Why can't the Itô integral ∫₀ᵀ H(t) dW(t) be defined as a pathwise Riemann-Stieltjes integral?
ABecause W(t) is not measurable with respect to the Borel sigma-algebra
BBecause the Riemann-Stieltjes integral ∫f dg requires g to have bounded variation, but Brownian paths have infinite variation on every interval
CBecause H(t) may be negative, and Riemann-Stieltjes integration only works for positive integrands
DBecause Brownian motion is not continuous, and Riemann-Stieltjes requires continuity of the integrator
The Riemann-Stieltjes integral ∫f dg is well-defined when g has bounded (finite) variation. Brownian paths are almost surely of infinite variation on every interval, so the classical construction breaks down — the approximating sums don't converge. Itô's construction bypasses this by defining the integral as an L² limit: first for simple (step function) integrands where the integral is just a finite sum, then extending by density to the closure in L². The resulting integral is fundamentally a probabilistic object, not a pathwise one.
Question 2 Multiple Choice
The Itô integral ∫₀ᵀ W(t) dW(t) equals (1/2)W(T)² − (1/2)T, not (1/2)W(T)² as ordinary calculus would suggest. The extra −(1/2)T term arises because:
AThe Itô integral uses left-endpoint evaluation, which introduces a systematic bias equal to half the quadratic variation
BBrownian motion has negative drift that accumulates over time
CThe integral is computed incorrectly; the true answer is (1/2)W(T)²
DThe factor −(1/2)T is a normalization constant required to make the integral a martingale
Left-endpoint evaluation means the integrand is evaluated at tᵢ₋₁, not at the midpoint or right endpoint. In the Riemann sum Σ W(tᵢ₋₁)(W(tᵢ) - W(tᵢ₋₁)), expanding W(tᵢ) = W(tᵢ₋₁) + ΔWᵢ and using the fact that Σ(ΔWᵢ)² → T (quadratic variation) produces the correction term −T/2. If we used midpoint evaluation (Stratonovich convention), the correction disappears and we get (1/2)W(T)². The Itô choice is preferred in probability because it makes the integral a martingale — E[∫₀ᵀ W dW] = 0 — which is essential for stochastic analysis.
Question 3 Short Answer
The Itô isometry states that E[(∫₀ᵀ H(t) dW(t))²] = E[∫₀ᵀ H(t)² dt]. In your own words, explain why this is the fundamental computational tool for Itô integrals.
Think about your answer, then reveal below.
Model answer: The Itô isometry converts a question about the variance of a stochastic integral (an L² norm in probability space) into an ordinary Lebesgue integral of H² over time. This means you can compute second moments of Itô integrals without working with the Brownian integrator directly — you just integrate the square of the integrand against dt. It also shows that the Itô integral is an isometry from L²(Ω × [0,T]) to L²(Ω), which is what allows the extension from simple processes to general adapted processes by L² approximation. Without it, the construction of the integral would not close.
The isometry E[(∫H dW)²] = E[∫H² dt] is a direct consequence of independent increments: cross-terms E[H(tᵢ)ΔWᵢ · H(tⱼ)ΔWⱼ] vanish for i ≠ j because ΔWⱼ is independent of everything up to time tⱼ₋₁. Only the diagonal terms survive, giving Σ E[H(tᵢ)²] · (tᵢ₊₁ - tᵢ), which in the limit is E[∫H² dt]. This structural feature — that the L² norm of the stochastic integral depends only on the integrand — is the foundation of everything that follows.
Question 4 True / False
Every Itô integral ∫₀ᵗ H(s) dW(s), where H is adapted and square-integrable, is a martingale.
TTrue
FFalse
Answer: True
The martingale property of Itô integrals follows from left-endpoint evaluation: H(s) is adapted (known at time s), and dW(s) is independent of the filtration up to time s. This means E[H(s)dW(s) | ℱₛ] = H(s)E[dW(s)] = 0, so the integral has zero expected increment at every time. The martingale property is one of the main reasons the Itô convention is preferred over the Stratonovich convention in probability theory — it preserves the 'fair game' structure that makes martingale techniques available.