Stochastic Foundations for Finance
Fall 2024
If Z \sim \mathcal{N}(0, 1), we have that: \operatorname{\Phi}(z) =\operatorname{P}(Z \leq z) = \int_{-\infty}^{z} \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} \, dx.
Since the integral cannot be solved in closed-form, the probability must then be obtained from a table or using a computer.
Example 1 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that X \leq 0? \operatorname{P}(X \leq 0) = \operatorname{P}\left( Z \leq \tfrac{0 - 10}{25} \right) = \operatorname{\Phi}(-0.40) = 0.3446.
Example 2 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that X > 12? \operatorname{P}(X \leq 12) = \operatorname{P}\left( Z \leq \tfrac{12 - 10}{25} \right) = \operatorname{\Phi}(0.08) = 0.5319. Therefore, \operatorname{P}(X > 12) = 1 - 0.5319 = 0.4681.
Example 3
Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. What is the probability that 2 < X \leq 14? \begin{aligned} \operatorname{P}(X \leq 14) & = \operatorname{P}\left( Z \leq \tfrac{14 - 10}{25} \right) = \operatorname{\Phi}(0.16) = 0.5636, \\ \operatorname{P}(X \leq 2) & = \operatorname{P}\left( Z \leq \tfrac{2 - 10}{25} \right) = \operatorname{\Phi}(-0.32) = 0.3745. \end{aligned}Therefore, \operatorname{P}(2 < X \leq 14) = 0.5636 - 0.3745 = 0.1891.
For a standard normal variable Z, a right-tail percentile is the value z_{\alpha} above which we obtain a certain probability \alpha.
Mathematically, this means finding z_{\alpha} such that: \operatorname{P}(Z > z_{\alpha}) = \alpha \Leftrightarrow \operatorname{P}(Z \leq z_{\alpha}) = 1 - \alpha
The following table shows common values for z_{\alpha}:
\alpha | 0.05 | 0.025 | 0.01 | 0.005 |
---|---|---|---|---|
z_{\alpha} | 1.64 | 1.96 | 2.33 | 2.58 |
Example 4 Since z_{2.5\%} = 1.96, the 95% confidence interval of Z is [-1.96, 1.96]. This means that if we randomly sample this variable 100,000 times, approximately 95,000 observations will fall inside this interval.
Example 5 Suppose that X \sim \mathcal{N}(\mu, \sigma^{2}) with \mu = 10 and \sigma = 25. Since z_{2.5\%} = 1.96, the 95% confidence interval of X is: [10-1.96(25), 10+1.96(25)] = [-39, 59].
Example 6 Let Y = e^{4 + 1.5 Z} where Z \sim \mathcal{N}(0, 1). What is the probability that Y \leq 100? \begin{aligned} \operatorname{P}(Y \leq 100) & = \operatorname{P}(e^{X} \leq 100) \\ & = \operatorname{P}(X \leq \ln(100)) \\ & = \operatorname{P}\left(Z \leq \tfrac{\ln(100) - 4}{1.5}\right) \\ & = \operatorname{\Phi}(0.4034) \\ & = 0.6567 \end{aligned}
Example 7 Let Y = e^{4 + 1.5 Z} where Z \sim \mathcal{N}(0, 1). The expectation and standard deviation of Y are: \begin{aligned} \operatorname{E}(Y) & = e^{4 + 0.5(1.5^{2})} = 168.17 \\ \operatorname{SD}(Y) & = 168.17 \sqrt{e^{1.5^{2}} - 1} = 489.95 \end{aligned}