# Given Brownian motion $B$ and bounded measures on $[0,1]$, $X^\mu(\omega) = \int_0^1 B_s(\omega) d\mu(s)$ is Gaussian.

This is part (3) from Exercise 1.12 in Revuz and Yor's Continuous Martingales and Brownian Motion.

Let $$B$$ be a standard Brownian Motion, $$\mu$$ and $$\nu$$ two bounded measures associated as in 2) with $$h$$ and $$g$$. Prove that

$$X^\mu(\omega) = \int_0^1 B_s(\omega) d\mu(s) \; \text{and} \; X^\nu(\omega) = \int_0^1 B_s(\omega)d\nu(s)$$ are random variables, that the pair $$(X^\mu, X^\nu)$$ is Gaussian and that $$E[X^\mu X^\nu] = \int_0^1 \int_0^1 \inf(s,t) d\mu(s)d\nu(t) = (h,g).$$

$$h,g$$ are given as in below. $$X^\nu$$ and $$X^\mu$$ are random vairables by Fubini's Theorem. But how can we show that the pair is Gaussian and the covariance is given by $$\int_0^1 \int_0^1 \inf(s,t) d\mu(s) d\nu(t) = (h,g)$$? I would greatly appreciate any help.

Because $$t\mapsto B_t(\omega)$$ is continuous, you have the approximation $$X^\mu(\omega)=\lim_n\sum_{k=1}^{n}B_{k/n}(\omega)\cdot \mu((k-1)/n,k/n]$$ for each $$\omega$$. The sum on the right, call it $$X^\mu_n$$, is a linear combination of jointly Gaussian random variables, and so it is Gaussian. It's not hard to compute the mean and variance of $$X^\mu_n$$, and even the covariance of $$X^\mu_n$$ with $$X^\nu_n$$. The last step is to pass to the limit as $$n\to\infty$$.

• The last step requires a proof: Given a sequence of random variables $(X_n)$, where each $X_n$ is normally distributed and suppose that$X_n\rightarrow X$ pointwisely. It is not immediately clear why $X$ is also normally distributed. – Danny Pak-Keung Chan Oct 9 at 19:09
• Yes, a proof is needed, but if you use characteristic functions, it's not too difficult. – John Dawkins Oct 10 at 17:29

Define $$A(t)=\mu\left((0,t]\right)$$. We prove under the assumption that $$A$$ is continuous. (That is, $$\mu$$ is a continuous measure. However, we do not require that $$\mu$$ is absolutely continuous with respect to the Lebesgue measure.)

Fact 1: For any $$T\in[0,\infty)$$ and any Borel functions $$f,g:[0,T]\rightarrow\mathbb{R}$$ such that $$\int_{0}^{T}f^{2}dt<\infty$$, $$\int_{0}^{T}g^{2}dt<\infty$$, the random vector $$(\int_{0}^{T}f(s)dB(s),\int_{0}^{T}g(s)dB(s))$$ is a jointly normal with mean $$(0,0)$$ and covariance matrix $$\begin{pmatrix}\int_{0}^{T}f^{2}(s)ds & \int_{0}^{T}f(s)g(s)ds\\ \int_{0}^{T}f(s)g(s)ds & \int_{0}^{T}g^{2}(s)ds \end{pmatrix}$$. If I have time, I will go back to prove Fact 1 (by considering Fourier transform).

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For your problem. Note that $$A=\{A(t)\mid t\in[0,1]\}$$ is an adapted process (of course because it is deterministic) with continuous sample paths. Apply Ito product rule on $$A(t)B(t)$$, we have:

$$A(1)B(1)-A(0)B(0)=\int_{0}^{1}B(t)dA(t)+\int_{0}^{1}A(t)dB(t)+[A,B](1)$$. Clearly $$A$$ is a process with local finite variation, so the quadratic covariation $$[A,B]$$ is zero. It follows that $$\begin{eqnarray*} \int_{0}^{1}B(t)d\mu(t) & = & \int_{0}^{1}B(t)dA(t)\\ & = & A(1)B(1)-\int_{0}^{1}A(t)dB(t)\\ & = & \int_{0}^{1}A(1)dB(t)-\int_{0}^{1}A(t)dB(t).\\ & = & \int_{0}^{1}\left(A(1)-A(t)\right)dB(t) \end{eqnarray*}$$ Since the integrand is deterministic, $$\int_{0}^{1}B(t)d\mu(t)$$ is normal.

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Suppose that $$\mu$$ and $$\nu$$ are continuous, finite measures on $$\mathcal{B}([0,1])$$. Define $$A_{1}(t)=\mu\left((0,t]\right)$$ and $$A_{2}(t)=\nu\left((0,t]\right)$$. By the above discussion, $$\int_{0}^{1}B(t)d\mu(t)=\int_{0}^{1}\left(A_{1}(1)-A_{1}(t)\right)dB(t)$$ and $$\int_{0}^{1}B(t)d\nu(t)=\int_{0}^{1}\left(A_{2}(1)-A_{2}(t)\right)dB(t)$$. Note that the integrands are deterministic, so from Fact 1, $$\left(\int_{0}^{1}B(t)d\mu(t),\int_{0}^{1}B(t)d\nu(t)\right)$$ is jointly normal and their covariance and be computed directly.

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Proof of Fact 1: Let $$(\Omega,\mathcal{F},P)$$ be a probability space. Let $$\mathbb{F}=\{\mathcal{F}_{t}\mid t\in[0,\infty)\}$$ be a filtration such that $$\mathcal{F}_{0}$$ contains all $$P$$-null sets. Let $$W=\{W_{t}\mid t\in[0,\infty)\}$$ be a standard Wiener process adapted to $$\mathbb{F}$$. (i.e., $$W_{t}$$ is $$\mathcal{F}_{t}$$-measurable and for any $$0\leq s, $$\mathcal{F}_{s}$$and $$W_{t}-W_{s}$$ are independent.)

Let $$T\in[0,\infty)$$. Let $$f,g:[0,T]\rightarrow\mathbb{R}$$ be Borel functions such that $$\int_{0}^{T}f^{2}(t)dt<\infty$$ and $$\int_{0}^{T}g^{2}(t)dt<\infty$$. Define random variables $$\xi=\int_{0}^{T}f(t)dW(t)$$ and $$\eta=\int_{0}^{T}g(t)dW(t)$$. Then $$(\xi,\eta)$$ is jointly normal with mean $$(0,0)$$ and covariance matrix $$\begin{pmatrix}\int_{0}^{T}f^{2} & \int_{0}^{T}fg\\ \int_{0}^{T}fg & \int_{0}^{T}g^{2} \end{pmatrix}.$$

Proof: Define processes $$X$$, $$Y$$ by $$X_{t}=\int_{0}^{t}f(s)dW(s)$$ and $$Y_{t}=\int_{0}^{t}g(s)dW(s)$$. Since $$E\left[\int_{0}^{T}f^{2}(t)dt\right]<\infty$$ and $$E\left[\int_{0}^{T}g^{2}(t)dt\right]<\infty$$, $$X=\{X(t)\mid t\in[0,T]\}$$ and $$Y=\{Y(t)\mid t\in[0,T]\}$$ are $$L^{2}$$-bounded continuous martingales. Let $$\theta_{1},\theta_{2}\in\mathbb{R}$$ be arbitrary. Define process $$Z=\{Z(t)\mid t\in[0,T]\}$$ by $$Z(t)=\exp\left(i(\theta_{1}X(t)+\theta_{2}Y(t))\right)$$. By Ito's lemma, we have: $$\begin{eqnarray*} dZ(t) & = & Z(t)\left(i(\theta_{1}dX_{t}+\theta_{2}dY_{t})\right)+\frac{1}{2}Z(t)\left((i(\theta_{1}dX_{t}+\theta_{2}dY_{t}))^{2}\right)\\ & = & Z(t)\left\{ i(\theta_{1}f(t)+\theta_{2}g(t))dW(t)-\frac{1}{2}\left(\theta_{1}^{2}f^{2}(t)+2\theta_{1}\theta_{2}f(t)g(t)+\theta_{2}g^{2}(t)\right)dt\right\} . \end{eqnarray*}$$ That is, $$Z(t)-1=\int_{0}^{t}Z(s)\cdot i(\theta_{1}f(s)+\theta_{2}g(s))dW(s)-\frac{1}{2}\left\{ \theta_{1}^{2}\int_{0}^{t}f^{2}(s)Z(s)ds+2\theta_{1}\theta_{2}\int_{0}^{t}f(s)g(s)Z(s)ds+\theta_{2}^{2}\int_{0}^{t}g^{2}(s)Z(s)ds\right\} .$$ Now, denote $$m(s)=E\left[Z(s)\right]$$. Observe that $$|Z(s)|=1$$, so $$|m(s)|\leq1$$.

Taking expectation on both sides. Note that $$E\left[\int_{0}^{T}\left\{ Z(s)\cdot i(\theta_{1}f(s)+\theta_{2}g(s))\right\} ^{2}ds\right]<\infty$$, so $$\left(\int_{0}^{t}Z(s)\cdot i(\theta_{1}f(s)+\theta_{2}g(s))dW(s)\right)_{t\in[0,T]}$$ is a martingale and hence its expectation equals 0. Therefore, we obtain:

$$m(t)-1=-\frac{1}{2}\left\{ \theta_{1}^{2}\int_{0}^{t}f^{2}(s)m(s)ds+2\theta_{1}\theta_{2}\int_{0}^{t}f(s)g(s)m(s)ds+\theta_{2}^{2}\int_{0}^{t}g^{2}(s)m(s)ds\right\} .$$ This implies that $$m'(t)=m(t)H(t),$$ where $$H(t)=-\frac{1}{2}\left(\theta_{1}^{2}f^{2}(t)+2\theta_{1}\theta_{2}f(t)g(t)+\theta_{2}^{2}g^{2}(t)\right)$$. Note that $$m(0)=1$$. Solving the differential equations yield: $$\begin{eqnarray*} m(t) & = & m(0)\int_{0}^{t}H(s)ds\\ & = & \int_{0}^{t}H(s)ds. \end{eqnarray*}$$ In particular, $$m(T)=\int_{0}^{T}H(s)ds.$$ Recall that $$(\theta_{1},\theta_{2})\mapsto E\left[\exp\left(i(\theta_{1}\xi+\theta_{2}\eta)\right)\right]=m(T)$$ is the characteristic function for the random vector $$(\xi,\eta)$$.

Recall that a random vector $$(\xi_{1},\xi_{2})$$ is jointly normal with mean $$(\mu_{1},\mu_{2})$$ and covariance matrix $$\Sigma$$ iff for any $$\theta_{1},\theta_{2}\in\mathbb{R}$$, $$E\left[\exp\left(i(\theta_{1}\xi_{1}+\theta_{2}\xi_{2})\right)\right]=\exp\left(i(\mu_{1}\theta_{1}+\mu_{2}\theta_{2})-\frac{1}{2}(\theta_{1},\theta_{2})\Sigma\begin{pmatrix}\theta_{1}\\ \theta_{2} \end{pmatrix}\right).$$ By writing $$m(T)$$ out, the results follow.

• For the most general case, for example, if $A$ is not continuous, it is outside my capability. I have just studied a little bit about stochastic calculus. Sorry... – Danny Pak-Keung Chan Oct 9 at 18:33