Gaussian process & Brownian motion

[shan]

Homework Statement

Let $$\{ B(t) \}_{t \geq 0}$$ be a standard Brownian motion and $$U \sim U[0,1]$$ and $${Y(t)}_t\geq0$$ be defined by $$Y(t) = B(t) + I_{t=U}$$. Verify that Y(t) is a Gaussian process and state its mean and covariance functions. Is Y(t) a standard Brownian motion?

Homework Equations

The Attempt at a Solution

The way I thought about doing this is look at the increments of Y(t), show that they are normally distributed, show that a random vector of increments of Y(t) is multivariate normal, conclude that Y(t) is Gaussian. If the mean is 0 and variance is t, then Y(t) is standard Brownian.

So for the increment between time r and r-1:
$$Y(t_r) - Y(t_{r-1}) = B(t_r) + I_{t_r=U} - B(t_{r-1}) - I_{t_{r-1}=U} =B(t_r) - B(t_{r-1}) + I_{t_r=U} - I_{t_{r-1}=U}$$

Where the first two terms are Normal(0,1) and the last two terms have this distribution:
$$0 P(t_r \ne U \cap t_{r-1} \ne U) -1 P(t_r \ne U \cap t_{r-1} = U) 1 P(t_r = U \cap t_{r-1} \ne U)$$

I know that the above is still normally distributed but I don't know how to work out the mean and variance. Am I on the right track?

 

[mighty2000]

Yes, you are on the right track. To find the mean and variance of the increment of Y(t), we can use the properties of Brownian motion and the uniform distribution.

First, we know that the increment of a standard Brownian motion is normally distributed with mean 0 and variance t. Therefore, the first two terms in your expression, B(t_r) - B(t_{r-1}), have a normal distribution with mean 0 and variance t.

Next, we can look at the probabilities in the last two terms. We can see that the probability of t_r = U and t_{r-1} = U is the same as P(t_r = U)P(t_{r-1} = U) since U is uniformly distributed. This probability is equal to 1/2 * 1/2 = 1/4.

Similarly, the probability of t_r = U and t_{r-1} ≠ U is equal to P(t_r = U)P(t_{r-1} ≠ U) = 1/2 * 1/2 = 1/4. The same applies for the probability of t_r ≠ U and t_{r-1} = U.

Using these probabilities, we can calculate the mean and variance of the last two terms:
Mean = 0 * P(t_r \ne U \cap t_{r-1} \ne U) + (-1) * P(t_r \ne U \cap t_{r-1} = U) + (1) * P(t_r = U \cap t_{r-1} \ne U)
= 0 * (1/4) + (-1) * (1/4) + (1) * (1/4)
= 0

Variance = (0 - 0)^2 * P(t_r \ne U \cap t_{r-1} \ne U) + (0 - (-1))^2 * P(t_r \ne U \cap t_{r-1} = U) + (0 - 1)^2 * P(t_r = U \cap t_{r-1} \ne U)
= 0 * (1/4) + 1^2 * (1/4) + (-1)^2 * (1/4)
= 1/2

Therefore, the increment of Y(t) has a normal distribution with mean 0