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A mean derivative that is not continuous in the tangent norm

Example

Let $Ω = N$ with $F = 2^{N}$ , and define $X : Ω \to [0, \infty)$ by $X (k) = k$ . Let $M_{X}$ be the model of probability laws $Q$ on $(Ω, F)$ satisfying $E_{Q} X < \infty$ , and define $ψ : M_{X} \to R$ by $ψ (Q) = E_{Q} X$ . Take $P {X = k} = c k^{- 3}$ , where $c = (\sum_{k \geq 1} k^{- 3})^{- 1}$ . Then $E_{P} X < \infty$ but $E_{P} X^{2} = \infty$ . At $P$ , designate the bounded-score one-sided paths $d P_{t, s} = (1 + t s) d P$ with $E_{P} s = 0$ and sufficiently small $t \geq 0$ as the allowed paths. Their tangent space is $L_{0}^{2} (P)$ . The right derivative on bounded scores is $s \mapsto E_{P} [X s]$ , but this map is not continuous in the $L^{2} (P)$ norm. Thus $ψ$ is not pathwise differentiable relative to this tangent space and has no finite-variance influence curve there.

Facts & Assumptions

Given: The discrete probability law and the bounded-score path class in the example.

[F1]

Pathwise differentiability and parameter influence curves: DQM uses a common-measure square-root $L^{2}$ expansion; pathwise differentiability requires the derivative to extend continuously to the tangent-space closure, and an influence curve is a square-integrable representer.

[F2]

Mean-zero square-integrable functions: $L_{0}^{2} (P)$ consists of square-integrable real functions modulo almost-sure equality with mean zero.

[F3]

Probability measures and probability spaces: a probability law has total mass one.

[F4]

Expectation of a nonnegative or integrable random variable: the expectation of a nonnegative or integrable function is its Lebesgue integral.

[F5]

For measurable representatives $f, g$ of real $L^{2} (P)$ classes, Cauchy-Schwarz inequality for $L^{2}$ gives $\int ∣ f g ∣ d P \leq ∥ f ∥_{2} ∥ g ∥_{2}$ , and The modulus of an integral is bounded by the integral of the modulus then gives $∣ E_{P} [f g] ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ .

[F6]

Dominated convergence: pointwise convergence under one integrable majorant implies convergence of the integrals.

[F7]

The Lebesgue integral is linear on $L^{1} (μ)$ : the integral is linear on integrable functions.

[F8]

Integrating against a density agrees with integrating the product: integration against a measure with density $q$ is integration of the product against $P$ .

[F9]

The measure with density $f$ relative to $μ$ : the measure with density $q$ relative to $P$ assigns $A$ the mass $\int_{A} q d P$ .

Verification

Given: $P {X = k} = c k^{- 3}$ and the bounded-score paths above. Put $μ = E_{P} X$ and $H_{m} = \sum_{k = 1}^{m} k^{- 1}$ .

1.1givenF3F4algebra

For $r > 1$ and $k \geq 2$ , monotonicity gives $k^{- r} \leq \int_{k - 1}^{k} x^{- r} d x$ , so $\sum_{k \geq 1} k^{- r} \leq 1 + \int_{1}^{\infty} x^{- r} d x < \infty$ . Taking $r = 3$ shows $c$ is well-defined and the displayed masses sum to one; taking $r = 2$ gives $μ = c \sum_{k \geq 1} k^{- 2} < \infty$ , so $P \in M_{X}$ . But $E_{P} X^{2} = c \sum_{k \geq 1} k^{- 1} = \infty$ , because each dyadic block $2^{j - 1} < k \leq 2^{j}$ contributes at least $1 / 2$ .

1.2givenF1F2F3F5F6F7F9

For bounded centered $s$ , choose $t \geq 0$ small enough that $1 + t s \geq 0$ . By [F7], $E_{P} (1 + t s) = 1$ , so [F9] gives a probability law $P_{t, s}$ with density $1 + t s$ . The identity $\sqrt{1 + u} - 1 - u / 2 = - u^{2} / [2 (1 + \sqrt{1 + u})^{2}]$ bounds the square-root remainder for $u = t s$ in $L^{2} (P)$ by $t^{2} ∥ s^{2} ∥_{2} / 2 = o (t)$ , since $s$ is bounded. Thus these paths are DQM with score $s$ under [F1]. To see their scores are dense in $L_{0}^{2} (P)$ , take $g \in L_{0}^{2} (P)$ and truncate to $g_{m} = \max (- m, \min (g, m))$ . Dominated convergence gives $g_{m} \to g$ in $L^{2} (P)$ because $∣ g_{m} - g ∣^{2} \leq g^{2}$ and the difference tends pointwise to zero. Cauchy--Schwarz and $∥ 1 ∥_{2} = 1$ give $E_{P} g_{m} \to 0$ , so $g_{m} - E_{P} g_{m}$ are bounded centered scores converging to $g$ . Hence the tangent space is $L_{0}^{2} (P)$ .

2.1givenstep 1.1F4F7F8F9algebra

Step 1.1 gives $P \in M_{X}$ and $E_{P} X < \infty$ ; since $s$ is bounded, [F8] gives the finite expectation $E_{P_{t, s}} X = E_{P} [X (1 + t s)]$ , so every designated path lies in $M_{X}$ . Linearity [F7] then gives $ψ (P_{t, s}) = E_{P_{t, s}} X = μ + t E_{P} [X s]$ , so the right derivative on each allowed score is ${\dot{ψ}}_{P} (s) = E_{P} [X s]$ .

3.1givenF1F4step 1.1step 2.1algebra

For $m \geq 2$ let $s_{m} = (X \land m) - E_{P} [X \land m]$ . These are bounded centered scores. Their norms are positive because $X \land m$ takes the distinct values $1$ and $2$ on the positive-mass points $k = 1$ and $k = 2$ . Direct summation gives $E_{P} [X s_{m}] = c H_{m} + c m \sum_{k > m} k^{- 2} - μ E_{P} [X \land m] \geq c H_{m} - μ^{2}$ , since $0 \leq X \land m \leq X$ . Also $∥ s_{m} ∥_{2}^{2} = E_{P} [(X \land m)^{2}] - (E_{P} [X \land m])^{2} \leq c H_{m} + c m^{2} \sum_{k > m} k^{- 3} \leq c H_{m} + c / 2,$ where the last bound follows from $\sum_{k > m} k^{- 3} \leq \int_{m}^{\infty} x^{- 3} d x = 1 / (2 m^{2})$ . The dyadic-block argument in step 1.1 shows $H_{m} \to \infty$ , so $∣ E_{P} [X s_{m}] ∣ / ∥ s_{m} ∥_{2} \to \infty$ . The derivative on the dense score set is therefore unbounded in the tangent norm and cannot extend continuously, contrary to the requirement in [F1].

4.1F1F5step 3.1∎

If a finite-variance influence curve $ϕ \in L_{0}^{2} (P)$ represented this derivative, then for every $m \geq 2$ Cauchy--Schwarz would give $∣ E_{P} [X s_{m}] ∣ = ∣ E_{P} [ϕ s_{m}] ∣ \leq ∥ ϕ ∥_{2} ∥ s_{m} ∥_{2}$ , contradicting step 3.1. Thus no such influence curve exists.

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