Fixed Income Markets

2026-01-12

Definition of Fixed Income

All these objects are affected by interest rates and default scenarios. We will first study fixed-income from given interest rates, then we will assume default can occur. We will not allow for interest rates to change with time.

Continuously Compounded Interest Rates

Assume $X$ dollars are deposited in two banks, Bank 1 with a rate of $j$ per year and Bank 2 with a risk of $r$ compounded every 6 months.

They should grow to the same amount after one year:

$$(1+j)X=(1+r)(1+r)X$$

If this were not true, then arbitrage would exist in the market.

Solving for $r$ we find that interest rate over the period of six months should be

$$r=(1+j{)}^{\frac{1}{2}}-1$$

Since there are two $6$ month periods in a year, and the interest rate at Bank 2 has to be quoted per year, we define

$$j_2:=2r=2((1+j{)}^{\frac{1}{2}}-1)$$

In general, we define $j_n$ as the interest rate $j$ compounded $n$ times.

2026-01-14

Conventions for This Course

We will reserve the variable $r:=j_1$ the simple rate per year. That is, any amount $X$ will grow to $(1+r)X$ after one year.

Thus, $j_n$ is the interest rate per year compounded $n$ times per year. That is, any amount $X$ will grow to $(1+\frac{j_n}{n}{)}^{n}X$ after one year. Note, this means that $j_n$ is not the interest rate per period, but the interest rate per year compounded $n$ times. The interest rate per period is $\frac{j_n}{n}$.

$R:=j_{\infty }$ is the force of interest. Any amount $X$ will grow to $e^{R}X$ after one year.

notice, we will always compare interest rates on a per year basis

In general, for a period $\Delta T$:

$$(1+r)=e^{R\Delta T}$$

Formally

• In general, $j_1=j=r$ is the interest rate per period $\Delta T$ compounded once, while $R=j_{\infty }$ is interest rate for the same period compounded continuously. These are the two most important rates: $j_1=r$ and $R=j_{\infty }$
• In the description before, we related $r$ and $R$ for various $\Delta T$ at a fixed moment in time, i.e. today $t$. These rates are likely to change with time $t$. Hence as can denote them as functions, for instance $r(t,\Delta T)=e^{(R(t)\Delta T)}-1$, leading to a whole area of modelling rates.
• Moreover, in reality, we actually observe at every $t$ the interest rate $r$ for a large variety of periods $\Delta T$. The observable $r(t,\Delta T)$ is called the Term Structure of interest rates. As there might be no single $R(t)$ capable of matching the observable rates, we need more complexity in the modelling of interest rates.
• Knowing the whole term structure $r(t,\Delta T)$ permits concept like forward rates, this is, the interest rate between times $t_i$ and $t_{i+1}$ as seen from $t<t_{i<}<t_{i+1}$, denoted $r(t,t_i,t_{i+1})$.

First-order Difference Equations

notice, these are difference equations, not differential equations.

Proposition 4

Let $X_k$ be the Initial Value Problem (IVP):

$$X_{k+1}=\alpha X_k+\beta ,X_0=1$$

Where $\alpha ,\beta$ are fixed constants. Then:

$$X_k={\alpha }^k+\frac{\beta }{1-\alpha }(1-{\alpha }^k)$$

Proof

We proceed by analogy to DE; we first divide it into a homogeneous denoted $H_k$ and a particular solution denoted $P_k$:

$$X_k=H_k+P_k$$

The homogeneous problem satisfies $H_{k+1}=\alpha H_k$, the solution is simply $H_k={\alpha }^k{H}_0$, where $H_0$ is a constant.

In finance, we aren’t mathematicians. Instead of formally deriving a solution, we often simply guess and check. In this case we can guess the solution is in this form:

$$X_k=a+b{c}^k$$

Assumption 1

• We borrow \ X$attime$t=0$andrepayitbypaying$$ A$at$N$

Mathematical Representation

Let $D_k$ be the amount we owe after making the $k$-th payment (at time $k\Delta T$ plus one instant), e.g.

$$D_1=(1+r)D_0-A=(1+r)X-A$$

Then, $D_k$ grows with interest rate $r$, compounded simply, applied to $D_k$. But it is diminished by the payment $A$ at time $(k+1)\Delta T$. This is:

$$D_{k+1}=(1+r)D_k-A,D_0=X$$

We know that $D_0=X$ and that $D_N=0$. So, we now have a difference equation with initial and final values specified.

Proposition 5

In the problem above, the discrete payments shall be:

$$A=\frac{r(1+r{)}^N}{(1+r{)}^N-1}X$$

Cases:

• Case 1. What happens if the number of payment periods $N$ approaches infinity? That is, we end up with a perpetual loan. Then:

$$A=\underset{N\to \infty }{\lim }\frac{r(1+r{)}^N}{(1+r{)}^N-1}X=rX$$

• Case 2. What happens if the number of payment periods $N=1$? then

$$A=\frac{r(1+r{)}^1}{(1+r{)}^1-1}X=(1+r)X$$

• If the investor agrees to paying fraction $e$ of the loan, then after maturity we can “reconsider”, then:

$$A=rX\frac{(1+r{)}^N-(1-e)}{(1+r{)}^N-1}$$

• Case 3. What happens if the interest rate $r=0$? then

$$A=\frac{X}{N}$$

Replaying the Loan Continuously

We take the loan of size $X$ now and want to repay it continuously between now, time $0$, and time $T$.

Approach I: Take limit as $\Delta T\to 0$ first, solve differential equation second. The interest on the loan is compounded continuously with force $R$ and we repay the loan continuously at rate $B$.

Our goal is to find the constant $B$ so that the loan is repaid by time $T$.

In human terms, $B$ is in units, dollars per time, e.g. dollars per year. This problem asks us, if I was paying this loan back every millisecond or less, how much would I have to pay per year to ensure the loan is repaid by time $T$?

Proposition 6

The rate to repay the loan continuously shall be:

$$B=\frac{RX{e}^{RT}}{e^{RT}-1}$$

Proof

Let $D(t)$ be the size of the loan at time $t\in [0,T]$. Since $T$ is the time when the loan is completely repaid, we have the boundary conditions

$$D(0)=X,D(T)=0$$

The loan grows continuously at force of interest $R$, as is repaid continuously at the constant rate $B$. Therefore, over an infinitesimal time interval $dt$,

• interest increases the balance by $RD(t)dt$,
• repayment decreases the balance by $Bdt$.

Thus the balance satisfies the differential equation

$$\frac{dD}{dt}=RD(t)-B$$

This is the first-order linear differential solution. Solving it using the integrating factor.

$$I=e^{\int -Rdt}=e^{-Rt}$$

Multiplying the ODE by $I$ we have

$$e^{-Rt}\frac{dD}{dt}-R{e}^{-Rt}D=-B{e}^{-Rt}$$

By the product rule, the left-hand side is

$$\frac{d}{dt}(e^{-Rt}D)=-B{e}^{-Rt}$$

Integrating both sides from $0$ to $t$ yields

$$e^{-Rt}D-D(0)=-B{\int }_0^t{e}^{-Rs}ds$$

Substituting the initial condition $D(0)=X$ gives

$$e^{-Rt}D-X=-B{\int }_0^t{e}^{-Rs}ds$$

Evaluating the integral on the right-hand side gives

$$e^{-Rt}D-X=-B{\left[-\frac{1}{R}{e}^{-Rs}\right]}_0^t=\frac{B}{R}(1-e^{-Rt})$$

Summary

In conclusion, we have 4 cases we considered

Types of Loans	Constant Payments	Monotone Payments
Discrete Time	Prop. 5	Prop. 7
Continuous Time	Prop. 6	Prop. 8