Question

Let \(\\{Z(t), t \geqslant 0\\}\) denote a Brownian bridge process. Show that if $$ Y(t)=(t+1) Z(t /(t+1)) $$ then \(\\{Y(t), t \geqslant 0\\}\) is a standard Brownian motion process.

Short answer

We can show that the given process \(Y(t) = (t+1)Z(t/(t+1))\) is a standard Brownian motion process by verifying the following properties: 1. \(Y(0) = 0\) 2. The increments of the process are independent and normally distributed with mean 0 and variance equal to the increment length (stationary increments). 3. The process has continuous paths. Using the properties of the Brownian bridge process, we show that \(Y(0) = 0\), the increments of \(Y(t)\) are independent and normally distributed with mean 0 and variance equal to the increment length, and \(Y(t)\) has continuous paths. Thus, the process \(Y(t)\) is indeed a standard Brownian motion process.

Step-by-step solution

Check if \(Y(0) = 0\)

As per the definition of \(Y(t)\), substitute \(t=0\): $$ Y(0) = (0 + 1) Z\left(\frac{0}{0+1}\right) = Z(0) $$ Using property A of the Brownian bridge process, we have \(Z(0) = 0\). Thus, \(Y(0) = 0\).

Check if the increments of the process are independent and normally distributed with mean 0 and variance equal to the increment length

Consider any two time points \(0 \leq s < t\). Let the increment of the process \(Y\) be denoted as \(\Delta Y(t,s)\), and the increment of the process \(Z\) be denoted as \(\Delta Z(t,s)\). Then: $$ \Delta Y(t,s) = Y(t) - Y(s) = (t+1)Z\left(\frac{t}{t+1}\right) - (s+1)Z\left(\frac{s}{s+1}\right) $$ Applying the properties of the Brownian Bridge process, we know that the increments \(\Delta Z(t,s)\) are independent and normally distributed with: $$ E[\Delta Z(t,s)] = 0,\ \text{and}\ \text{Var}[\Delta Z(t,s)] = (1-\frac{s}{t+1})\left(\frac{t}{t+1}-\frac{s}{t+1}\right) $$ Therefore, using the properties of linear combinations of normal variables, we can deduce that the increments of the process \(Y\) are independent and normally distributed with: $$ E[\Delta Y(t,s)] = 0,\ \text{and}\ \text{Var}[\Delta Y(t,s)] = (t-s) $$

Check if the process has continuous paths

Given that the Brownian bridge process \(Z\) is a continuous process (property C), we can see that the process \(Y\) also has continuous paths. The continuous paths of the Brownian bridge process ensure that the given transformation does not introduce any discontinuities in the process \(Y(t)\). Since all three properties have been verified for the process \(Y(t) = (t+1)Z(t/(t+1))\), we can conclude that it is indeed a standard Brownian motion process.