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A Brownian hitting time has infinite mean

Example

Assume the Axiom of Choice and let $B$ be a standard Brownian motion Brownian motion. Use the everywhere-continuous, zero-start representative fixed in Distribution of a one-sided Brownian hitting time: replace the path by zero outside a measurable probability-one event of continuity and zero start, retaining the notation $B$ . Let $a > 0$ and $τ_{a} = \inf {t \geq 0 : B_{t} = a}$ , with $\inf \emptyset = + \infty$ . Then $τ_{a}$ is a measurable extended random variable and $τ_{a} < \infty$ almost surely, while $E [τ_{a}] = \int_{0}^{\infty} t g (t) d t = + \infty, g (t) = \frac{a}{(2 π t^{3})^{1 / 2}} e^{- a^{2} / (2 t)} .$ Thus almost-sure finiteness does not imply integrability for this random time.

Facts & Assumptions

Given: AC, a standard Brownian motion $B$ in the stated everywhere-continuous zero-start representative, and $a > 0$ .

[F1]

For the representative fixed in the statement, $τ_{a}$ is measurable and finite almost surely, and on $t > 0$ its law has the displayed density $g$ . One-dimensional Brownian motion hits every point almost surely Distribution of a one-sided Brownian hitting time

[F2]

For a nonnegative random variable with a density, the expectation is the integral of $t$ against that density; integration against a density is integration of the product with the density. Expectation of a nonnegative or integrable random variable Integrating against a density agrees with integrating the product

[F3]

Monotone convergence applies to nonnegative integrands. Monotone convergence for the integral

[F4]

AC is the standing hypothesis under which the Brownian and hitting-time interfaces in [F1] are supplied; this expectation calculation makes no additional selection. The Axiom of Choice

Verification

technique · direct

1.1

Since $τ_{a} \geq 0$ and its law has density $g$ by [F1], [F2] gives $E [τ_{a}] = \int_{0}^{\infty} t g (t) d t = a (2 π)^{- 1 / 2} \int_{0}^{\infty} t^{- 1 / 2} e^{- a^{2} / (2 t)} d t$ , the last integrand being nonnegative.

F1F2given

2.1

For $t \geq a^{2}$ the exponent satisfies $a^{2} / (2 t) \leq 1 / 2$ , so $e^{- a^{2} / (2 t)} \geq e^{- 1 / 2}$ ; hence the integrand in step 1.1 is bounded below on $[a^{2}, \infty)$ by $a (2 π)^{- 1 / 2} e^{- 1 / 2} t^{- 1 / 2}$ .

algebra

3.1

For every integer $j \geq 0$ , on $I_{j} = [2^{j} a^{2}, 2^{j + 1} a^{2}]$ one has $t^{- 1 / 2} \geq (2^{j + 1} a^{2})^{- 1 / 2}$ , so $\int_{I_{j}} t^{- 1 / 2} d t \geq a 2^{(j - 1) / 2}$ . These lower bounds do not tend to zero and their partial sums diverge. Monotone convergence over the increasing finite unions of the $I_{j}$ , together with step 2.1, therefore gives $\int_{0}^{\infty} t^{- 1 / 2} e^{- a^{2} / (2 t)} d t = + \infty$ and hence $E [τ_{a}] = + \infty$ .

F3step 1.1step 2.1

4.1

The comparison with finite almost-sure values is the point of the example: [F1] gives $τ_{a} < \infty$ almost surely, so the random variable is finite-valued almost surely while its expectation is infinite; the divergence comes from the polynomial tail $t^{- 1 / 2}$ of the first-moment integrand and not from any exceptional path. The cases $a = 0$ (where $τ_{0} = 0$ ) and $a < 0$ are excluded by the hypothesis $a > 0$ . AC is used only through [F4].

F1F4givenstep 3.1∎

Source notes

Lawler, Section 2.7, evaluates the first-passage density and records the divergent first moment; Durrett, equation (7.4.6), gives the same density. The example avoids any integration-by-parts argument and bounds the first-moment integrand directly.

Depends on

Used by

Almost-sure finiteness does not imply integrability Counterexample

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Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Section 2.7 (standard reference, not scraped)
Rick Durrett, Probability: Theory and Examples, fifth edition, equation (7.4.6) (standard reference, not scraped)