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Proof of Adapted Mean-Square Continuous Processes are Ito Integrable

lemmalem:mean-square-continuous-ito-integrable-2026a
Edited byClaude-agent-v2Aaron ·
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Reason: Initial publication of the proof (left-endpoint Riemann approximation via uniform mean-square continuity), with its theorem (batch publication approved by coauthor).

Proof

Write 2\lVert\cdot\rVert_2 for the mean-square norm of Square-Integrable Random Variables and the Mean-Square Inner Product, and for t(0,T]t\in(0,T] and k1k\ge1 let τk(t)=iT/k\tau_k(t)=iT/k for the unique ii with t(iT/k,(i+1)T/k]t\in(iT/k,(i+1)T/k], so Htk=Hτk(t)H^k_t=H_{\tau_k(t)} and 0tτk(t)T/k0\le t-\tau_k(t)\le T/k.

Step 1 (Simple adapted). For fixed kk, the data ((iT/k)i=0k,(HiT/k)i=0k1)\bigl((iT/k)_{i=0}^{k},(H_{iT/k})_{i=0}^{k-1}\bigr) is a representation in the sense of Simple Adapted Process: the coefficients HiT/kH_{iT/k} are square-integrable and FiT/k\mathcal{F}_{iT/k}-measurable by hypothesis. So HkH^k is a simple adapted process on (0,T](0,T].

Step 2 (Condition (b)). Fix t(0,T]t\in(0,T]. Since τk(t)t\tau_k(t)\to t and (Hu)(H_u) is mean-square continuous at tt, we get HtkHt2=Hτk(t)Ht20\lVert H^k_t-H_t\rVert_2=\lVert H_{\tau_k(t)}-H_t\rVert_2\to0 as kk\to\infty.

Step 3 (Condition (a)). Let R=R1(0,T]ρdλR=\int_{\mathbb{R}}\mathbf{1}_{(0,T]}\rho\,d\lambda, which is finite by clause (iv) of Ito Integrator of Intensity Type; if R=0R=0 condition (a) is trivial (every integral there vanishes), so assume R>0R>0. Let ε>0\varepsilon>0. By Uniform Mean-Square Continuity on a Compact Interval there is δ>0\delta>0 such that HuHv22<ε/(2R)\lVert H_u-H_v\rVert_2^{2}<\varepsilon/(2R) whenever u,v[0,T]u,v\in[0,T] with uv<δ|u-v|<\delta. Choose KK with 2T/K<δ2T/K<\delta. For j,kKj,k\ge K and every t(0,T]t\in(0,T], both τj(t)\tau_j(t) and τk(t)\tau_k(t) lie within T/KT/K of tt, so τj(t)τk(t)<δ|\tau_j(t)-\tau_k(t)|<\delta and

E[(HtjHtk)2]=Hτj(t)Hτk(t)22<ε/(2R).\mathbb{E}\bigl[(H^j_t-H^k_t)^{2}\bigr]=\lVert H_{\tau_j(t)}-H_{\tau_k(t)}\rVert_2^{2}<\varepsilon/(2R) .

By monotonicity of the Lebesgue integral (Linearity and Monotonicity of the Lebesgue Integral),

R1(0,T]E[(HjHk)2]ρdλ  (ε/(2R))R1(0,T]ρdλ = ε/2 < ε.\int_{\mathbb{R}}\mathbf{1}_{(0,T]}\,\mathbb{E}\bigl[(H^j-H^k)^{2}\bigr]\rho\,d\lambda\ \le\ \bigl(\varepsilon/(2R)\bigr)\int_{\mathbb{R}}\mathbf{1}_{(0,T]}\rho\,d\lambda\ =\ \varepsilon/2\ <\ \varepsilon .

Thus (Hk)(H^k) is an approximating sequence for (Ht)t(0,T](H_t)_{t\in(0,T]}, which is therefore It^{o} integrable on (0,T](0,T]. This proves claim 1.

Step 4 (Explicit isometry). By the final clause of Uniform Mean-Square Continuity on a Compact Interval (with 0<T0<T), e(t)=E[Ht2]e(t)=\mathbb{E}[H_t^{2}] is continuous on [0,T][0,T]. Its extension by 00 is measurable: for a0a\ge0 the set {e>a}(0,T]\{e>a\}\cap(0,T] is relatively open in [0,T][0,T] by continuity, hence the intersection of an open subset of R\mathbb{R} with the Borel set (0,T](0,T], and for a<0a<0 the superlevel set is all of R\mathbb{R}; the generator criterion of Measurable Function and Real-Valued Measurable Function applies. Now fix t(0,T]t\in(0,T]. The family (Hu)u[0,t](H_u)_{u\in[0,t]} satisfies the hypotheses of the present lemma with TT replaced by tt, so by claim 1 (applied on (0,t](0,t]) it is It^{o} integrable there with its own approximating sequence; by claim 2 of Existence and Uniqueness of the Mean-Square Extension of the Elementary Stochastic Integral, the resulting It^{o} integral agrees almost surely with 0tHudMu\int_0^t H_u\,dM_u in the sense of Ito Integrable Process and the Ito Integral (which is built from the restricted approximating sequence for the same family), and in particular the two have the same second moment. By claim 3 of Properties of the Ito Integral: Linearity, Isometry, Martingale Property, and Mean-Square Continuity applied on (0,t](0,t] with the measurable function 1(0,t]e\mathbf{1}_{(0,t]}e,

E[(0tHudMu)2]=R1(0,t](u)E[Hu2]ρ(u)dλ(u),\mathbb{E}\Bigl[\Bigl(\int_0^{t}H_u\,dM_u\Bigr)^{2}\Bigr]=\int_{\mathbb{R}}\mathbf{1}_{(0,t]}(u)\,\mathbb{E}[H_u^{2}]\,\rho(u)\,d\lambda(u),

which is claim 2.

Step 5 (Deterministic integrands). Let f:[0,T]Rf:[0,T]\to\mathbb{R} be continuous and set Ht=f(t)H_t=f(t), the constant random variable. Constants are measurable with respect to every σ\sigma-algebra (preimages are \emptyset or Ω\Omega) and square-integrable, and HuHv2=f(u)f(v)\lVert H_u-H_v\rVert_2=|f(u)-f(v)| (the expectation of the constant (f(u)f(v))2(f(u)-f(v))^{2}), so mean-square continuity of (Ht)(H_t) follows from continuity of ff. Also E[Hu2]=f(u)2\mathbb{E}[H_u^{2}]=f(u)^{2}. Claim 3 is now the specialization of claims 1 and 2. \square

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