Question

Let \(\\{B(t) ; 0 \leq t \leq 1\\}\) be a standard Brownian motion process and let \(B(I)=B(t)-B(s)\), for \(I=(s, t], 0 \leq s \leq t \leq 1\) be the associated Gaussian ran dom measure. Validate the assertion that \(U=\int_{0}^{1} f(s) d B(s)\) and \(V=\int_{0}^{1} g(s)\). \(d B(s)\) are independent random variables whenever \(f\) and \(g\) are bounded continuous functions satisfying \(\int_{0}^{1} f(s) g(s) d s=0\).

Short answer

To validate the assertion that random variables \(U\) and \(V\) are independent when the given conditions are satisfied, i.e., when \(f\) and \(g\) are bounded continuous functions that satisfy the integral condition \(\int_{0}^{1} f(s)g(s)ds=0\), we calculate their covariance and show that it is equal to zero. Using the properties of Brownian motion and Itô integrals, we find that \(E[U] = E[V] = 0\) and \(E[UV] = \int_{0}^{1} f(s)g(s)ds\). Therefore, \(Cov(U, V) = \int_{0}^{1} f(s)g(s)ds = 0\), proving that \(U\) and \(V\) are independent random variables under the given conditions.

Step-by-step solution

Recall that the covariance of two random variables \(X\) and \(Y\) is defined as: \[Cov(X, Y) = E[(X - E[X])(Y - E[Y])] = E[XY] - E[X]E[Y]\] Applying the definition of covariance to the given random variables \(U\) and \(V\), we have: \[Cov(U, V) = E[UV] - E[U]E[V]\] #Step 2: Calculate the Expectation of U, V, and UV#

To calculate \(E[U], E[V]\) and \(E[UV]\), we need to use the properties of Brownian motion and Itô integrals. Recall that for Itô integrals, \(E[\int_{0}^{1} f(s)dB(s)] = 0\) and \(E[\int_{0}^{1} f(s)dB(s) \int_{0}^{1} g(s)dB(s)] = \int_{0}^{1} f(s)g(s)ds\). Apply these properties to calculate the expectations: \[E[U] = E[\int_{0}^{1} f(s)dB(s)] = 0\] \[E[V] = E[\int_{0}^{1} g(s)dB(s)] = 0\] \[E[UV] = E[\int_{0}^{1} f(s)dB(s) \int_{0}^{1} g(s)dB(s)] = \int_{0}^{1} f(s)g(s)ds\] #Step 3: Substitute Expectations into Covariance#

Now substitute the expectations from step 2 into the covariance formula from step 1: \[Cov(U, V) = E[UV] - E[U]E[V] = \int_{0}^{1} f(s)g(s)ds - 0 \cdot 0 = \int_{0}^{1} f(s)g(s)ds\] #Step 4: Validate Assertion and Conclude#

We are given the condition that \(\int_{0}^{1} f(s)g(s)ds = 0\). Therefore, based on the covariance from step 3: \[Cov(U, V) = \int_{0}^{1} f(s)g(s)ds = 0\] Since the covariance of \(U\) and \(V\) is zero, we can conclude that \(U\) and \(V\) are independent random variables whenever \(f\) and \(g\) are bounded continuous functions that satisfy the given integral condition.

Key concepts

Stochastic Processes

A stochastic process is a mathematical object that describes a system subjected to random changes over time. One typical example of a stochastic process is Brownian motion, named after botanist Robert Brown, who observed the seemingly random movement of pollen grains in water.

In formal terms, a stochastic process is a collection of random variables indexed by time. The Brownian motion process \( \{B(t) ; 0 \leq t \leq 1\} \) is a specific kind of stochastic process where \( B(t) \) represents the random position of a particle at time \( t \) with \( B(0) = 0 \) and the increments \( B(t) - B(s) \) are normally distributed with mean zero for \( s < t \) and independent of the past events.

The path of such a particle is continuous but nowhere differentiable, which means that the particle's future trajectory cannot be predicted with certainty, reflecting the essence of a stochastic process. Understanding these processes is crucial in fields like physics, finance, and many others where random fluctuations are an inherent feature.

Gaussian Random Measure

In the context of stochastic processes, a Gaussian random measure is a way to assign random values to subsets of the real line, typically to intervals in a way that generalizes the idea of a random variable. For the standard Brownian motion, the Gaussian random measure \( B(I) \) for an interval \( I=(s, t] \) is just the difference \( B(t)-B(s) \), representing the 'net' movement of the Brownian particle between times \( s \) and \( t \).

This measure has two key properties: Firstly, the values it takes on disjoint intervals are independent and, secondly, for any fixed interval \( I \), the measure \( B(I) \) is normally distributed with mean zero and variance equal to the length of the interval. This validates the definition of randomness: each interval’s outcome is unpredictable and unaffected by others, unless the intervals overlap.

Measures like these are essential when working with integrals involving stochastic processes (such as Itô integrals) because they give us a way to integrate 'randomness' over an interval, giving rise to random variables.

Ito Integrals

The Itô integral is a cornerstone of stochastic calculus, which is the mathematics underlying the analysis of stochastic processes. Named after Kiyosi Itô, this integral allows us to integrate with respect to a Brownian motion, which is nontrivial due to the motion’s erratic behavior over infinitesimally short intervals.

When we write \( U=\int_{0}^{1} f(s) dB(s) \) and \( V=\int_{0}^{1} g(s) dB(s) \) as in the original exercise, what we're doing is summing up many small increments of the Brownian motion, each weighted by the corresponding value of \( f(s) \) or \( g(s) \) at time \( s \). These integrals are random variables themselves.

One non-intuitive property of the Itô integral is that, under certain regularity conditions on the functions involved, the expected value of an Itô integral is zero. This leads to the conclusion that two Itô integrals are independent if the integral of the product of their respective functions over the given interval is zero. This is a profound result that underpins much of modern financial mathematics, as it allows for the modeling of independent risk factors and their impact on the pricing and risk management of financial derivatives.