Discount functions

Define a discount function $v(t_1, t_2)$ is interpreted as the time $t_1$ value of a payment of $1$ at time $t_2$ .

0 <= t1, t2
0 < v(t1, t2)

def v(t1: float, t2: float) -> float:

If $t_1 < t_2$ , then values are discounted to $t_1$ and likely $v(t_1, t_2) < 1$
If $t_1 > t_2$ , then values are accumulated to $t_2$ and likely $v(t_1, t_2) > 1$

For discount functions we assume that

\begin{align} v(t_1, t_2) &> 0 \\ v(t_1, t_2) \cdot v(t_2, t_3) &= v(t_1, t_3) \quad \forall t_1, t_2, t_3 \in \mathbb{R}_{\geq 0} \end{align}

Arbitrage with forward rates

Arbitrage opportunities are when you can create profit with 0 initial investment.

Our definition of a discount function $v(t_1, t_2) \cdot v(t_2, t_3) = v(t_1, t_3)$ avoids arbitrage opportunities.

If we had

v(t_1, t_2) \cdot v(t_2, t_3) > v(t_1, t_3)

We can buy a payment of $1$ at time $t_3$ for $v(t_1, t_3)$ at time 0.

We borrow this initial payment and owe $v(t_1, t_3) \cdot v(t_2, t_1)$ at time $t_2$ . To repay this at time $t_2$ we borrow again and repay $(v(t_1, t_3) \cdot v(t_2, t_1)) \cdot v(t_3, t_2)$ . So our $t_1$ net cashflow is 0, $t_2$ net cashflow is 0, and our $t_3$ net cashflow is positive. This is an arbitrage opportunity.

1 - (v(t_1, t_3) \cdot v(t_2, t_1)) \cdot v(t_3, t_2) = \\ 1 - \frac{v(t_1, t_3)}{v(t_1, t_2) \cdot v(t_2, t_3)} > 0 \\

A similar argument applies if $v(t_1, t_2) \cdot v(t_2, t_3) < v(t_1, t_3)$ .

Formulas

$v(t, t) = 1$

\begin{align*} v(t, t) \cdot v(t, t) &= v(t, t) \implies \\ v(t, t) \cdot v(t, t) - v(t, t) &= 0 \implies \\ v(t, t) \cdot (v(t, t) - 1) &= 0 \implies \\ v(t, t) &= 0 \text{ or } v(t, t) = 1 \end{align*}

We defined $v(t,t) > 0$ , so $v(t, t) = 1$ .

$v(t_1, t_2) = \frac{1}{v(t_2, t_1)}$

Follows from $v(t_1, t_2)v(t_2, t_1) = v(t_1, t_1) = 1$

$\prod_{i=1}^n v(t_i, t_{i+1}) = v(t_1, t_{n+1})$

Follows from induction.

Discrete and computational aspects

In practice, we may define discount rates over some discrete range ending at the last timestep $N$ in a simulation.

0 <= t1, t2 <= N
0 < v(t1, t2)

def v(t1: int, t2: int) -> float:

In this case we must consider how to specify the discount rate, and see that the vector

[v(0, t) for t in range(N)]

Allows for $O(1)$ lookups of the discount rate for any t1, t2 pair

Present value

We can define a 1-parameter version of our discount function

$v(t) = v(0, t)$

This is convenient because we are often interested in the present value, the value at time 0.

Exponential discounting

An exponential function will satisfy our discount function definition

e^{-k(t_2 - t_1)}e^{-k(t_3 - t_2)} = e^{-k(t_3 - t_1)}

To take the present value, $v(t) = e^{-kt}$ . Note that for something like an annual interest rate of 5%, $k = \ln(1.05)$ .

Code challenge

Return an array arr of length n where v0[i] = $v(0, i)$ for a constant interest_rate

Given an array annual_interest_rate[i] = $v(i, i+1)^{-1}-1$ , compute an array pv[i] = $v(0, i)$

Implement a function that takes an array cashflows and a discount array pv[i] = $v(0, i)$ and compute the following

\sum_{i=0}^N \text{cashflows}[i] \cdot v(k, i)

We want to know all values of a discount function $v(t_1, t_2)$ where

0 <= t1, t2 < N

We know some values of $v(t_1, t_2)$ , we need to compute the rest of them. The known values of $v(t_1, t_2)$ come in the format (t1, t2, v(t1, t2)).

Under what conditions do we violate $v(t_1, t_2) \cdot v(t_2, t_3) = v(t_1, t_3)$
Under what conditions is it possible to compute all values of $v(t_1, t_2)$ ?
If possible, construct a data structure in $O(N)$ that provides $O(1)$ lookup of $v(t_1, t_2)$ for any t1, t2 pair.