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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 enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here 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.

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    $\begingroup$ "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? $\endgroup$ – Did Oct 11 '14 at 10:59
  • $\begingroup$ @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. :-) $\endgroup$ – Venus Oct 11 '14 at 12:07
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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

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

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

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