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} So the random series inside $\mathbb{E}$ becomes \begin{align*} \sum_{i=0}^\infty \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \sum_{i=0}^\infty \langle S(\pi x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} The right-hand side is the definition of the constant-weight signature kernel $k^{\pi x, y}(s,t)$ — that is, the $\phi$-kernel with $\phi \equiv 1$ — between paths $\pi x$ and $y$. Hence \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}_\pi\!\left[ k^{\pi x, y}(s,t) \right]. \end{align*} The interpretation: the weighted kernel is an average (over the law of $\pi$) of unweighted kernels of randomly-rescaled inputs.[/guided]

[step:Derive the symmetric representation by rescaling the second slot]The same homogeneity argument applies to the second slot. By multilinearity of the inner product and $S(\pi y)^i_{a,t} = \pi^i\,S(y)^i_{a,t}$, \begin{align*} \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(x)^i_{a,s},\, \pi^i\,S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(x)^i_{a,s},\, S(\pi y)^i_{a,t}\rangle_{V^{\otimes i}}. \end{align*} Substituting into the formula at the end of Step 2, \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}\!\left[\sum_{i=0}^\infty \langle S(x)^i_{a,s},\, S(\pi y)^i_{a,t}\rangle_{V^{\otimes i}}\right] = \mathbb{E}_\pi\!\left[ k^{x, \pi y}(s,t) \right]. \end{align*} Combining with the previous step, \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}_\pi\!\left[ k^{\pi x, y}(s,t) \right] = \mathbb{E}_\pi\!\left[ k^{x, \pi y}(s,t) \right], \end{align*} which is the claimed identity. The representation formula and well-definedness of $k_\phi^{x,y}$ are established.[/step]

[guided]The argument is symmetric to Step 3. The signature is homogeneous in its driving path (Step 3 derivation), and the inner product on $V^{\otimes i}$ is bilinear, so the scalar $\pi^i$ is free to migrate to either slot. Migrating it to the *second* slot rather than the first: \begin{align*} \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(x)^i_{a,s},\, \pi^i\,S(y)^i_{a,t}\rangle_{V^{\otimes i}} = \langle S(x)^i_{a,s},\, S(\pi y)^i_{a,t}\rangle_{V^{\otimes i}}, \end{align*} where the last equality is the homogeneity $S(\pi y)^i_{a,t} = \pi^i\,S(y)^i_{a,t}$ applied to $y$ (this works pathwise: for each realization of $\pi$, $\pi y$ is a well-defined bounded variation path on $[a,t]$, and the level-$i$ signature scales by $\pi^i$). Substituting into the post-Fubini formula from Step 2, \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}\!\left[\sum_{i=0}^\infty \pi^i\,\langle S(x)^i_{a,s},\, S(y)^i_{a,t}\rangle_{V^{\otimes i}}\right] = \mathbb{E}\!\left[\sum_{i=0}^\infty \langle S(x)^i_{a,s},\, S(\pi y)^i_{a,t}\rangle_{V^{\otimes i}}\right]. \end{align*} The inner sum, by definition of the constant-weight signature kernel ($\phi \equiv 1$), is $k^{x, \pi y}(s,t)$: \begin{align*} \sum_{i=0}^\infty \langle S(x)^i_{a,s},\, S(\pi y)^i_{a,t}\rangle_{V^{\otimes i}} = k^{x, \pi y}(s,t). \end{align*} Hence \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}_\pi\!\left[ k^{x, \pi y}(s,t) \right]. \end{align*} Combining with Step 3, \begin{align*} k_\phi^{x,y}(s,t) = \mathbb{E}_\pi\!\left[ k^{\pi x, y}(s,t) \right] = \mathbb{E}_\pi\!\left[ k^{x, \pi y}(s,t) \right]. \end{align*} The two representations are equally valid because $\pi$ is a *scalar* — it commutes through the bilinear inner product. They are not in general both equal to $\mathbb{E}_\pi[k^{\pi x, \pi y}(s,t)]$: pushing $\pi$ into both slots simultaneously would yield $\pi^{2i}$ instead of $\pi^i$ at level $i$, corresponding to the kernel weighted by $\mathbb{E}[\pi^{2i}] = \psi(2i)$, not $\phi(i)$. The PDE-averaging interpretation in [Weighted Signature Kernel as Averaged PDE Solution](/theorems/???) exploits exactly this — solving a single signature kernel PDE on the averaged input is enough; one does not need to average over both slots.[/guided]