Stochastic Foundations for Finance
Fall 2024
Example 1 (Computing Stock Price Probabilities) Time is measured in years. Suppose that the price S_{t} of a stock is such that log-changes are i.i.d. normally distributed such that \operatorname{E}_{t} \Delta \ln(S_{t}) = 20\% and \sigma_{t} \Delta \ln(S_{t}) = 30\% per year. If the stock price today is $100, what is the probability that the stock price is greater than $110 in 5 years from now? \begin{aligned} \operatorname{P}(S_{5} \leq 110) & = \operatorname{P}(\ln(S_{5}) \leq \ln(110)) \\ & = \operatorname{\Phi}\left(\frac{\ln(110) - (\ln(100) + 0.2 \times 5)}{0.3 \sqrt{5}} \right) \\ & = 8.87 \%. \end{aligned} Thus, \operatorname{P}(S_{5} > 110) = 1 = 0.0887 =91.13 \%.
Example 2 (Calculating a Confidence Interval on the Stock Price) Consider a stock whose price follows a lognormal distribution. The expected return of the stock price returns is 12% per year and the volatility is 25% per year. The current stock price is $25. If T = 0.5, we have that: \begin{aligned} \operatorname{E}(\ln(S_{T})) & = \ln(25) + \left(0.12 - 0.5(0.25)^{2}\right)(0.5) = 3.2633, \\ \operatorname{SD}(\ln(S_{T})) & = 0.25 \sqrt{0.5} = 0.1768. \end{aligned} Using the 95% confidence interval for \ln(S_{T}) we can compute [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.
Example 3 Consider a stock whose price follows a lognormal distribution. The expected return of the stock price returns is 12% per year and the volatility is 25% per year. If the current stock price is $25, the expected price and standard deviation 6 months from now are: \begin{align*} \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{align*}