Proof on brownian motion

Question

If B is a standard brownian motion, then

$$X_t = 3t - 2B_t$$ is also Brownian motion. And what is mean and variance of this $X_t$.

I would like to prove it. But I can’t. Help me to do please.

Answer 1

This $X_t$ is not a Brownian motion, but it is a Gaussian process. The mean is $\mathbb{E}[X_t] = \mathbb{E}[3t - 2B_t] = 3t - \mathbb{E}[2B_t] = 3t$ and the variance is $$\mathbb{E}[(X_t-3t)^2] = \mathbb{E}[(2B_t)^2] = 4 \mathbb{E}[B_t^2] = 4t.$$

It isn't a Brownian motion because its mean is not equal to $0$.

Answer 2

Note that this is the solution to diffusion equation (with $X_0=0$): $$ X_t = \int_{0}^{t}\sigma(x)dB_t + \int_{0}^{t}\mu(x)dt $$ If you equate the first stochastic integral to the second term in your solution, and the second integral to the first term, you get $\sigma(x) = -2, \mu(x) = 3. $