Question

Show that the process X = (Xₜ)ₜ≥0 defined by Xₜ = -Wₜ is also a standard Brownian motion.

Answer 1

To show that the process X = (Xₜ)ₜ≥0 defined by Xₜ = -Wₜ is also a standard Brownian motion, we need to verify if it satisfies the properties of a standard Brownian motion. By checking the properties of the process, we can conclude that it is indeed a standard Brownian motion.

Step-by-step explanation:

In order to show that the process X = (Xₜ)ₜ≥0 defined by Xₜ = -Wₜ is also a standard Brownian motion, we need to check if it satisfies the properties of a standard Brownian motion:

1. The initial value X₀ = 0
2. The increments Xₜ - Xₛ, for t > s, are normally distributed with mean 0 and variance t-s
3. The increments Xₜ - Xₛ, for t > s, are independent of the past values Xᵣ for r ≤ s

Let's verify each of these properties for the given process:

1. X₀ = -W₀ = -0 = 0
2. For t > s, Xₜ - Xₛ = -Wₜ + Wₛ = -(Wₜ - Wₛ). We know that Wₜ - Wₛ is also a standard Brownian motion, so its increments are normally distributed with mean 0 and variance t-s. Hence, Xₜ - Xₛ is also normally distributed with mean 0 and variance t-s.
3. The independence of Xₜ - Xₛ and Xᵣ for r ≤ s follows directly from the independence of Wₜ - Wₛ and Wᵣ for r ≤ s.

Therefore, the process X = (Xₜ)ₜ≥0 defined by Xₜ = -Wₜ is indeed a standard Brownian motion.