Stochastic Foundations for Finance
Fall 2024
Definition 1 The derivative of f(x) at the point x is denoted by f'(x) and is defined as: \begin{aligned} f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}. \end{aligned}
Example 1 (Derivative of a Linear Function) Let’s first apply Definition 1 to compute the derivative of f(x) = x. Intuitively the rate of change of x with respect to x is one-to-one, which suggest that this derivative should be 1. To see this formally, note that: \begin{aligned} \frac{(x + h) - x}{h} = 1, \end{aligned} which shows that f'(x) = 1.
Example 2 (Derivative of a Square Function) We can apply Definition 1 to compute the derivative of f(x) = x^{2}. We first note that: \begin{aligned} \frac{(x + h)^{2} - x^{2}}{h} = \frac{x^{2} + 2 xh + h^{2} - x^{2}}{h} = 2x + h, \end{aligned} which allows us to compute: \begin{aligned} f'(x) = \lim_{h \rightarrow 0} 2x + h = 2x. \end{aligned}
Name | Formula |
---|---|
Sum Rule | (f + g)' = f' + g' |
Product Rule | (f g)' = f' g + f g' |
Inverse Rule (f \neq 0) | (1/f)' = - 1/(f')^{2} |
Chain Rule | (g \circ f)' = (g' \circ f) f' |
Example 3 (Scalar Multiplication) Consider an arbitrary differentiable function f and let g = c, where c \in \mathbb{R} is a constant. Then, g' = 0 and the product rule implies that: \begin{aligned} (f g)' = f' g + f g' = c f'. \end{aligned} Therefore, (c f)' = c f'.
Example 4 (Derivative of a Power Function) We can use the product rule to compute the derivative of x^{3} as follows: \begin{aligned} (x^{3})' & = (x x^{2})' \\ & = x^{2} + x (2 x) \\ & = 3 x^{2}. \end{aligned}
Example 5 (Derivative of the Exponential Function) Let f(x) = e^{x} and note that: \frac{e^{x + h} - e^{x}}{h} = e^{x} \left( \frac{e^{h} - 1}{h} \right). We can use the Taylor expansion of the exponential function to write: \begin{aligned} \frac{e^{h} - 1}{h} & = \frac{1}{h} \left( 1 + \frac{h}{1!} + \frac{h^{2}}{2!} + \frac{h^{3}}{3!} + ... - 1 \right) \\ & = 1 + \frac{h}{2!} + \frac{h^{2}}{3!} + \frac{h^{3}}{4!} + ..., \end{aligned} which shows that: f'(x) = e^{x} \lim_{h \rightarrow 0} \frac{e^{h} - 1}{h} = e^{x}.
Example 6 (Derivative of the Logarithm Function) Define f(x) = \exp(\ln(x)) = x. The chain rule then implies that: \begin{aligned} f'(x) & = \ln'(x) \exp(\ln(x)) \\ & = \ln'(x) x \\ & = 1. \end{aligned} Therefore, \ln'(x) = 1 / x.
Example 7 Consider a power function such that f(x) = x^{\alpha} where x > 0 and \alpha \in \mathbb{R}. Since f(x) = e^{\alpha \ln(x)}, the chain rule allows us to compute: \begin{aligned} f'(x) = \alpha \ln'(x) e^{\alpha \ln(x)} = \alpha x^{-1} x^{\alpha} = \alpha x^{\alpha - 1}. \end{aligned}
Function | Derivative |
---|---|
e^{x} | e^{x} |
\ln(x) | x^{-1} |
x^{\alpha} | \alpha x^{\alpha - 1} |
Definition 2 The differential of y = f(x) at the point x is denoted by dy and is defined as: dy = f'(x) dx, where dx \in \mathbb{R} is an arbitrary quantity.
Property 1 (First Order Approximation) Take y = f(x) and define \Delta y = f(x + \Delta x) - f(x) for a change \Delta x in x. If \Delta x is small, then we have that: \Delta y \approx f'(x) \Delta x
Property 2 (Second Order Approximation) Take y = f(x) and define \Delta y = f(x + \Delta x) - f(x) for a change \Delta x in x. If \Delta x is small, then we have that: \Delta y = f(x + \Delta x) - f(x) \approx f'(x) \Delta x + \frac{1}{2} f''(x) (\Delta x)^{2}
Property 3 (Important Properties of the Definite Integral) If f(x) and g(x) are two integrable functions over [a,b], then
Definition 3 Any function F such that F'(x) = f(x) is called the antiderivative or indefinite integral of f.
Function | Antiderivative |
---|---|
e^{x} | e^{x} |
1/x | \ln(x) |
x^{\alpha} | \dfrac{x^{\alpha + 1}}{\alpha + 1} |
Theorem 1 Let f be a continuous function on [a,b]. Let F be a function defined for all x \in [a,b] by: F(x) = \int_{a}^{x} f(t) \, dt. Then F is uniformly continuous on [a,b], differentiable on (a,b), and F'(x) = f(x) for all x \in (a,b).
Theorem 2 Let f be a function defined on [a,b] and F an antiderivative of f in (a,b). If f is Riemann integrable on [a,b] then \int_{a}^{b} f(x) \, dx = F(b) - F(a).
Example 8 (Present Value of a Continuous-Time Annuity) The most common use of integrals in finance is to compute the present value of continuous cash flows. At each time t \in [0, T], a security pays a cash flow of c \, dt. If the interest rate is r expressed with continuous compounding, the present value of these cash flows is: V = \int_{0}^{T} c e^{-r t} \, dt = c \left.\left( - \frac{e^{-r t}}{r} \right) \right|_{0}^{T} = \frac{c}{r} \left( 1 - e^{-r T} \right). This is the continuous-time analog of the present value of an annuity.
Example 9 Consider the continuous annuity presented in Example 8 when T \rightarrow \infty. This is what we called a perpetuity, that in this case pays at each time t a cashflow of c \, dt. The value of this instrument is then: V = \int_{0}^{\infty} c e^{-r t} \, dt = \lim_{T \rightarrow \infty} \int_{0}^{T} c e^{-r t} \, dt = \lim_{T \rightarrow \infty} \frac{c}{r} \left( 1 - e^{-r T} \right) = \frac{c}{r}.