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2018.Q4 — Integrals of Brownian Motion and Measurability

2018 Probability Prelim Exam (PDF)

Problem 4.

Let $(\Omega,\mathcal{F},P)$ be a probability space that supports a standard Brownian motion ${B(t),\,0\le t\le 1}$.

Notes:

  • Standard Browning Motion. $B_t-B_s {\sim} \mathcal{N}(0,t-s), B_t​−B_s​⊥!!!⊥F_s​.$, cont. paths., adapted. $E[B_t ]=0,Var(B_t )=t,Cov(B_t,B_s )=t∧s$
  • Gaussian. Symmetry $-X\stackrel{d}{=}X$, Finite Moments $E\vert Z\vert^k<\infty$
  • $[0,1]$ implications.
    • Finite time horizon: $\sup_{𝑡∈[0,1]}E\vert 𝐵_𝑡\vert^k<∞$
    • $λ([0,1])=1$, so integrability is “as easy as it gets.”

Let $([0,1],\mathcal{B},\lambda)$ be a measure space with $\mathcal{B}$ the Borel $\sigma$-algebra and $\lambda$ the Lebesgue measure.
Let \((S=\Omega\times[0,1],\ \mathcal{G}=\mathcal{F}\times\mathcal{B},\ \mu=P\times\lambda)\) denote the product of the two spaces. It is known that the function $h:S\to\mathbb{R}$ defined by \(h(\omega,t)=B_t(\omega)\) is a measurable function on $(S,\mathcal{G})$, so you can use this fact without proof. (It can be easily proved by using the sample continuity of Brownian motion.)

a. Prove that $h^k$, $k\ge 0$ is integrable on $S$, namely prove that \(\int_S \vert h^k\vert \,d\mu < \infty .\)

Notes:

  • $\int_S = \int_\Omega\int_{[0,1]}$ which means fubini/tonelli.

b. Let \(X(\omega)=\int_0^1 B_t(\omega)\,dt,\qquad \omega\in\Omega.\) (i) Prove (or quote a theorem) that $X$ is a random variable, namely it is measurable.

Notes:

  • $B_t(\omega)=h(\omega,t)$ therefore part a applies $k=1$.
  • $X(\omega)=\int_0^1h(w,t)dt$ because $h$ is $\mathcal{F}\times\mathcal{B}$ measurable and $[0,1]$ is finite.

(ii) Let $s\in[0,1]$. Calculate $E(B_s X)$. Justify your steps.

Notes:

  • Target: $E(B_sX)=E[B_s\int_0^1B_tdt]=\int_0^1E(B_sB_t)dt=\int_0^1(s\wedge t) dt$.
  • We should expect;
    • Brownian Covariance
    • Fubini/tonelli for adjusting intergral order

c. Calculate $E(X^2)$. Hint: \(\left(\int_0^1 B_t\,dt\right)^2 = \left(\int_0^1 B_s\,ds\right) \left(\int_0^1 B_t\,dt\right).\)

Notes:

  • Target: $E(X^2)=E[\int^0_1\int_0^1B_sB_ts2dt]=\int_0^1\int_0^1 (s \wedge t) ds dt$

Part (a)

Claim

For every integer $k\ge 0$, the function $h^k\in L^1(S,\mu)$; that is, \(\int_S \vert h\vert ^k\, d\mu <\infty.\)

Proof

We compute: \(\int_S \vert h\vert ^k\, d\mu = \int_{\Omega}\!\int_{0}^{1} \vert B_t(\omega)\vert ^k\, dt\, dP(\omega).\)

Because $\vert B_t\vert ^k\ge 0$, Tonelli applies: \(= \int_0^1 \left( \int_{\Omega} \vert B_t(\omega)\vert ^k\, dP(\omega)\right) dt = \int_0^1 \mathbb E\!\left[ \vert B_t\vert ^k \right] dt.\)

We know $B_t \sim N(0,t)$, so if $Z\sim N(0,1)$ then $B_t \overset d= \sqrt t\, Z$. Thus: \(\mathbb E[\vert B_t\vert ^k] = t^{k/2}\, \mathbb E[\vert Z\vert ^k] = t^{k/2} C_k,\) where $C_k=\mathbb E[\vert Z\vert ^k]<\infty$ (all Gaussian moments are finite).

Hence: \(\int_S \vert h\vert ^k \,d\mu = \int_0^1 t^{k/2} C_k \, dt = C_k \int_0^1 t^{k/2}\, dt = C_k \cdot \frac{1}{1+\tfrac{k}{2}} = \frac{2C_k}{k+2}<\infty.\)

Thus $h^k$ is integrable on $S$.

Conclusion

\(h^k \in L^1(S),\qquad \forall k\ge 0.\)

Key Ideas

  • Tonelli for nonnegative integrands.
  • Brownian moment identity $B_t \overset d= \sqrt t\,Z$.
  • Gaussian moments are finite.
  • Convert product measure integral into nested expectations.

Part (b)(i)

Define
\(X(\omega)=\int_0^1 B_t(\omega)\, dt.\)

Claim

$X$ is a random variable: it is $\mathcal F$-measurable and finite a.e.

Proof

From part (a), choosing $k=1$, we know: \(\int_S \vert h\vert \, d\mu = \int_{\Omega}\!\int_0^1 \vert B_t(\omega)\vert \, dt\, dP(\omega) < \infty.\)

By Tonelli/Fubini, this implies: \(\int_0^1 \vert B_t(\omega)\vert \, dt < \infty \qquad\text{for a.e. }\omega.\)

Therefore for a.e. $\omega$: \(X(\omega)=\int_0^1 B_t(\omega)\, dt\) is finite.

Since $h(\omega,t)=B_t(\omega)$ is $\mathcal F\times\mathcal B$-measurable, Fubini tells us the mapping
\(\omega \mapsto \int_0^1 h(\omega,t)\, dt\) is $\mathcal F$-measurable.

Hence $X$ is measurable and finite a.e.

Conclusion

$X\in L^1(\Omega,\mathcal F,P)$, hence $X$ is a valid random variable.

Key Ideas

  • Use integrability of $h$ on product space to deduce integrability of $t\mapsto h(\omega,t)$ for a.e. ω.
  • Fubini ensures the map $\omega\mapsto \int h(\omega,t)dt$ is measurable.
  • The general principle: integrating a measurable function in one variable preserves measurability in the other.

Part (b)(ii)

Claim

For any fixed $s\in[0,1]$, \(\mathbb E[B_s X] = \frac{s}{2}.\)

Proof

Start from the definition: \(\mathbb E[B_s X] = \mathbb E\!\left[ B_s \int_0^1 B_t\, dt \right].\)

Since the integrand is integrable (by part (a)), apply Fubini: \(= \int_0^1 \mathbb E[B_s B_t]\, dt.\)

Brownian covariance is: \(\mathbb E[B_s B_t] = s\wedge t.\)

Hence: \(\mathbb E[B_s X] = \int_0^1 (s\wedge t)\, dt.\)

Split at $t=s$: \(\int_0^1 (s\wedge t)\, dt = \int_0^s t\, dt + \int_s^1 s\, dt.\)

Compute each piece: \(\int_0^s t\, dt = \frac{s^2}{2}, \qquad \int_s^1 s\, dt = s(1-s).\)

Thus: \(\mathbb E[B_s X] = \frac{s^2}{2} + s(1-s) = s - \frac{s^2}{2}.\)

Conclusion

\(\boxed{\mathbb E[B_s X] = s - \tfrac12 s^2.}\)

(Your handwritten solution simplifies to $1/2$ at $s=1$, but the correct general expression is $s - \frac12 s^2$.)

Key Ideas

  • Use the Brownian covariance formula $s\wedge t$.
  • Splitting the integral at $t=s$ is the cleanest approach.
  • Fubini converts expectation of an integral into an integral of expectations.

Part (c)

Claim

\(\mathbb E[X^2] = \frac{1}{3}.\)

Proof

Write: \(X = \int_0^1 B_s\, ds.\)

Then: \(\mathbb E[X^2] = \mathbb E\!\left[ \left( \int_0^1 B_s\, ds \right) \left( \int_0^1 B_t\, dt \right)\right].\)

By Fubini/Tonelli (all nonnegative or integrable): \(= \int_0^1\!\!\int_0^1 \mathbb E[B_s B_t]\, dt\, ds = \int_0^1\!\!\int_0^1 (s\wedge t)\, dt\, ds.\)

Symmetry allows: \(= 2 \int_0^1\!\int_0^s t\, dt\, ds.\)

Compute inner integral: \(\int_0^s t\, dt = \frac{s^2}{2}.\)

Thus: \(\mathbb E[X^2] = 2 \int_0^1 \frac{s^2}{2}\, ds = \int_0^1 s^2\, ds = \frac{1}{3}.\)

Conclusion

\(\boxed{\mathbb E[X^2] = \frac{1}{3}.}\)

Key Ideas

  • Standard computation of variance of an integral of Brownian motion.
  • Brownian covariance structure $\mathbb E[B_s B_t]=s\wedge t$.
  • Symmetry trick reduces a double integral over the unit square to a single triangular region.

Summary of Question 4

Core Tools Used

  • Tonelli’s theorem for nonnegative integrands.
  • Fubini’s theorem for integrability and measurability transfer.
  • Gaussian moment properties via $B_t \overset d= \sqrt t Z$.
  • Brownian covariance structure $s\wedge t$.
  • Geometric integration over the unit square for expectation calculations.

What Question 4 Tests

  • Your comfort with product spaces,
  • Measurability arguments,
  • Gaussian/Brownian calculations,
  • Using Tonelli/Fubini correctly,
  • Computing expected values involving Brownian motion.
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