# Introductory Brownian motion questions

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?

• From the formula for variance $E(X-EX)$ its clear that adjusting $X$ by a constant won't affect the variance (for b) – Calvin Khor Mar 4 at 11:52

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,

$$\Bbb P (X(3) > 30) = 1 - \Phi (\frac{30 - 27.5}{\sqrt{24}}).$$

• Hi Cettt, in the same problem, there is a part $(d)$ that asks how to find $P(X(0.5) > 10)$. How can I do this? I found that $\mathbb{E}[X(0.5)] = 21.25$, and using your method of rewriting $X$ as $X(t) = 20 + 2.5t + \sqrt{8}B(t)$, I got $\text{Var}(X(0.5)) = 8 \cdot 0.5 = 4$. Is this correct? – user614735 Mar 6 at 12:35