**2. Results on** L*p***-calculus**

In this section, we state the preliminary results on L*p*-calculus needed for the following sections. Proposition 1 is the chain rule theorem in L*p*-calculus, which was first proved in [8] (Theorem 3.19) in the setting of mean square calculus (*p* = 2). Both Lemma 1 and Lemma 2 provide conditions under which the product of three stochastic processes is L*p*-continuous or L*p*-differentiable. Proposition 2 is a result concerning L*p*-differentiation under the L*p*-Riemann integral sign, when the interval of integration is fixed. These four results have been already used and stated in the recent contribution [17], and will be required through our forthcoming exposition.

For the sake of completeness, we demonstrate Proposition 2 with an alternative proof to [17], based on Fubini's theorem for L*p*-Riemann integration. In the random framework, Fubini's theorem has not been tackled yet in the recent literature. It states that, if a stochastic process depending on two variables is L*p*-continuous, then the two iterated L*p*-Riemann integrals can be interchanged.

We present a new result, Proposition 3, in which we put conditions in order to L*p*-differentiate an L*p*-Riemann integral whose interval of integration depends on *t*. This proposition supposes the extension of the so-called Leibniz's rule for integration to the random scenario. The proof relies on a new chain rule theorem.

**Proposition 1** (Chain rule theorem ([17] Proposition 2.1))**.** *Let* {*X*(*t*) : *t* ∈ [*a*, *b*]} *be a stochastic process, where* [*a*, *b*] *is any interval in* R*. Let f be a deterministic C*<sup>1</sup> *function on an open set that contains X*([*a*, *b*])*. Fix* 1 ≤ *p* < ∞*. Let t* ∈ [*a*, *b*] *be any point such that:*


*Then f* ◦ *X is* <sup>L</sup>*p-differentiable at t and* (*<sup>f</sup>* ◦ *<sup>X</sup>*) (*t*) = *f* (*X*(*t*))*X* (*t*)*.*

**Lemma 1** ([17] Lemma 2.1)**.** *Let Y*1(*t*,*s*)*, Y*2(*t*,*s*) *and Y*3(*t*,*s*) *be three stochastic processes and fix* 1 ≤ *p* < ∞*. If <sup>Y</sup>*<sup>1</sup> *and <sup>Y</sup>*<sup>2</sup> *are* <sup>L</sup>*q-continuous for all* <sup>1</sup> <sup>≤</sup> *<sup>q</sup>* <sup>&</sup>lt; <sup>∞</sup>*, and <sup>Y</sup>*<sup>3</sup> *is* <sup>L</sup>*p*+*η-continuous for certain <sup>η</sup>* <sup>&</sup>gt; <sup>0</sup>*, then the product process Y*1*Y*2*Y*<sup>3</sup> *is* L*p-continuous.*

*On the other hand, if Y*<sup>1</sup> *and Y*<sup>2</sup> *are* L∞*-continuous, and Y*<sup>3</sup> *is* L*p-continuous, then the product process Y*1*Y*2*Y*<sup>3</sup> *is* L*p-continuous.*

**Lemma 2** ([17] Lemma 2.2)**.** *Let Y*1(*t*)*, Y*2(*t*) *and Y*3(*t*) *be three stochastic processes, and* 1 ≤ *p* < ∞*. If <sup>Y</sup>*<sup>1</sup> *and <sup>Y</sup>*<sup>2</sup> *are* <sup>L</sup>*q-differentiable for all* <sup>1</sup> <sup>≤</sup> *<sup>q</sup>* <sup>&</sup>lt; <sup>∞</sup>*, and <sup>Y</sup>*<sup>3</sup> *is* <sup>L</sup>*p*+*η-differentiable for certain <sup>η</sup>* <sup>&</sup>gt; <sup>0</sup>*, then the product process Y*1*Y*2*Y*<sup>3</sup> *is* L*p-differentiable and* <sup>d</sup> <sup>d</sup>*t*(*Y*1(*t*)*Y*2(*t*)*Y*3(*t*)) = *Y* <sup>1</sup>(*t*)*Y*2(*t*)*Y*3(*t*) + *Y*1(*t*)*Y* <sup>2</sup>(*t*)*Y*3(*t*) + *Y*1(*t*)*Y*2(*t*)*Y* <sup>3</sup>(*t*)*.*

*Additionally, if Y*<sup>1</sup> *and Y*<sup>2</sup> *are assumed to be* L∞*-differentiable, and Y*<sup>3</sup> *is* L*p-differentiable, then Y*1*Y*2*Y*<sup>3</sup> *is* L*p-differentiable, with* <sup>d</sup> <sup>d</sup>*t*(*Y*1(*t*)*Y*2(*t*)*Y*3(*t*)) = *Y* <sup>1</sup>(*t*)*Y*2(*t*)*Y*3(*t*) + *Y*1(*t*)*Y* <sup>2</sup>(*t*)*Y*3(*t*) + *Y*1(*t*)*Y*2(*t*)*Y* <sup>3</sup>(*t*)*.*

**Lemma 3** (Fubini's theorem for iterated L*p*-Riemann integrals)**.** *Let <sup>H</sup>*(*t*,*s*) *be a process on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*. If H is* L*p-continuous, then b a d <sup>c</sup> <sup>H</sup>*(*t*,*s*) <sup>d</sup>*<sup>s</sup>* <sup>d</sup>*<sup>t</sup>* <sup>=</sup> *d c b <sup>a</sup> H*(*t*,*s*) d*t* d*s, where the integrals are regarded as* L*p-Riemann integrals.*

**Proof.** The proof is a variation of Fubini's theorem for Itô stochastic integration with respect to the standard Brownian motion ([32] Theorem 2.10.1). The stochastic processes *H*(*t*,*s*), *d <sup>c</sup> H*(*t*,*s*) d*s* and *b <sup>a</sup> <sup>H</sup>*(*t*,*s*) <sup>d</sup>*<sup>t</sup>* are L*p*-continuous, so the iterated integrals exist. Let {*Pn*}<sup>∞</sup> *<sup>n</sup>*=<sup>1</sup> be a sequence of partitions of [*a*, *b*] with mesh tending to 0. Write *Pn* = {*a* = *t n* <sup>0</sup> < *t n* <sup>1</sup> < ··· < *t n <sup>n</sup>* <sup>=</sup> *<sup>b</sup>*}, and let *<sup>r</sup><sup>n</sup> <sup>i</sup>* ∈ [*t n <sup>i</sup>*−1, *<sup>t</sup> n i* ], <sup>1</sup> <sup>≤</sup> *<sup>i</sup>* <sup>≤</sup> *<sup>n</sup>*, *<sup>n</sup>* <sup>≥</sup> 1. Consider the processes *Gn*(*t*,*s*) = <sup>∑</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *H*(*r<sup>n</sup> <sup>i</sup>* ,*s*)[*t<sup>n</sup> <sup>i</sup>*−1,*t<sup>n</sup> <sup>i</sup>* ](*t*) (here denotes the characteristic function of a set) and *Fn*(*s*) = *b <sup>a</sup> Gn*(*t*,*s*) <sup>d</sup>*<sup>t</sup>* <sup>=</sup> <sup>∑</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *H*(*r<sup>n</sup> <sup>i</sup>* ,*s*)(*t n <sup>i</sup>* − *t n <sup>i</sup>*−1). Notice that, by definition of L*p*-Riemann integral, lim*n*→<sup>∞</sup> *Fn*(*s*) = *b <sup>a</sup> <sup>H</sup>*(*t*,*s*) <sup>d</sup>*<sup>t</sup>* in L*p*.

By definition of L*p*-Riemann integral,

$$\int\_{\mathcal{L}}^{d} F\_{\mathfrak{n}}(\mathbf{s}) \, \mathrm{d}\mathbf{s} = \sum\_{i=1}^{n} \left( \int\_{\mathcal{c}}^{d} H(r\_{i}^{\mathfrak{n}}, \mathbf{s}) \, \mathrm{d}\mathbf{s} \right) (t\_{i}^{\mathfrak{n}} - t\_{i-1}^{\mathfrak{n}}) \stackrel{\mathfrak{n} \to \infty}{\longrightarrow} \int\_{\mathcal{a}}^{b} \int\_{\mathcal{c}}^{d} H(t, \mathbf{s}) \, \mathrm{d}\mathbf{s} \, \mathrm{d}\mathbf{t}$$

in L*p*. On the other hand,

$$\begin{aligned} \left\| \left| \int\_{\mathfrak{c}}^{d} \int\_{a}^{b} H(t,s) \, \mathrm{d}t \, \mathrm{d}s - \int\_{\mathfrak{c}}^{d} F\_{\mathfrak{n}}(s) \, \mathrm{d}s \right| \right\|\_{p} &= \left\| \int\_{\mathfrak{c}}^{d} \left( \int\_{a}^{b} H(t,s) \, \mathrm{d}t - F\_{\mathfrak{n}}(s) \right) \, \mathrm{d}s \right\|\_{p} \\ &\leq \int\_{\mathfrak{c}}^{d} \left\| \int\_{a}^{b} H(t,s) \, \mathrm{d}t - F\_{\mathfrak{n}}(s) \right\|\_{p} \, \mathrm{d}s, \end{aligned}$$

where the last inequality is due to a property of L*p*-integration ([5] p. 102). As *H*(*t*,*s*) and *Fn*(*s*) are <sup>L</sup>*p*-bounded on [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*] and [*c*, *<sup>d</sup>*], respectively (uniformly on *<sup>n</sup>* <sup>≥</sup> 1), the dominated convergence theorem allows concluding that lim*n*→<sup>∞</sup> *d <sup>c</sup> Fn*(*s*) <sup>d</sup>*<sup>s</sup>* <sup>=</sup> *d c b <sup>a</sup> <sup>H</sup>*(*t*,*s*) <sup>d</sup>*<sup>t</sup>* <sup>d</sup>*<sup>s</sup>* in L*p*.

**Proposition 2** (L*p*-differentiation under the L*p*-Riemann integral sign)**.** *Let F*(*t*,*s*) *be a stochastic process on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*. Fix* <sup>1</sup> <sup>≤</sup> *<sup>p</sup>* <sup>&</sup>lt; <sup>∞</sup>*. Suppose that <sup>F</sup>*(*t*, ·) *is* <sup>L</sup>*p-continuous on* [*c*, *<sup>d</sup>*]*, for each <sup>t</sup>* <sup>∈</sup> [*a*, *<sup>b</sup>*]*, and that there exists the* L*p-partial derivative <sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>t</sup>* (*t*,*s*) *for all* (*t*,*s*) <sup>∈</sup> [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*, which is* <sup>L</sup>*p-continuous on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*. Let <sup>G</sup>*(*t*) = *d <sup>c</sup> <sup>F</sup>*(*t*,*s*) <sup>d</sup>*<sup>s</sup> (the integral is understood as an* <sup>L</sup>*p-Riemann integral). Then <sup>G</sup> is* L*p-differentiable on* [*a*, *b*] *and G* (*t*) = *d c ∂F <sup>∂</sup><sup>t</sup>* (*t*,*s*) d*s.*

**Proof.** We present an alternative and simpler proof to ([17] Proposition 2.2), based on Fubini's theorem (Lemma 3). Since *<sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>t</sup>* is L*p*-continuous, by Barrow's rule ([5] p. 104) we can write *G*(*t*) = *d <sup>c</sup> <sup>F</sup>*(*a*,*s*) <sup>d</sup>*<sup>s</sup>* <sup>+</sup> *d c t a ∂F <sup>∂</sup><sup>t</sup>* (*τ*,*s*) <sup>d</sup>*<sup>τ</sup>* <sup>d</sup>*<sup>s</sup>* <sup>=</sup> *d <sup>c</sup> <sup>F</sup>*(*a*,*s*) <sup>d</sup>*<sup>s</sup>* <sup>+</sup> *t a d c ∂F <sup>∂</sup><sup>t</sup>* (*τ*,*s*) d*s* d*τ*. The stochastic process *<sup>τ</sup>* <sup>∈</sup> [*a*, *<sup>b</sup>*] → *d c ∂F <sup>∂</sup><sup>t</sup>* (*τ*,*s*) <sup>d</sup>*<sup>s</sup>* is L*p*-continuous; therefore, *<sup>G</sup>* (*t*) = *d c ∂F <sup>∂</sup><sup>t</sup>* (*t*,*s*) <sup>d</sup>*<sup>s</sup>* in L*p*, as a consequence of the fundamental theorem for L*p*-calculus; see ([5] p. 103).

**Lemma 4** (Version of the chain rule theorem)**.** *Let G*(*t*,*s*) *be a stochastic process on* [*a*, *b*] × [*c*, *d*]*. Let <sup>u</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> [*c*, *<sup>d</sup>*] *be a differentiable deterministic function. Suppose that <sup>G</sup>*(*t*,*s*) *has* <sup>L</sup>*p-partial derivatives, with ∂G <sup>∂</sup><sup>t</sup>* (*t*,*s*) *being* <sup>L</sup>*p-continuous on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*, and <sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>s</sup>* (*t*, ·) *being* <sup>L</sup>*p-continuous on* [*c*, *<sup>d</sup>*]*, for each <sup>t</sup>* <sup>∈</sup> [*a*, *<sup>b</sup>*]*. Then* <sup>d</sup> <sup>d</sup>*<sup>t</sup> <sup>G</sup>*(*t*, *<sup>u</sup>*(*t*)) = *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*t*, *u*(*t*)) + *u* (*t*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>s</sup>* (*t*, *<sup>u</sup>*(*t*)) *in* <sup>L</sup>*p.*

**Proof.** For *h* = 0, by the triangular inequality,

$$\leq \underbrace{\left\| \frac{\mathbb{G}(t+h, u(t+h)) - \mathbb{G}(t, u(t))}{h} - \frac{\partial \mathbb{G}}{\partial t}(t, u(t)) - u'(t) \frac{\partial \mathbb{G}}{\partial s}(t, u(t)) \right\|}\_{I\_{1}(t, h)} $$

$$+ \underbrace{\left\| \frac{\mathbb{G}(t+h, u(t+h)) - \mathbb{G}(t, u(t+h))}{h} - \frac{\partial \mathbb{G}}{\partial t}(t, u(t)) \right\|\_{p}}\_{I\_{2}(t, h)} - u'(t) \frac{\partial \mathbb{G}}{\partial s}(t, u(t)) \right\|\_{p}}\_{I\_{3}(t, h)}.$$

*p*

By Barrow's rule ([5] p. 104) and an inequality from ([5] p. 102),

$$\begin{split} I\_{1}(t,h) &= \left\| \frac{1}{h} \int\_{t}^{t+h} \frac{\partial G}{\partial t}(\tau, u(t+h)) \, \mathrm{d}\tau - \frac{\partial G}{\partial t}(t, u(t)) \right\|\_{p} \\ &= \left\| \frac{1}{h} \int\_{t}^{t+h} \left( \frac{\partial G}{\partial t}(\tau, u(t+h)) - \frac{\partial G}{\partial t}(t, u(t)) \right) \, \mathrm{d}\tau \right\|\_{p} \\ &\leq \frac{1}{|h|} \left\| \int\_{t}^{t+h} \left\| \frac{\partial G}{\partial t}(\tau, u(t+h)) - \frac{\partial G}{\partial t}(t, u(t)) \right\|\_{p} \, \mathrm{d}\tau \right\|. \end{split}$$

The process *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*t*, *<sup>u</sup>*(*r*)) is L*p*-uniform continuous on [*a*, *<sup>b</sup>*] <sup>×</sup> [*a*, *<sup>b</sup>*]; therefore,

$$I\_1(t,h) \le \sup\_{\tau \in [t,t+h] \cup [t+h,t]} \left\| \frac{\partial G}{\partial t}(\tau, u(t+h)) - \frac{\partial G}{\partial t}(t, u(t)) \right\|\_p \stackrel{h \to 0}{\longrightarrow} 0.$$

On the other hand, let *<sup>Y</sup>*(*r*) = *<sup>G</sup>*(*t*,*r*), for *<sup>t</sup>* ∈ [*a*, *<sup>b</sup>*] fixed. To conclude that lim*h*→<sup>0</sup> *<sup>I</sup>*2(*t*, *<sup>h</sup>*) = 0, we need (*Y* ◦ *u*) (*t*) = *Y* (*u*(*t*))*u* (*t*). We have that *Y* is L*p*-*C*1([*c*, *d*]) and that *u* is differentiable on [*a*, *b*], so the following existing version of the chain rule theorem applies: ([33] Theorem 2.1).

**Remark 1.** *Although not needed in the subsequent development, Lemma 4 gives in fact a general multidimensional chain rule theorem for* L*p-calculus, for the composition of a stochastic process G*(*t*,*s*) *and two deterministic functions* (*v*(*r*), *u*(*r*))*. This is the generalization of ([33] Theorem 2.1) to several variables. Indeed, let <sup>G</sup>*(*t*,*s*) *be a stochastic process on an open set* <sup>Λ</sup> <sup>⊆</sup> <sup>R</sup>2*, with* <sup>L</sup>*p-partial derivatives, <sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*t*,*s*) *and <sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>s</sup>* (*t*,*s*)*, being* <sup>L</sup>*p-continuous on* <sup>Λ</sup>*. Let <sup>v</sup>*, *<sup>u</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> <sup>R</sup> *be two <sup>C</sup>*<sup>1</sup> *deterministic functions with* (*v*(*r*), *<sup>u</sup>*(*r*)) <sup>∈</sup> <sup>Λ</sup>*. Then* d <sup>d</sup>*<sup>r</sup> G*(*v*(*r*), *u*(*r*)) = *v* (*r*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*v*(*r*), *u*(*r*)) + *u* (*r*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>s</sup>* (*v*(*r*), *u*(*r*))*. For the proof, just define G*(*t*,*r*) = *G*(*v*(*t*),*r*)*. By ([33] Theorem 2.1), <sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*t*,*r*) = *v* (*t*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*v*(*t*),*r*)*, which is* <sup>L</sup>*p-continuous on* (*t*,*r*)*. Additionally, <sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>r</sup>* (*t*,*r*) = *∂G <sup>∂</sup><sup>s</sup>* (*v*(*t*),*r*) *is* <sup>L</sup>*p-continuous. Then <sup>G</sup>*(*v*(*r*), *<sup>u</sup>*(*r*)) = *<sup>G</sup>*(*r*, *<sup>u</sup>*(*r*)) *can be* <sup>L</sup>*p-differentiated at each r, by our Lemma 4:* <sup>d</sup> <sup>d</sup>*<sup>r</sup> <sup>G</sup>*(*v*(*r*), *<sup>u</sup>*(*r*)) = *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*r*, *u*(*r*)) + *u* (*r*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>r</sup>* (*r*, *u*(*r*)) = *v* (*r*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*v*(*r*), *u*(*r*)) + *u* (*r*) *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>s</sup>* (*v*(*r*), *u*(*r*))*.*

**Proposition 3** (Random Leibniz's rule for L*p*-calculus)**.** *Let <sup>F</sup>*(*t*,*s*) *be a stochastic process on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*. Let <sup>u</sup>*, *<sup>v</sup>* : [*a*, *<sup>b</sup>*] <sup>→</sup> [*c*, *<sup>d</sup>*] *be two differentiable deterministic functions. Suppose that <sup>F</sup>*(*t*, ·) *is* <sup>L</sup>*p-continuous on* [*c*, *<sup>d</sup>*]*, for each <sup>t</sup>* <sup>∈</sup> [*a*, *<sup>b</sup>*]*, and that <sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>t</sup>* (*t*,*s*) *exists in the* <sup>L</sup>*p-sense and is* <sup>L</sup>*p-continuous on* [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*]*. Then H*(*t*) = *v*(*t*) *<sup>u</sup>*(*t*) *<sup>F</sup>*(*t*,*s*) <sup>d</sup>*s is* <sup>L</sup>*p-differentiable and*

$$H'(t) = v'(t)F(t, v(t)) - u'(t)F(t, u(t)) + \int\_{u(t)}^{v(t)} \frac{\partial F}{\partial t}(t, s) \, \mathrm{d}s$$

*(the integral is considered as an* L*p-Riemann integral).*

**Proof.** First, notice that *<sup>H</sup>*(*t*) is well-defined, since *<sup>F</sup>*(*t*, ·) is L*p*-continuous. Decompose *<sup>H</sup>*(*t*) as *H*(*t*) = *v*(*t*) *<sup>a</sup> <sup>F</sup>*(*t*,*s*) <sup>d</sup>*<sup>s</sup>* <sup>−</sup> *u*(*t*) *<sup>a</sup> <sup>F</sup>*(*t*,*s*) <sup>d</sup>*s*. Let *<sup>G</sup>*(*t*,*r*) = *r <sup>a</sup> F*(*t*,*s*) d*s*, *t* ∈ [*a*, *b*], *r* ∈ [*c*, *d*]. We have *H*(*t*) = *G*(*t*, *v*(*t*)) − *G*(*t*, *u*(*t*)).

Let us check the conditions of Lemma 4. By Lemma 2, *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>t</sup>* (*t*,*r*) = *r a ∂F <sup>∂</sup><sup>t</sup>* (*t*,*s*) d*s*, which is <sup>L</sup>*p*-continuous on [*a*, *<sup>b</sup>*] <sup>×</sup> [*c*, *<sup>d</sup>*] as a consequence of the L*p*-continuity of *<sup>∂</sup><sup>F</sup> <sup>∂</sup><sup>t</sup>* (*t*,*s*). On the other hand, *∂G <sup>∂</sup><sup>r</sup>* (*t*,*r*) = *<sup>F</sup>*(*t*,*r*), by the fundamental theorem of L*p*-calculus ([5] p. 103), with *<sup>∂</sup><sup>G</sup> <sup>∂</sup><sup>r</sup>* (*t*, ·) = *F*(*t*, ·) being L*p*-continuous. Thus, by Lemma 4,

$$\begin{split}H'(t) &= \frac{\partial G}{\partial t}(t, v(t)) + v'(t)\frac{\partial G}{\partial r}(t, v(t)) - \frac{\partial G}{\partial t}(t, u(t)) - u'(t)\frac{\partial G}{\partial r}(t, u(t)) \\ &= v'(t)F(t, v(t)) - u'(t)F(t, u(t)) + \int\_{u(t)}^{v(t)} \frac{\partial F}{\partial t}(t, s) \, \mathrm{d}s. \end{split}$$

**Remark 2** (Proposition 3 against another proof of the random Leibniz's rule)**.** *In [10, Proposition 6], a result pointing towards the conclusion of Proposition 3 was stated (in the mean square case p* = 2*, with v*(*t*) = *t, u*(*t*) = 0 *and* [*c*, *d*]=[*a*, *b*]*). However, the proof presented therein is not correct. In the notation therein, the authors proved an inequality of the form*

$$\|\|K(t,\Delta t)\|\|\_{2} \le (t-a) \max\_{x \in [a,t]} \|\|K\_1(x,t,\Delta t)\|\|\_{2} + \max\_{x \in [t,t+\Delta t]} \|\|K\_2(x,t,\Delta t)\|\|\_{2}.$$

*The authors justified correctly that K*1(*x*, *t*, Δ*t*)<sup>2</sup> →0 *and K*2(*x*, *t*, Δ*t*)<sup>2</sup> →0 *as* Δ*t* → 0*, for each x* ∈ [*a*, *b*]*. However, this fact does not imply*

$$\max\_{\mathbf{x}\in[a,t]} \|\mathsf{K}\_1(\mathbf{x},t,\Delta t)\|\_2 \stackrel{\Delta t\to\mathbb{0}}{\longrightarrow} 0, \quad \max\_{\mathbf{x}\in[t,t+\Delta t]} \|\mathsf{K}\_2(\mathbf{x},t,\Delta t)\|\_2 \stackrel{\Delta t\to\mathbb{0}}{\longrightarrow} 0,$$

*as they stated at the end of their proof. For K*1*, one has to utilize the dominated convergence theorem. For K*2*, one should use uniform continuity.*

**Remark 3** (Random Leibniz's rule cannot be proved with a mean value theorem)**.** *In the deterministic setting, both Proposition 2 and Proposition 3 can be proven with the mean value theorem. However, such proofs do not work in the random scenario, as there is no version of the stochastic mean value theorem. In previous contributions (see [15] Lemma 2.4, Corollary 2.5; [34] Lemma 3.1, Theorem 3.2), there is an incorrect version of it. For instance, if <sup>U</sup>* <sup>∼</sup> *Uniform*(0, 1) *and <sup>Y</sup>*(*t*) = {*t*>*U*}(*t*)*, <sup>t</sup>* <sup>∈</sup> [0, 1]*, then <sup>Y</sup> is mean square continuous on* [0, 1] *(notice that Y*(*t*) <sup>−</sup> *<sup>Y</sup>*(*s*)<sup>2</sup> <sup>2</sup> <sup>=</sup> <sup>|</sup>*<sup>t</sup>* <sup>−</sup> *<sup>s</sup>*|*). Suppose that there exists <sup>η</sup>* <sup>∈</sup> [0, 1] *such that* - 1 <sup>0</sup> *Y*(*s*) d*s* = *Y*(*η*) *almost surely. Then Y*(*η*) = 1 − *U almost surely. But this is not possible, since* 1 − *U* ∈ (0, 1) *and Y*(*η*) ∈ {0, 1}*. Thus, Y does not satisfy any mean square mean value theorem.*
