# The Brownian motion process in Sheldon M. Ross

Today I study Brownian Motion and Geometric Brownian Motion using textbook: An Elementary Introduction to Mathematical Finance, Third Edition by Sheldon M. Ross but I missed the class because I was ill. My professor gave us a few problems as an exercise to understand this subject. Here is one of them where I am totally clueless.

Question

Let $X(t)$, $t ≥ 0$ be a Brownian motion process with drift parameter $μ = 3$ and variance parameter $σ^2 = 9$. If $X(0) = 10$ and $\Delta = 0.1$ in the approximation model to the Brownian motion process. For this approximation model, find

(a) $E[X(1)]\,$;

(b) $Var(X(1))$;

(c) $P(X(.5) > 10)$.

Solution

According the solution manual, the solution is

(a) With $2p - 1 = 0.3162$, $E[X(1)] = 10 + 10(0.9487)(0.3168) = 13.005$

(b) $Var[X(1)] = 9(1 - 0.09998) \approx 8.10018$

(c) $P[X(0.5) > 10] = (0.6581)^5 + 5(0.6581)^4(0.3419) + 10(0.6581)^3(0.3419)^2 = 0.7773$

Background          Using formula $p=\frac{1}{2}\left(1+\frac{\mu}{\sigma}\sqrt{\Delta}\right)$, I know how to obtain $2p - 1 = 0.3162$. But how to obtain the other numbers? I really don't understand where they come from. Would you help me, please? Any help would be appreciated. Thanks in advance.

• "For this approximation model..." Surely you realize that you nowhere told us what this model was, and that, without this information, it is impossible to answer your question? – Did Oct 11 '14 at 10:59
• @Did Sorry for that. I'm new for this stuff. I thought it can be answered easily without any extra info. I added some information to my OP. I hope it's enough. Thanks for your concern. :-) – Venus Oct 11 '14 at 12:07

## 1 Answer

There are some typos/rounding errors in the solution provided in the manual. For question a, the correct solution is

$$E[X(1)]=X(0)+\sigma \sqrt{\Delta} \frac{t}{\Delta} \frac{\mu}{\sigma} \sqrt{\Delta}=\\ 10 + 10 \cdot 0.9487 \cdot 0.3162 = 13$$

The meaning of this solution is as follows: starting from time zero, where we have an initial value of $X(0)$, we have to take $\frac{t}{\Delta}=\frac{1}{0.1}=10$ steps, each of absolute amplitude $\sigma \sqrt{\Delta}=3 \sqrt{0.1}=0.9487$, and where the difference between the probability $p$ of a positive step and the probability $1-p$ of a negative step is $2p-1=\frac{\mu}{\sigma} \sqrt{\Delta}=0.3162$. It is unclear why the solution in the manual gives for this last value the number $0.3168$, thus yielding $E[X(1)]=13.005$ (probably a typo). Also note that the correct result of $E[X(1)]=13$ can be obtained more simply by the formula

$$E[X(t)]=X(0)+\mu t$$

which in this case directly gives $E[X(1)]=10+3 \cdot 1=13$.

As regards question b, taking into account that in this case $2p-1=\sqrt{0.1}$, the correct answer is given by the formula

$$Var E[X(t)-X(0)]=\sigma^2 t [1-(2p-1)^2]= \\ 9 \cdot 1 \cdot (1-0.1)=8.1$$

Again, it is curious to note that the answer provided in the manual, i.e. a value of $8.10018$, is incorrect. In particular, it seems to depend on a rounding error: the value of the term $[1-(2p-1)^2]$ is assumed to be $[1-0.09888]$ instead of $[1-0.1]$, probably because the authors computed $(2p-1)^2$ not using the exact value of $(2p-1)=\sqrt{0.1}$ $=0.31622776...$, but the approximated value of $0.3162$ truncated at the fourth decimal digit, finally leading to $0.3162^2=0.09888$.

As regards question c, it can be solved by summing the probabilities relative to all combinations where, among the $5$ steps considered, the positive steps are more than the negative ones. This is given by the formula

$$\sum_{k=\lceil n/2 \rceil}^n {n \choose k} p^k (1-p)^{1-k}$$

In our case, we have $p=\frac{1}{2}(1+\sqrt{0.1})$ $\approx 0.6581$ and $1-p\approx 1-0.6581$ $\approx 0.3419$. Also, the summation includes only the values of $k=3,4,5$. Thus, we obtain

$$P[X(0.5) > 10] \approx (0.6581)^5 + 5(0.6581)^4(0.3419) + 10(0.6581)^3(0.3419)^2 \approx 0.7773$$

• Thanks for your answer. I really appreciate it. Let me study it first. +1 for now – Venus Oct 17 '14 at 9:21