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An inverse-CDF singularity can be integrable but have infinite variation

Example

Let $U$ have the uniform law on $[0, 1)$ and define $g (u) = (1 - u)^{- 1 / 2}$ for $0 \leq u < 1$ . Extend $g$ to the closed cube by setting $g (1) = 0$ ; this endpoint convention does not change $g (U)$ . The map $g$ has distribution function $F_{X} (x) = 0$ for $x < 1$ and $F_{X} (x) = 1 - x^{- 2}$ for $x \geq 1$ , and satisfies $F_{X} (g (u)) = u$ for all $0 \leq u < 1$ (the inverse-CDF property in this example). Its expectation is finite, but $V_{H K, 1} (g) = \infty$ , so the Koksma–Hlawka bound gives no finite error guarantee for this integrand.

Facts & Assumptions

Given: A real random variable $U$ with the uniform probability law on $[0, 1)$ , represented by the identity map on $([0, 1), B ([0, 1)), P_{U})$ .

[A1]

For $0 \leq a < 1$ , $P_{U} (U \leq a) = a$ ; thus $P_{U} (U > a) = 1 - a$ .

[F1]

A real random variable is a measurable map into $R$ with its Borel sigma-algebra. (Random elements and real random variables, The Borel sigma-algebra of a topological space)

[F2]

A map between measurable spaces is measurable when the preimage of every measurable target set is measurable. (A measurable function between measurable spaces)

[F3]

The half-lines $(- \infty, x]$ generate the Borel sigma-algebra of $R$ . (Seven generating families for the Borel sigma-algebra on the real line)

[F4]

A measurable function composed with a random element is measurable. (Composition with a Borel measurable outer map preserves measurability)

[F5]

The cumulative distribution function of a real random variable is $F_{X} (x) = P (X \leq x)$ . (Cumulative distribution function of a real random variable)

[F6]

For $a > 0$ , $a^{- 1 / 2} = 1 / \sqrt{a}$ , the square root is the unique nonnegative root, squaring preserves nonstrict order on nonnegative reals, and taking reciprocals reverses positive order; the equality case follows from trichotomy. (Rational powers $a^{r}$ of a positive base, Laws of rational exponents, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives, Inverses of positives are positive, and reciprocation reverses order, Ordered field)

[F7]

For every positive integer $n$ , $0 < (1 / 2)^{n} < 1$ , and $(1 / 2)^{n} \to 0$ . (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ )

[F8]

The expectation of a nonnegative random variable is its extended Lebesgue integral. (Expectation of a nonnegative or integrable random variable)

[F9]

For a measurable $h \geq 0$ , $E [h (U)] = \int h d P_{U}$ . (Change of variables for expectation)

[F10]

For nonnegative $X$ , $E [X] = \int_{0}^{\infty} P (X > t) d t$ . (Layer-cake formulas for random variables)

[F11]

For rational $p > 1$ , $\int_{1}^{\infty} t^{- p} d t = 1 / (p - 1)$ . (The improper $p$ -test for rational exponents)

[F12]

In dimension one, upper-anchor Hardy–Krause variation is ordinary total variation. (Hardy–Krause variation anchored at the upper corner)

[F13]

Koksma–Hlawka applies to a finite nonempty point multiset and requires finite upper-anchor Hardy–Krause variation. (Koksma–Hlawka integration bound with upper-anchor variation)

Proof

1.1givenF1F2F3F4F6construct

On $[0, 1)$ define $g (u) = (1 - u)^{- 1 / 2}$ . Since $0 < 1 - u \leq 1$ , [F6] gives $g (u) \geq 1$ . For $x < 1$ , the sublevel set ${u : g (u) \leq x}$ is empty. For $x \geq 1$ , [F6] gives $g (u) \leq x$ exactly when $x^{- 2} \leq 1 - u$ , so this sublevel set is $[0, 1 - x^{- 2}]$ (with the endpoint interpreted inside $[0, 1)$ ). Each such set is Borel. Since the half-lines in [F3] generate the target Borel sigma-algebra and preimages commute with complements and countable unions, [F2] shows that $g$ is measurable. Thus $X : = g (U)$ is a real random variable by [F1] and [F4].

1.2F6F7F12givenchoose

For $n \geq 1$ put $u_{n} = 1 - 2^{- 2 n}$ . By [F7], $0 < u_{n} < 1$ . The unique nonnegative square root of $(2^{- n})^{2}$ is $2^{- n}$ , so [F6] gives $g (u_{n}) = 2^{n}$ . Given $M > 0$ , choose $n$ with $2^{- n} < 2 / (M + 1)$ using [F7]; then $g (u_{n}) > (M + 1) / 2$ . The grid with knots $0 < u_{n} < 1$ has variation at least $∣ g (u_{n}) - g (0) ∣ + ∣ g (1) - g (u_{n}) ∣ = 2 g (u_{n}) - 1 > M$ . Hence $V_{H K, 1} (g) = \infty$ by [F12].

2.1A1F5F6step 1.1algebra

By [F5] and [A1], the distribution function of $X$ is $F_{X} (x) = 0$ for $x < 1$ and $F_{X} (x) = 1 - x^{- 2}$ for $x \geq 1$ . For each $u < 1$ , $F_{X} (g (u)) = 1 - (g (u))^{- 2} = u$ , so $g$ is the inverse-CDF transform on $[0, 1)$ .

2.2A1F6F8F10F11step 1.1

For $0 \leq t < 1$ , $X \geq 1$ gives $P (X > t) = 1$ . For $t \geq 1$ , [F6] gives $X > t$ exactly when $U > 1 - t^{- 2}$ , and [A1] then gives $P (X > t) = t^{- 2}$ . By [F8], [F10], and [F11] with $p = 2$ , $E [X] = \int_{0}^{1} 1 d t + \int_{1}^{\infty} t^{- 2} d t = 1 + 1 = 2$ .

3.1F8F9step 1.1step 2.2

By [F9], change of variables for expectation applied to the measurable nonnegative $g$ and random element $U$ identifies $E [X]$ with $\int g d P_{U}$ . By [F8] and step 2.2 this value is finite and equals $2$ .

4.1F13step 3.1step 1.2∎

The expectation is finite by step 3.1, but the integrand has infinite upper-anchor variation by step 1.2. Therefore the finite-variation hypothesis in [F13] fails, so Koksma–Hlawka yields no finite error bound for this integrand.

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Sources

Art B. Owen, Monte Carlo Book: The Quasi-Monte Carlo Parts, Chapter 15 §15.11 (standard reference, not scraped)
Harald Niederreiter, Random Number Generation and Quasi-Monte Carlo Methods, Chapter 2 §2.2 (standard reference, not scraped)