Let $X(t), t \geq 0$ be a Brownian motion process with drift parameter $\mu = 2.5$ and variance $\sigma^{2} = 8$. If $X(0) = 20$, find
(a) $E(X(3))$
(b) $\mathrm{Var}(X(3))$
(c) $P(X(3) > 30)$
I am reading a book, and I am stuck on that exercise.
$X(3) - X(0)$ has a drift parameter of $7.5$ and variance of $24$ and it is normally distributed. Then, we can add $20$ to each side to get that $X(3)$ has an expectation of $27.5$? Is that right?
Then, $X(3) - X(0)$ has a variance of $24$. I don't know if I can just add $8$ to each side to separate the variances though, since it's not linear. If I can, it would just be $32$. Is that right?
I have no clue how to do $(c)$. I think it will have to do something with the CDF of the normal distribution though. Can someone please help me?
your first calculation is correct.
For the second one it could be helpful to write $X$ like this $$ X(t) = 20 + 2.5 \cdot t + \sqrt{8} \cdot B(t), $$ where $B$ is a standard Brownian motion. We know that $Var(B(t)) = t$ for all $t \geq 0$ hence $$ Var(X(3)) = Var(\sqrt{8} \cdot B(3)) = 8 \cdot 3 = 24. $$
For your third question just note that $X(3)$ has a normal distribution and according to the first two exercises we must have $X(3) \sim N(27.5, 24)$
Therefore,
$$ \Bbb P (X(3) > 30) = 1 - \Phi (\frac{30 - 27.5}{\sqrt{24}}). $$