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A finite stochastic adjoint gives the discrete gradient under domination

Statement

Gradients of scalar functions are Euclidean column vectors, with $D g (θ) [v] = ⟨ \nabla g (θ), v ⟩$ under The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ; vector expectations are taken coordinatewise.

Let $(Ω, F, P)$ be a probability space, let $(E_{k}, E_{k})$ be measurable spaces for $0 \leq k \leq N$ , and let each $ξ_{k} : Ω \to E_{k}$ be a measurable random element; write $ξ = (ξ_{0}, \dots, ξ_{N})$ . Let $Θ \subseteq R^{d}$ be open, fix $θ_{0} \in Θ$ , and choose an open ball $U$ with $θ_{0} \in U \subseteq Θ$ . Let $d, m \geq 1$ , let $N \geq 0$ be a finite integer, and let the states lie in $R^{m}$ . The maps are $u_{0} : Θ \times E_{0} \to R^{m}$ , $Ψ_{k} : R^{m} \times Θ \times E_{k + 1} \to R^{m}$ , $ℓ_{k} : R^{m} \times Θ \to R$ , and $ϕ : R^{m} \times Θ \to R$ . Gradients of scalar functions are columns; $D_{u} Ψ_{k}$ is an $m \times m$ Jacobian and $D_{θ} Ψ_{k}$ is an $m \times d$ Jacobian. Let $u_{0} = u_{0} (θ, ξ_{0}), u_{k + 1} = Ψ_{k} (u_{k}, θ, ξ_{k + 1}) (0 \leq k < N) .$ Assume the displayed maps are measurable for the corresponding product sigma-algebras (Borel on Euclidean factors and $E_{k}$ on each input factor), and that there is a common measurable null set outside which their sections are continuously differentiable in their state and parameter variables on $U$ . Define $F_{N} (θ, ξ) : = ϕ (u_{N}, θ) + \sum_{k = 0}^{N - 1} ℓ_{k} (u_{k}, θ), J_{N} (θ) : = E [F_{N} (θ, ξ)] .$ Assume $F_{N} (θ, ξ)$ is integrable for every $θ \in U$ , and that there is a nonnegative integrable random variable $M$ with $∥ \nabla_{θ} F_{N} (θ, ξ) ∥_{2} \leq M$ for all $θ \in U$ almost surely, while $E ∥ \nabla_{θ} F_{N} (θ_{0}, ξ) ∥_{2}^{2} < \infty$ . Put $S_{0} = D_{θ} u_{0}$ , propagate $S_{k + 1} = D_{u} Ψ_{k} S_{k} + D_{θ} Ψ_{k},$ and define reverse variables by $p_{N} = \nabla_{u} ϕ (u_{N}, θ), p_{k} = \nabla_{u} ℓ_{k} (u_{k}, θ) + (D_{u} Ψ_{k})^{T} p_{k + 1} .$ Then the discrete adjoint integrand $Γ_{N} = \nabla_{θ} ϕ + S_{0}^{T} p_{0} + \sum_{k = 0}^{N - 1} (\nabla_{θ} ℓ_{k} + (D_{θ} Ψ_{k})^{T} p_{k + 1})$ satisfies $\nabla J_{N} (θ_{0}) = E [Γ_{N} (θ_{0}, ξ)]$ .

For a family of discretizations indexed by $h$ , if each discrete objective satisfies the preceding pathwise interchange hypotheses, $E ∥ Γ_{h} (θ_{0}) - Γ (θ_{0}) ∥_{2} \to 0$ , and the continuous-limit objective has derivative $\nabla J (θ_{0}) = E [Γ (θ_{0})]$ under the same pathwise interchange conditions, then $\nabla J_{h} (θ_{0}) \to \nabla J (θ_{0})$ . This conclusion gives no rate unless a quantitative bound on $E ∥ Γ_{h} (θ_{0}) - Γ (θ_{0}) ∥_{2}$ is also supplied. Euler–Maruyama is the special case $Ψ_{k} (u, θ, Δ W_{k}) = u + b (u, θ) h_{k} + σ (u, θ) Δ W_{k}$ , with $Δ W_{k}$ the simulated Brownian increment.

All tangent, reverse and adjoint derivatives are evaluated on the common full-probability event where the sample maps are $C^{1}$ . Define $Γ_{N} = 0$ on its measurable null complement, and use this version in expectations.

Facts & Assumptions

Given: The finite simulation graph, its explicitly measurable input random elements and product-measurable maps, its differentiability and pathwise envelope hypotheses, and, for the final convergence assertion, $L^{1}$ convergence of the discrete adjoint integrands and the stated derivative representation of the continuous-limit objective.

[F1]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : derivatives of compositions of differentiable Euclidean maps are obtained by composing their total derivatives.

[F2]

The principle of mathematical induction: a property true at the initial index and inherited at each successor holds at every finite index.

[F3]

A dominated pathwise simulation gradient is unbiased: an integrable common envelope for a continuous pathwise gradient allows differentiation of its expected objective, with derivative equal to the expected pathwise derivative.

[F4]

Random elements and real random variables: each $ξ_{k}$ is a measurable random element. A constant parameter map is measurable because each measurable-set preimage is either empty or the whole probability space.

[F5]

The reverse recursion is the exact gradient of a finite state-step objective: for a finite deterministic $C^{1}$ state-step graph, the tangent and reverse recursions in the Statement give its exact objective gradient.

[F6]

Composition with a Borel measurable outer map preserves measurability: the composite of two measurable maps is measurable once the inner map is measurable into the outer map's measurable domain.

[F7]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ and Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ , and $∥ z ∥_{2} = \sup_{∥ v ∥_{2} \leq 1} ⟨ v, z ⟩$ (equality is attained by $v = z / ∥ z ∥_{2}$ when $z \neq 0$ ).

[F8]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear for integrable scalar random variables.

[F9]

Coordinate maps and permutations are measurable on finite products: a finite tuple of measurable coordinate maps is measurable into the finite product sigma-algebra.

[F10]

Arithmetic and lattice operations preserve measurability whenever they are defined: finite sums, differences, real scalar multiples and products of finite real measurable functions are measurable.

[F11]

Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable: a pointwise limit of a sequence of measurable functions is measurable.

Proof

1.1givenF1F2F3F4F6F9F10F11

Fix $ϑ \in U$ . By [F4] the constant map $ω \mapsto ϑ$ and each $ξ_{k}$ are measurable. The finite-tuple lemma [F9] first makes $ω \mapsto (ϑ, ξ_{0} (ω))$ measurable; joint measurability of the initial-state map and composition [F6] then make $u_{0} (ϑ, ξ_{0})$ measurable. If $u_{k}$ is measurable, [F9] makes $ω \mapsto (u_{k} (ω), ϑ, ξ_{k + 1} (ω))$ measurable into the step map's finite product domain, and [F6] makes $u_{k + 1}$ measurable. Finite induction [F2] gives every state. Applying [F9] and [F6] to $(u_{k}, ϑ)$ in each payoff map and then [F10] to the finite sum proves measurability of $F_{N} (ϑ, ξ)$ , also when $N = 0$ . Let $A$ be the common measurable null set excluded in the Statement. Fix a coordinate $j$ and put $h_{n} = 1 / (n + 1)$ for $n \geq 0$ . If $ϑ + h_{n} e_{j} \in U$ , form the measurable difference quotient $Q_{j, n} = (F_{N} (ϑ + h_{n} e_{j}, ξ) - F_{N} (ϑ, ξ)) / h_{n}$ ; otherwise put $Q_{j, n} = 0$ . Measurability follows from the fixed-parameter argument and [F10]. Replace each $Q_{j, n}$ by zero on $A$ : equivalently multiply it by the measurable indicator of $A^{c}$ , which remains measurable by [F10]. Since $U$ is open, these modified quotients converge at every outcome to the version of $\partial_{j} F_{N} (ϑ, ξ)$ given by its derivative on $A^{c}$ and zero on $A$ . By [F11] that version is measurable. This works for every fixed $ϑ \in U$ and every coordinate, as required by [F3], without assuming the probability space is complete. The initial state is differentiable with sensitivity $S_{0} = D_{θ} u_{0}$ . If $u_{k}$ is differentiable, [F1] applied to the parameter-dependent composition $u_{k + 1} (θ) = Ψ_{k} (u_{k} (θ), θ, ξ_{k + 1})$ , then evaluated at $θ_{0}$ , gives $D_{θ} u_{k + 1} = D_{u} Ψ_{k} D_{θ} u_{k} + D_{θ} Ψ_{k}$ . Thus the stated tangent recursion holds for every $k < N$ by [F2].

2.1givenF1step 1.1

Applying the chain rule to $F_{N}$ at $θ_{0}$ gives its pathwise gradient $\nabla_{θ} F_{N} = \nabla_{θ} ϕ + S_{N}^{T} p_{N} + \sum_{k = 0}^{N - 1} (\nabla_{θ} ℓ_{k} + S_{k}^{T} \nabla_{u} ℓ_{k}),$ where $p_{N} = \nabla_{u} ϕ$ .

3.1givenF5step 1.1step 2.1algebra

Fix an input realization $ξ$ outside the common null set and set $x_{0}^{ξ} (θ) = u_{0} (θ, ξ_{0})$ , $Ψ_{k}^{ξ} (u, θ) = Ψ_{k} (u, θ, ξ_{k + 1})$ , $ℓ_{k}^{ξ} (u, θ) = ℓ_{k} (u, θ)$ , and $ϕ^{ξ} (u, θ) = ϕ (u, θ)$ . For $N \geq 1$ this is a finite deterministic $C^{1}$ state-step graph on $U$ , so [F5] applies. For $N = 0$ , $p_{0} = p_{N} = \nabla_{u} ϕ$ and step 2.1 directly gives the derivative $\nabla_{θ} ϕ + S_{0}^{T} p_{0} = Γ_{0}$ . For each $k < N$ , the definitions of $p_{k}$ and $S_{k + 1}$ imply $S_{k}^{T} p_{k} = S_{k}^{T} \nabla_{u} ℓ_{k} + S_{k + 1}^{T} p_{k + 1} - (D_{θ} Ψ_{k})^{T} p_{k + 1} .$ Rearrange and sum these identities over $k$ : the $S_{k}^{T} p_{k}$ terms telescope, and $p_{N} = \nabla_{u} ϕ$ . Thus $S_{N}^{T} p_{N} + \sum_{k = 0}^{N - 1} S_{k}^{T} \nabla_{u} ℓ_{k} = S_{0}^{T} p_{0} + \sum_{k = 0}^{N - 1} (D_{θ} Ψ_{k})^{T} p_{k + 1},$ so the gradient in step 2.1 is exactly $Γ_{N}$ , agreeing with [F5] when $N \geq 1$ .

4.1givenF3step 3.1

The pathwise derivative at $θ_{0}$ is $Γ_{N} (θ_{0}, ξ)$ by step 3.1. The Statement supplies integrability of $F_{N}$ , the common integrable derivative envelope on $U$ , and square-integrability at $θ_{0}$ . Thus [F3] gives $\nabla J_{N} (θ_{0}) = E [Γ_{N} (θ_{0}, ξ)]$ .

5.1givenF7F8step 4.1algebra∎

In the discretization family, step 4.1 gives $\nabla J_{h} (θ_{0}) = E [Γ_{h} (θ_{0})]$ . For any integrable vector $Z$ , [F7] and [F8] give, for every $∥ v ∥_{2} \leq 1$ , $⟨ v, E Z ⟩ = E ⟨ v, Z ⟩ \leq E ∥ Z ∥_{2} .$ Taking the supremum over such $v$ and using [F7] yields $∥ E Z ∥_{2} \leq E ∥ Z ∥_{2}$ . Applying this to $Z = Γ_{h} (θ_{0}) - Γ (θ_{0})$ yields $∥ \nabla J_{h} (θ_{0}) - \nabla J (θ_{0}) ∥_{2} = ∥ E [Γ_{h} (θ_{0}) - Γ (θ_{0})] ∥_{2} \leq E ∥ Γ_{h} (θ_{0}) - Γ (θ_{0}) ∥_{2} ⟶ 0$ by the assumed $L^{1}$ convergence. No state-payoff weak-convergence premise appears in this argument.

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Sources

Michael C. Fu, Stochastic Gradient Estimation, preprint of Handbook of Simulation Optimization, §3 (standard reference, not scraped)
Paul Glasserman, Monte Carlo Methods in Financial Engineering, §§6–7 (standard reference, not scraped)