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First- and second-order delta method for random vectors

Statement

Let d,k≥1, let rn>0 be deterministic with rn→∞, and let Tn be Rd-valued random vectors with target θ∈Rd. Suppose Zn=rn(Tn−θ)⇒Z for an Rd-valued random vector Z. All vector norms below are Euclidean.

If g:Rd→Rk is Borel and there is a linear map A:Rd→Rk such that g(θ+h)−g(θ)=Ah+ρ1(h),∥ρ1(h)∥∥h∥⟶0(h→0, h≠0), then rn{g(Tn)−g(θ)}⇒AZ.

If, in addition, A=0 and there is a continuous symmetric bilinear map H:Rd×Rd→Rk such that g(θ+h)−g(θ)=12H[h,h]+ρ2(h),∥ρ2(h)∥∥h∥2⟶0(h→0, h≠0), then rn2{g(Tn)−g(θ)}⇒12H[Z,Z]. For scalar-valued g, H is the usual Hessian bilinear form. The second-order premise is a pointwise Peano expansion; continuous second derivatives on a neighborhood are not required.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

Stochastic order and convergence notation for finite-dimensional random vectors: OP(1) means bounded in probability by tail probabilities, and oP(1) means convergence in probability to zero.

[F2]

Finite-dimensional stochastic-order calculus and tightness: finite-dimensional weak convergence implies OP(1).

[F3]

Finite-dimensional stochastic-order calculus and tightness: a fixed continuous bilinear map sends an oP(1) factor times an OP(1) factor to oP(1).

[F4]

Vector Slutsky theorem: if Xn⇒X and Yn−c=oP(1), then (Xn,Yn)⇒(X,c), and continuous maps preserve this limit.

[F5]

The real Frechet derivative on real and complex matrix spaces with the Frobenius norm: on finite-dimensional real matrix spaces, differentiability means a real-linear first-order approximation with a remainder whose norm divided by the input norm tends to zero. Identify Rd with real column matrices and use the Euclidean/Frobenius norm equality.

[F6]

Convergence in distribution of random elements: Xn⇒X means weak convergence of their laws in the common metric target; the random elements need not share probability spaces.

Proof

Given: The objects and hypotheses in the statement.

1.1F2F6given

By [F6], Zn⇒Z is weak convergence in Rd. By [F2], this implies Zn=OP(1).

2.1F1F3step 1.1givenalgebra

Since rn−1→0, the deterministic sequence rn−1 is oP(1). Applying the bilinear product rule in [F3] to scalar multiplication gives Tn−θ=rn−1Zn=oP(1).

3.1F1F2step 1.1step 2.1givenchoosealgebraF5

Fix ε,η>0. By [F2] choose M>0 with lim sup⁡nP(∥Zn∥>M)≤η. The first-order remainder limit in the statement gives δ>0 such that ∥ρ1(h)∥≤(ε/M)∥h∥ when 0<∥h∥<δ. Step 2.1 gives P(∥Tn−θ∥≥δ)→0, so the limiting upper probability that rn∥ρ1(Tn−θ)∥>ε is at most η. Letting η↓0 proves rnρ1(Tn−θ)=oP(1). Substitution in the first expansion yields rn{g(Tn)−g(θ)}=AZn+oP(1).

3.2F1F2step 1.1step 2.1givenchoosealgebra

Fix ε,η>0. By [F2] choose M>0 with lim sup⁡nP(∥Zn∥>M)≤η. The second-order remainder limit in the statement gives δ>0 such that ∥ρ2(h)∥≤(ε/M2)∥h∥2 for 0<∥h∥<δ. Step 2.1 puts Tn−θ inside this neighborhood with probability tending to one. Outside the two tail events, rn2∥ρ2(Tn−θ)∥≤ε because rn∥Tn−θ∥=∥Zn∥. Thus rn2ρ2(Tn−θ)=oP(1); since A=0, substitution in the second expansion gives rn2{g(Tn)−g(θ)}=12H[Zn,Zn]+oP(1).

4.1F4givenstep 3.1

The map (x,e)↦Ax+e is continuous. Apply [F4] to Zn⇒Z and the oP(1) error in step 3.1 to obtain the first-order limit AZ.

5.1F4givenstep 3.2∎

The quadratic map z↦12H[z,z] is continuous. Apply [F4] to its value at Zn and the error in step 3.2 to obtain the second-order limit 12H[Z,Z].

Remarks

The first-order result is choice-free and does not require Z to be Gaussian. If AZ=0 almost surely, its first-order limit is degenerate; the second-order statement gives a different scale only when its stated expansion holds.

Depends on

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Sources