Modeling Stock Prices in Continuous-Time
Stochastic Processes
A stochastic process describes the evolution of a random variable over time. In finance, we use stochastic processes to model the evolution of stock prices, interest rates, volatility, foreign exchange rates, and commodity prices. We distinguish between:
- Discrete-time processes: The values of the process \left\{ S_{n} \right\} are allowed to change only at discrete time intervals, i.e., n \in \{ 0, 1, 2, \ldots, N \} or n \in \mathbb{N}.
- Continuous-time processes: The stochastic process \left\{ S_{t} \right\} is defined for all t \in [0, T].
We will now consider several stochastic processes commonly used to model the future evolution of the price of an asset such as a stock. We start by understanding discrete-time processes and then extend the analysis to include continuous-time processes. The analysis is informal, as the theory of stochastic process in continuous time requires advanced mathematical concepts, which is beyond the scope of these notes.
It is essential to realize that a stochastic process for a stock price is trying to model all possible histories between now and a specific time in the future. A sample path is one of the many possible histories generated using the stochastic process.
Random Walks
We will now study one of the simplest yet most intriguing stochastic processes defined in discrete time. A one-dimensional random walk \left\{ X_{n} \right\} is a stochastic process defined as \begin{aligned} X_{0} & = x_{0}, \\ X_{n + 1} & = X_{n} + e_{n + 1}, \end{aligned} where \left\{ e_{n} \right\} are independent and identically distributed (i.i.d.) random variables such that \operatorname{E}(e_{n}) = 0 for all n \geq 1. Note that e_{n} need not be normally distributed. For example, for each n, the variable e_{n} could take the values 1 and -1 with equal probability. A random walk only requires that the shocks e_{n} are independent.
An essential property of a random walk is that its sample paths diverge as n grows. Indeed, we have \begin{aligned} X_{n} & = X_{n-1} + e_{n} \\ & = X_{n-2} + e_{n-1} + e_{n} \\ & \hphantom{xxxxxx} \vdots \\ & = X_{0} + e_{1} + \cdots + e_{n-1} + e_{n} \\ & = X_{0} + \sum_{i=1}^{n} e_{i}. \end{aligned} Denoting \operatorname{Var}(e_{n}) = \sigma^{2}, and since we have that \{e_{n}\} are independent, we have \operatorname{Var}(X_{n}) = n \sigma^{2}. Therefore, the variance of X_{n} increases linearly with n as n \rightarrow \infty.
Intuitively, this is saying that if we simulate many different sample paths for n = 0, \ldots, N where N is very large, we should expect to see some values of X_{N} to be very high and positive whereas others will be significantly negative.
Brownian Motion
A very useful random walk can be defined as follows: W_{t + \Delta t} = W_{t} + \sqrt{\Delta t} e_{t + \Delta t}, where W_{0} = 0 and \left\{ e_{t} \right\} are i.i.d. such that e_{t} \sim N(0, 1). Note that here time increases each step by \Delta t. Letting \Delta t \rightarrow 0, the resulting process \left\{ W_{t} \right\} for t \in [0, T] is called a Brownian motion or Wiener process.
The Brownian motion has the following properties:
- The sample paths are continuous.
- For s < t, the increment W_{t} - W_{s} \sim N(0, t - s), i.e. is normally distributed with mean 0 and variance t - s.
- Increments are independent of each other.
- In particular, note that W_{t} \sim N(0, t) for 0 < t \leq T.
Geometric Brownian Motion
Now we turn our attention to modeling stock prices \left\{ S_{t} \right\}. We need to be careful, though, as stock prices cannot be negative. We also would like to allow the model to display a certain drift \mu and volatility \sigma.
To achieve this, we model the percentage change of a stock price between t and t + \Delta t as \frac{\Delta S_{t}}{S_{t}} = \mu \Delta t + \sigma \Delta W_{t}. Note that the percentage change in price over an interval \Delta t is normally distributed with mean \mu \Delta t and variance \sigma^{2} \Delta t. Letting \Delta t \rightarrow 0, the resulting process \left\{ S_{t} \right\} for t \in [0, T] is called a geometric Brownian motion (GBM).
Stochastic Calculus
Once we have defined how S_{t} behaves over time, we now turn our attention to model how a function of S_{t} behaves over time. The reason why we are interested in this is because we want to find a way to price derivatives as a function of the relevant state variables. We will see later that when the stock price is driven by a single source of uncertainty, then the value of a call or put option depends only on the stock price itself and time-to-maturity, i.e. the price of the derivative when the stock price is S and the time-to-maturity is T will be of the form F(S, T).
We will start studying how X_{t} = F(S_{t}) behaves over time and we will add later the time dimension to the problem. In what follows we assume that F(\cdot) is a smooth function such that its first and second derivatives exist.
Ito’s Lemma
Remember that the Wiener process increment is defined \Delta W_{t} = W_{t + \Delta t} - W_{t} = \sqrt{\Delta t} e_{t + \Delta t}.
Consider a GBM process \left\{S_{t}\right\} and a smooth function F(\cdot). A second order Taylor approximation around S_{t} implies F(S_{t} + \Delta S_{t}) \approx F(S_{t}) + F'(S_{t}) (\Delta S_{t}) + \frac{1}{2} F''(S_{t}) (\Delta S_{t})^{2}.
Using the results derived in the appendix, we have that \begin{aligned} (\Delta S_{t})^{2} & = (\mu S_{t} \Delta t + \sigma S_{t} \Delta W_{t})^{2} \\ & = (\mu S_{t})^{2} \underbrace{(\Delta t)^{2}}_{\approx 0} + 2 \mu \sigma (S_{t})^{2} \underbrace{(\Delta t)(\Delta W_{t})}_{\approx 0} + (\sigma S_{t})^{2} \underbrace{(\Delta W_{t})^{2}}_{\approx \Delta t} \\ & \approx \sigma^{2} S_{t}^{2} \Delta t. \end{aligned} We can finally conclude that \Delta F(S_{t}) \approx \left( \mu S_{t} F'(S_{t}) + \frac{1}{2} \sigma^{2} S_{t}^{2} F''(S_{t}) \right) \Delta t + \sigma S_{t} F'(S_{t}) \Delta W_{t}.
The continuous-time analog of the previous analysis is as follows.
It is usually more convenient to use the box calculus when working with stochastic processes defined through Brownian motions.
Solving for GBM
Define X = \ln(S), which implies S = e^{X}. We have that F_{S} = 1/S and F_{SS} = -1/S^{2}, which implies \begin{aligned} dX & = F_{S} dS + \frac{1}{2} F_{SS} (dS)^{2} \\ & = \frac{1}{S} \left( \mu S dt + \sigma S dW \right) + \frac{1}{2} \left( -\frac{1}{S^{2}} \right) \sigma^{2} S^{2} dt \\ & = \left( \mu dt + \sigma dW \right) - \frac{1}{2} \sigma^{2} dt \\ & = \left( \mu - \frac{1}{2} \sigma^{2} \right) dt + \sigma dW. \end{aligned} We can then solve for X_{T}: \begin{aligned} X_{T} - X_{0} & = \int_{0}^{T} dX = \int_{0}^{T} \left( \mu - \frac{1}{2} \sigma^{2} \right) dt + \int_{0}^{T} \sigma dW \\ & = \left( \mu - \frac{1}{2} \sigma^{2} \right) T + \sigma W_{T}, \end{aligned} and conclude that S_{T} = S_{0} \exp\left( \left( \mu - \frac{1}{2} \sigma^{2} \right) T + \sigma W_{T} \right). \tag{2}
Properties of Stock Prices Following a GBM
Equation (2) can be rewritten as: \ln(S_{T}) = \ln(S_{0}) + \left( \mu - \frac{1}{2} \sigma^{2} \right) T + \sigma W_{T}.
We can conclude that \ln(S_{T}) \sim N(m, s^{2}), where \begin{aligned} m & = \ln(S_{0}) + \left( \mu - \frac{1}{2} \sigma^{2} \right) T, \\ s & = \sigma \sqrt{T}. \end{aligned} In other words, S_{T} is lognormally distributed with mean m and variance s^{2}.
Example 1 Consider a stock whose price at time t is given by S_{t} and that follows a GBM. The expected return is 12% per year and the volatility is 25% per year. The current spot price is $25. If we denote X_{T} = \ln(S_{T}) and take T = 0.5, we have that: \begin{aligned} \operatorname{E}(X_{T}) & = \ln(25) + \left(0.12 - 0.5(0.25)^{2}\right)(0.5) = 3.2633, \\ \operatorname{SD}(X_{T}) & = 0.25 \sqrt{0.5} = 0.1768. \end{aligned} Hence, the 95% confidence interval for S_{T} is given by: [e^{3.2633 - 1.96(0.1768)}, e^{3.2633 + 1.96(0.1768)}] = [18.48, 36.96].
Therefore, there is a 95% probability that the stock price in 6 months will lie between $18.48 and $36.96.
Moments of the Stock Price
The fact that the stock price at time T is log-normally distributed allows us to compute the mean and standard deviation of S_{T}.
Proof
Since \ln(S_{T}) \sim \mathcal{N}(m, s^{2}), we can compute its moments using the results derived earlier so that \begin{aligned} \operatorname{E}(S_{T}) & = e^{m + \frac{1}{2} s^{2}} = e^{\ln(S_{0}) + \left(\mu - \frac{1}{2} \sigma^{2}\right) T + \frac{1}{2} \sigma^{2} T} = e^{\ln(S_{0})} e^{\mu T} = S_{0} e^{\mu T}. \end{aligned} In this model, the expected stock price at any point in the future is just the current stock price growing at the rate \mu for T years.Therefore, the expected stock price grows at a rate \mu. The variance of S_{T}, however, is large and increases exponentially with time.
Example 2 Consider a stock whose price at time t is given by S_{t} and that follows a GBM. The expected return is 12% per year and the volatility is 25% per year. The current spot price is $25. The expected price and standard deviation 6 months from now are: \begin{aligned} \operatorname{E}(S_{T}) & = 25 e^{0.12 (0.5)} = \$26.55, \\ \operatorname{SD}(S_{T}) & = 26.55 \sqrt{e^{0.25^{2} (0.5)} - 1} = \$4.73. \end{aligned}
A Generalized Form of Ito’s Lemma
Most derivatives not only depend on the underlying asset but also depend on time since they have fixed expiration dates. The analysis we did before for Ito’s Lemma generalizes easily to handle this case. Consider a non-dividend paying stock that follows a GBM: \begin{aligned} dS = \mu S dt + \sigma S dW, \end{aligned} and a smooth function F(S, t). Ito’s Lemma in this case applies in the following form: dF = F_{S} dS + \frac{1}{2} F_{SS} (dS)^{2} + F_{t} dt, where (dS)^{2} = \sigma^{2} S^{2} dt.
Appendix
Some Intuition on Brownian Motion
Remember that we defined the Brownian motion or Wiener process as a random walk driven by normally distributed shocks: W_{t + \Delta t} = W_{t} + \sqrt{\Delta t} e_{t + \Delta t}, where \{e_{t}\} is an i.i.d. sequence of random variables distributed \mathcal{N}(0, 1).
Let’s start by splitting the interval [0, T] into n intervals of length \Delta t = t_{i+1} - t_{i}.
Note that t_{i} = i \Delta t and T = t_{n} = n \Delta t. The Brownian motion increments are then defined as \Delta W_{t_{i}} = W_{t_{i+1}} - W_{t_{i}}.
The first question one might have is why using normally distributed increments. There are two answers for that. First, a sum of normally distributed random variables is also normal and in this case we have: W_{T} - W_{0} = \sum_{i=0}^{n-1} \Delta W_{t_{i}} = \sum_{i=0}^{n-1} \sqrt{\Delta t} e_{t + \Delta t}.
The variance of \sum_{i=0}^{n-1} \sqrt{\Delta t} e_{t + \Delta t} is given by \sum_{i=0}^{n-1} \Delta t = n \Delta t = T, which implies that W_{T} \sim \mathcal{N}(0, T). So by using normally distributed increments we guarantee that the resulting process for Brownian motion is also normal.
Second, imagine that we use a different distribution for the i.i.d. increments while still requiring \operatorname{E}(e_{t}) = 0 and \operatorname{Var}(e_{t}) = 1. For example, e_{t} could take the values 1 and -1 with equal probability. Nevertheless, the central limit theorem guarantees that: \sqrt{n} \left( \frac{1}{n} \sum_{i=0}^{n-1} \sqrt{\Delta t} e_{t + \Delta t} \right) \xrightarrow[]{d} \mathcal{N}(0, \Delta t).
In other words, even if we use a different distribution for the increments, as n \rightarrow \infty we have that W_{T} \sim \mathcal{N}(0, T). Therefore, there is no loss in generality in assuming normally distributed increments for the Brownian motion.
A second question that one might have, and one of the most puzzling facts in stochastic calculus in my opinion, is the fact that when we apply Ito’s lemma we use the fact that (dW_{t})^{2} = dt. Clearly, (\Delta W_{t})^{2} = \Delta t e_{t}^{2} \neq \Delta t where e_{t} \sim \mathcal{N}(0, 1). Indeed, if \Delta W_{t} is random, then (\Delta W_{t})^{2} must also be random. However, we will see in a moment that it is fine to say that (\Delta W_{t})^{2} \approx \Delta t as \Delta t \rightarrow 0.
Let’s start by computing the mean and variance of (\Delta W_{t})^{2}: \begin{aligned} \operatorname{E}\left[(\Delta W_{t})^{2}\right] & = \Delta t \\ \operatorname{Var}\left[(\Delta W_{t})^{2}\right] & = \operatorname{E}\left[(\Delta W_{t})^{4}\right] - \left(\operatorname{E}\left[(\Delta W_{t})^{2}\right]\right)^{2} \\ & = 3 (\Delta t)^{2} - (\Delta t)^{2} \\ & = 2 (\Delta t)^{2}. \end{aligned}
In computing the variance of (\Delta W_{t})^{2} we used the fact that if X \sim \mathcal{N}(0, \sigma^{2}), then \operatorname{E}(X^{4}) = 3\sigma^{4}. Since \Delta W_{t} \sim \mathcal{N}(0, \Delta t), we have that \operatorname{E}\left[(\Delta W_{t})^{4}\right] = 3 (\Delta t)^{2}.
Consider now the following sum: S_{n} = \sum_{i=0}^{n-1} (\Delta W_{t_{i}})^{2}.
Clearly, S_{n} is a sum of n independent random variables so its variance is the sum of the variance of each \Delta W_{t}: \begin{aligned} \operatorname{E}(S_{n}) & = n \Delta t = T \\ \operatorname{Var}(S_{n}) & = n (2 (\Delta t)^{2}) = \frac{2 T^{2}}{n}. \end{aligned}
Since \lim_{n \rightarrow \infty} \operatorname{Var}(S_{n}) = 0, we have that S_{n} \rightarrow T as n \rightarrow \infty in probability. Intuitively, the previous result is really the weak-law of large numbers since we can re-write it as \frac{S_{n}}{n} \rightarrow \Delta t as n \rightarrow \infty in probability. However, when you apply the weak-law of large numbers to an arbitrary sequence of i.i.d. random variables, you cannot say that you can approximate each random variable by its mean just because its average converges to their mean. In our case, since the variance of (\Delta W_{t})^{2} is so small compared to its mean, we can safely say that (\Delta W_{t})^{2} behaves as if (\Delta W_{t})^{2} = \Delta t as n \rightarrow \infty. In other words, we have that (\Delta W_{t})^{2} \approx \Delta t for small \Delta t.
We can apply the same analysis to study the behavior of (\Delta t)(\Delta W_{t}) as \Delta t \rightarrow 0. Since: \begin{aligned} \operatorname{E}\left[(\Delta t)(\Delta W_{t})\right] & = 0 \\ \operatorname{Var}\left[(\Delta t)(\Delta W_{t})\right] & = \operatorname{E}\left[((\Delta t)(\Delta W_{t}))^{2}\right] - (\operatorname{E}[(\Delta t)(\Delta W_{t})])^{2} \\ & = (\Delta t)^{2} \operatorname{E}[(\Delta W_{t})^{2}] - ((\Delta t) \operatorname{E}[\Delta W_{t}])^{2} \\ & = (\Delta t)^{3}. \end{aligned} Consider now the following sum: C_{n} = \sum_{i=0}^{n-1} (\Delta t) (\Delta W_{t_{i}}).
The mean and variance of C_{n} are given by: \begin{aligned} \operatorname{E}(C_{n}) & = 0 \\ \operatorname{Var}(C_{n}) & = \frac{T^{3}}{n^{2}}. \\ \end{aligned} Since \lim_{n \rightarrow \infty} \operatorname{Var}(C_{n}) = 0, we have that C_{n} \rightarrow 0 as n \rightarrow \infty in probability, implying that (\Delta t)(\Delta W_{t}) \approx 0 for small \Delta t.
Computing Partial Expectations
Since \ln(S_{T}) \sim \mathcal{N}(m, s^{2}), we can use the result introduced earlier about partial expectations to show the following property.
Proof
\begin{aligned} \operatorname{E}\left(S_{T} \large\mathbb{1}_{\{S_{T} > K\}}\right) & = e^{m + \frac{1}{2}s^{2}} \mathop{\Phi}\left(\frac{m + s^{2} - \ln(K)}{s}\right) \\ & = S_{0} e^{\mu T} \mathop{\Phi}\left( \frac{\ln(S_{0}/K) +(\mu + \frac{1}{2} \sigma^{2})T}{\sigma \sqrt{T}} \right), \\ \operatorname{E}\left(K \large\mathbb{1}_{\{S_{T} > K\}}\right) & = K \mathop{\Phi}\left(\frac{m - \ln(K)}{s}\right) \\ & = K \mathop{\Phi}\left( \frac{\ln(S_{0}/K) + (\mu - \frac{1}{2} \sigma^{2})T}{\sigma \sqrt{T}} \right). \end{aligned}It turns out that these results are everything we need in order to derive the Black-Scholes pricing formulas!
Martingales
A martingale is a process closely related to the random walk but slightly more general. A discrete-time martingale \left\{Z_{n}\right\}_{n \geq 0} is a stochastic process such that: \operatorname{E}\left(Z_{n+1} \;\middle|\; Z_{0}, Z_{1}, \ldots, Z_{n}\right) = Z_{n}.
Intuitively, the history of the process \{Z_{n}\} is irrelevant to forecast Z_{n+1}. The current value of Z_{n} is the only thing that matters. A random walk is a martingale, but note that a martingale need not be a random walk.
For example, consider the process \left\{ Z_{n} \right\}: Z_{n+1} = Z_{n} \varepsilon_{n+1}, where \left\{ \varepsilon_{n} \right\} is an i.i.d. sequence such that \operatorname{E}\left(\varepsilon_{n}\right) = 1 for all n \geq 0. It is a martingale since: \begin{aligned} \operatorname{E}\left(Z_{n+1} \;\middle|\; Z_{1}, Z_{2}, \ldots, Z_{n}\right) & = \operatorname{E}\left(Z_{n} \varepsilon_{n+1} \;\middle|\; Z_{n}\right), \\ & = Z_{n} \operatorname{E}\left(\varepsilon_{n+1} \;\middle|\; Z_{n}\right), \\ & = Z_{n}. \end{aligned}
Practice Problems
Problem 1 Consider a stock whose price at time t is given by S_{t} and that follows a geometric Brownian motion (GBM). The expected return is 18% per year and the volatility is 32% per year. The current spot price is $60.
- Compute the expected price 9 months from now.
- Compute the mean and standard deviation of the log-spot price 9 months from now.
- Compute the 95% confidence interval of \ln(S_{T}) 9-months from now, and report the corresponding values for S_{T}.
Solution
- \operatorname{E}(S_{T}) = 60 e^{0.18(9/12)} = \$68.67.
- The mean and standard deviation can be computed as follows: \begin{aligned} m & = \ln(60) + (0.18 - 0.5(0.32^{2}))(9/12) \\ & = 4.1909 \\ s & = 0.32 \sqrt{9/12} \\ & = 0.2771 \end{aligned}
- The confidence interval of \ln(S_{T}) 9-months from now satisfies: \begin{aligned} \operatorname{P}\left( -1.96 < \frac{\ln(S_{T}) - m}{s} \leq 1.96 \right) & = \operatorname{P}\left(m - 1.96 s < \ln(S_{T}) \leq m + 1.96 s \right) \\ & = \operatorname{P}(3.6478 < \ln(S_{T}) \leq 4.7341) \\ & = \operatorname{P}\left(e^{3.6478} < S_{T} \leq e^{4.7341}\right) \\ & = \operatorname{P}(38.39 < S_{T} \leq 113.76) \end{aligned} Therefore, the 95% confidence interval for \ln(S_{T}) is [3.6478, 4.7341] and the corresponding values for S_{T} are [38.39, 113.76].
Problem 2 Consider a stock whose price at time t is given by S_{t} and that follows a GBM. The expected return is 11% per year and the volatility is 27% per year. The current spot price is $60.
- Compute the expected price of S_{t} 1 year from now.
- Compute the expected price of 1 / S_{t} 1 year from now.
Solution
- \operatorname{E}(S_{T}) = 60 e^{0.11(1)} = \$66.98
- We have that \ln(S_{T}) \sim \mathcal{N}(m, s^{2}), where: \begin{aligned} m & = \ln(60) + (0.11 - 0.5(0.27^{2}))(1) \\ & = 4.1679 \\ s & = 0.27 \sqrt{1} \\ & = 0.27 \end{aligned} Using the fact that if X \sim \mathcal{N}(\mu, \sigma^{2}), then \alpha X \sim \mathcal{N}(\alpha\mu, \alpha^{2}\sigma^{2}) and \operatorname{E}(Y^{\alpha}) = \operatorname{E}(e^{\alpha X}) = e^{\alpha\mu+\frac{1}{2} \alpha^{2}\sigma^{2}}, we can take \alpha=-1 to conclude: \begin{aligned} \operatorname{E}(1/S_{T}) & = e^{(-1)4.1679 + 0.5(-1)^{2}(0.27)^{2}} \\ & = 0.0161 \end{aligned}
Problem 3 Consider a stock whose price at time t is given by S_{t} and that follows a GBM. The expected return is 12% per year and the volatility is 35% per year. The current spot price is $55. Let T=18 months.
- Compute \operatorname{E}(S_{T}).
- Compute the mean and standard deviation of the log-spot price at T.
- Find C such that \operatorname{P}(S_{T} \leq C)=0.01.
Solution
- \operatorname{E}(S_{T}) = 55 e^{0.12(18/12)} = \$65.85
- We have that \ln(S_{T}) \sim \mathcal{N}(m, s^{2}), where: \begin{aligned} m & = \ln(55) + (0.12 - 0.5(0.35^{2}))(18/12) \\ & = 4.0955 \\ s & = 0.35 \sqrt{18/12} \\ & = 0.4287 \end{aligned}
- We want to find C such that: \begin{aligned} \operatorname{P}(S_{T} \leq C) & = \operatorname{P}\left(Z \leq \tfrac{\ln(C) - m}{s}\right) \\ & = 0.01 \end{aligned} Hence, \dfrac{\ln(C) - 4.0955}{0.4287} = -2.326, which implies that: \begin{aligned} \ln(C) & = 4.0955 - 2.326(0.4287) \\ & = 3.0982 \end{aligned} Therefore, C = e^{3.0982} = \$22.16.
Problem 4 Consider a stock whose price at time T is given by S_{T} and that follows a GBM, i.e., \ln(S_{T}) \sim \mathcal{N}(\ln(S_{0})+(\mu-0.5\sigma^{2})T, \sigma^{2} T).
The expected return is 12% per year and the volatility is 35% per year. The current spot price is $100.
- Compute the expected price in 2 years from now.
- Compute the mean and standard deviation of the log-spot price in 2 years from now.
- Compute the probability that the spot price is less than $100 in 2 years from now.
- Compute the probability that the spot price is greater than $120 in 2 years from now.
Solution
- \operatorname{E}(S_{T}) = 100 e^{0.12(2)} = \$127.12
- We have that \ln(S_{T}) \sim \mathcal{N}(m, s^{2}), where: \begin{aligned} m & = \ln(100) + (0.12 - 0.5(0.35^{2}))(2) = 4.7227 \\ s & = 0.35 \sqrt{2} = 0.4950 \end{aligned}
- The probability that S_{T} \leq 100 is: \begin{aligned} \operatorname{P}(S_{T} \leq 100) & = \operatorname{P}(\ln(S_{T}) \leq \ln(100)) \\ & = \operatorname{P}\left(Z \leq \frac{\ln(100) - 4.7227}{0.4950}\right) \\ & = \operatorname{P}(Z \leq -0.24) \\ & = 0.4052 \end{aligned}
- The probability that S_{T} > 120 is: \begin{aligned} \operatorname{P}(S_{T} > 120) & = 1 - \operatorname{P}(S_{T} \leq 120) \\ \operatorname{P}(S_{T} \leq 120) & = \operatorname{P}(\ln(S_{T}) \leq \ln(120)) \\ & = \operatorname{P}\left(Z \leq \frac{\ln(120) - 4.7227}{0.4950}\right) \\ & = \operatorname{P}(Z \leq 0.13) \\ & = 0.5518 \end{aligned} Hence, \operatorname{P}(S_{T} > 120) = 1 - 0.5518 = 0.4482.
Problem 5 Suppose that the stock price follows a geometric Brownian motion (GBM) with drift \mu and instantaneous volatility \sigma, i.e., dS = \mu S dt + \sigma S dW. Show that Y = S^{\alpha} also follow a GBM and determine the drift and volatility as a function of \mu, \sigma, and \alpha.
Solution
We have that Y_{S} = \alpha S^{\alpha - 1} and Y_{SS} = \alpha (\alpha - 1) S^{\alpha - 2}. Therefore, \begin{aligned} dY & = Y_{S} (dS) + \frac{1}{2} Y_{SS} (dS)^{2} \\ & =\alpha S^{\alpha - 1} (\mu S dt + \sigma S dW) + \frac{1}{2} \alpha (\alpha - 1) S^{\alpha - 2} (\sigma^{2} S^{2} dt) \\ & = \left(\alpha \mu S^{\alpha} + \frac{1}{2} \alpha (\alpha - 1) \sigma^{2} S^{\alpha}\right) dt + \alpha \sigma S^{\alpha} dW \\ & = \left(\alpha \mu + \frac{1}{2} \alpha (\alpha - 1) \sigma^{2} \right) Y dt + \alpha \sigma Y dW \\ \end{aligned} Hence, Y follows a GBM with drift \alpha \mu + \frac{1}{2} \alpha (\alpha - 1) \sigma^{2} and volatility \alpha \sigma.Problem 6 Let S be the price of TESLA stock that follows a geometric Brownian motion such that dS = \mu S dt + \sigma S dW. Your sales team would like to launch a new product called TESLA Quadro that tracks the price of TESLA to the power 4. In other words, the value of this instrument is given by Y = S^{4}. What is the process followed by Y?
Solution
We saw in class that dY = Y_{S} dS + \frac{1}{2} Y_{SS} (dS)^{2}, where (dS)^{2} = \sigma^{2} S^{2} dt. If Y = S^{4} then Y_{S} = 4 S^{3} and Y_{SS} = 12 S^{2}. We can then write the previous expression as \begin{aligned} dY & = 4 S^{3} (\mu S dt + \sigma S dW) + \frac{1}{2} (12 S^{2}) (\sigma^{2} S^{2} dt) \\ & = 4 \mu S^{3} dt + 4 \sigma S^{4} dW + 6 \sigma^{2} S^{4} dt \\ & = (4 \mu + 6 \sigma^{2}) Y dt + 4 \sigma Y dW \end{aligned} The new product then follows a geometric Brownian motion with drift 4 \mu + 6 \sigma^{2} and instantaneous volatility 4 \sigma.Problem 7 GoingUp Corp. has been gaining a lot of attention in the media for its upside potential. Financial experts agree that the stock price follows a geometric Brownian motion with drift (\mu) equal to 20% per year and volatility of price returns (\sigma) of 73% per year. The current stock price is $220. Compute the probability that the stock price is greater than $233 in 10 months from now.
Solution
We know that \ln(S_{T}) \sim N(m, s^{2}) where \begin{aligned} m & = \ln(S_{T}) + (\mu - 0.5 \sigma^{2}) T \\ & = \ln(220) + (0.20 - 0.5 \times 0.73^{2}) \times (10/12) \\ & = 5.3383 \\ s & = \sigma \sqrt{T} \\ & = 0.73 \sqrt{10/12} \\ & = 0.6664 \end{aligned} Therefore, \begin{aligned} \operatorname{P}(S_{T} \leq 233) & = \operatorname{P}(\ln(S_{T}) \leq \ln(233)) \\ & = \operatorname{P}\left( Z \leq \frac{\ln(233) - 5.3383}{0.6664} \right) \\ & = \Phi(0.1692) \\ & = 0.5672 \end{aligned} Therefore, \operatorname{P}(S_{T} > 233) = 1 - 0.5672 = 0.4328.Problem 8 You would like to invest in ZigZag Inc. but you are concerned that the stock price might go down. You have been studying the dynamics of the stock price and concluded that the stock follows a geometric Brownian motion with drift (\mu) equal to 13% per year and volatility of price returns (\sigma) of 58% per year. The current stock price is $118. Compute the probability that the stock price is less than $98 in 12 months from now.
Solution
We know that \ln(S_{T}) \sim N(m, s^{2}) where \begin{aligned} m & = \ln(S_{T}) + (\mu - 0.5 \sigma^{2}) T \\ & = \ln(118) + (0.13 - 0.5 \times 0.58^{2}) \\ & = 4.7325 \\ s & = \sigma \sqrt{T} \\ & = 0.58 \end{aligned} Therefore, \begin{aligned} \operatorname{P}(S_{T} \leq 98) & = \operatorname{P}(\ln(S_{T}) \leq \ln(98)) \\ & = \operatorname{P}\left( Z \leq \frac{\ln(98) - 4.7325}{0.58} \right) \\ & = \Phi(-0.2543) \\ & = 0.3996 \end{aligned}Problem 9 You are analyzing BMX stock. You believe that it is accurate to model the price evolution of the stock as a geometric Brownian motion. Using historical data, you estimate that the drift (\mu) is 12.0% per year and the volatility of stock returns (\sigma) is 39% per year. The stock price just closed at $331. Compute the expected stock price in 9 months from now.
Solution
We saw in class that \operatorname{E}(S_{T}) = 331 e^{0.12 \times 9/12} = 362.17.Optional Practice Problems
These problems are not required to study for the exam.
Problem 10 Suppose that the stock price follows a geometric Brownian motion (GBM) with drift \mu and instantaneous volatility \sigma. Show that Y = S e^{-\mu t} also follow a GBM and determine the drift and volatility as a function of \mu and \sigma.
Solution
We have that Y_{S} = e^{-\mu t}, Y_{SS} = 0 and Y_{t} = -\mu S e^{-\mu t}. Therefore, \begin{aligned} dY & = Y_{S} (dS) + \frac{1}{2} Y_{SS} (dS)^{2} + Y_{t} (dt) \\ & = e^{-\mu t} (\mu S dt + \sigma S dW) -\mu S e^{-\mu t} dt \\ & = \sigma S e^{-\mu t} dW \\ & = \sigma Y dW \\ \end{aligned} Hence, Y follows a GBM with zero drift and volatility \sigma.Problem 11 Suppose that the stock price follows a geometric Brownian motion (GBM) with drift r and instantaneous volatility \sigma, where r is the risk-free rate. Consider the futures price of S at time t and expiring at T, given by f = S e^{r (T -t)}. Show that f has zero drift and hence is a martingale.
Solution
We have that f_{S} = e^{r (T - t)}, f_{SS} = 0 and f_{t} = -r S e^{r (T - t)}. Therefore, \begin{aligned} df & = f_{S} (dS) + \frac{1}{2} f_{SS} (dS)^{2} + f_{t} (dt) \\ & = e^{r (T - t)} (r S dt + \sigma S dW) - r S e^{r (T - t)} dt \\ & = \sigma S e^{r (T - t)} dW \\ & = \sigma f dW \\ \end{aligned} Hence, f follows a GBM with zero drift which proves that it is a martingale.Problem 12 Suppose that the stock price follows a geometric Brownian motion (GBM) with drift \mu = 10\% and instantaneous volatility \sigma = 25\%. Compute \operatorname{E}(S_{T} \large\mathbb{1}_{\{S_{T} > K\}}) and \operatorname{E}(\large\mathbb{1}_{\{S_{T} > K\}}) = \operatorname{P}(S_{T} > K) if S_{0} = 100, K = 95 and T = 2.