I am new to studying B.M., and I am having trouble understanding how the calculations below were derived. I'm not sure about the Var, Cov, but my guess for deriving the answer for E(Xt) requires you to take into consideration the fact that sigma and mu are constants. I understand the basic properties of B.M., but I am having a hard time understanding how to properly apply them.
First off, this is not BM but GBM (geometric Brownian Motion). Google that and you will find a lot. The way I think of it is to break it into multiplied terms. Then, $$X_t=e^{\mu t}e^{(-\frac{\sigma^2}{2})t)}e^{\sigma B_t} $$ First, imagine $\sigma=0$. Then, as usual for compounding, the first term makes sense, because a sum compounding continuously for t years at rate $\mu$ would be $X_0e^{\mu t}$
The problem is the solution to the SDE has that unfortunate $e^{\sigma B_t} $term multiplying the result. Since $e^q$ for any real $q$ is always positive, it imparts an upward bias on the no-volatility result. And, that amount is $e^{\frac{\sigma^2}{2}}$.
So, by having $\mu-\frac{\sigma^2}{2}$ as what you plug in for drift, terms 2 and 3 in the multiplicative form of the solution exactly cancel each other out in terms of the expected value. Compare this to arithmetic Brownian Motion, where the result with drift is $X_t=X_0+\mu t+N(0, \sigma^2\sqrt t$), and the stochastic term introduces only the uncertainty/variance, and does not impact the expected value (since $E[N(0,\sigma^2)]=0$)
It's a bit hard to think about $e^X$, where $X$ is a random variable, but perfectly OK. For one thing you could just expand it using the standard expansion for $e^X= 1 + \frac{X^1}{1!}+\frac{X^2}{2!}+\frac{X^3}{3!}+...$. But if you Google "lognormal distribution" you will see a lovely graphic about what $E^X$looks like if X is a normal random variable, and why a GBM with drift $\mu$ and variance $\sigma^2$ has an expected value of $e^{(\mu+\frac{\sigma^2}{2})t}$, so plugging in the $\frac{-\sigma^2}{2}$ offsets the $e^{B_t}$ term properly. It also shows how to get the higher order moments quite simply.