Let $X(t), t \geq 0$ be a Brownian motion process with drift parameter $\mu$ and variance parameter $\sigma^{2}$ for which $X(0) = 0$. Show that $-X(t), t \geq 0$ is a Brownian motion process with drift parameter -$\mu$ and parameter $\sigma^{2}$
I'm not too sure about how to approach this question. I think that we need to show that $-X(t)$ satisfies the definition of a Brownian motion, namely that $X(0)$ is a given constant, and for all positive $y$ and $t$, $X(t + y) - X(y)$ is independent of the process up to time $y$ and has a normal distribution with parameters $t\mu$ and $t\sigma^{2}$.
For the first point, you are indeed right, since Brownian motion with drift starts at $0$ with probability $1$.
As for the other properties, note that if we let $Y(t)=-X(t)$, then you have that
$$Y(t) - Y(s) = X(s)-X(t) = -(X(t)-X(s)) \quad \text{for} \quad t>s$$
Using the properties of Brownian motion (distribution, independence, stationarity of increments) and the fact that the transformation applied is linear, can you see what the distribution of this last quantity is?