M10 Simple Models of Rates of Return

Topics in Insurance, Risk, and Finance

Author

Affiliation

Yuyu Chen

Department of Economics, University of Melbourne

Published

2026

Introduction

Learning outcomes

Define deterministic model and stochastic model and understand their pros and cons.
Define fixed rate model and varying rate model.
Understand notations \(S_n\), \(V_n\), \(A_n\), and \(P_n\).
Calculate or derive an analytical expression of moments (especially the first two) of series of cash flows, not limited to \(S_n\), \(V_n\), \(A_n\), and \(P_n\).
Derive algebraically recursive relationships between cash flows.

Investment rates

Many financial contracts are long-term (e.g., annuities), and interest rates/investment rates play an important role in determining the values of these contracts.
In many contexts, rates of return are regarded as constant, mainly for mathematical convenience.
However, interest rates/investment rates can change drastically due to the changes of economic environment.
Rates of return are essentially random/stochastic.

Constant or random?

Q: What is wrong with models with constant rates of return?

Denote by \(i\) the rate of return, which is a random variable.
If we want to pick a number as the constant rate of return, a natural candidate is the mean rate of return, i.e., \(\E(i)\).
Let us compute the accumulated value of 1 and the present value of 1 using constant and random rates of returns, respectively.
Jensen’s inequality: if \(X\) is a random variable and \(f\) is a convex function, then \(\E(f(X))\ge f(\E(X)).\)

Investment rates

Using constant rates of return may underestimate risks.
Modelling investment rates is a crucial task for financial institutions.
We will study simple stochastic models for rates of return in this subject.
Advanced models will be covered in another subject.

\(S_n\) and \(V_n\)

Assume that the rate of return in period \([t-1,t]\) is \(i_t\), then

A single investment of 1 at time 0 will accumulate at time \(n\) to: \[S_n=\prod_{t=1}^n(1+i_t).\]
The present value of 1 at time \(n\) is worth at time \(0\): \[V_n=\prod_{t=1}^n(1+i_t)^{-1}.\]

\(A_n\) and \(P_n\)

A series of annual investments, each of amount 1, at times \(0,1,\dots,n-1\) will accumulate at time \(n\) to: \[A_n=\sum_{t=1}^{n} \prod_{k=t}^{n} (1+i_k).\]
The present value of a series of annual investments, each of amount 1, at times \(1,2,\dots,n\) is: \[P_n=\sum_{t=1}^{n}\prod_{k=1}^{t}(1+i_k)^{-1}.\]

Objectives

We will study the above quantities using:

Simple models for the rates of return,
Lognormal distributions to model the rates of return,
Numerical simulations in Excel.

Two types of models: deterministic model

Deterministic model: The investment rates are deterministic over the time period (note the difference between “deterministic” and “constant”). Pros:

easy to use
mathematically friendly

Cons:

hard to determine the rates
provides one single answer which is correct only if the rates are correct
not useful for risk management

Note that in a deterministic model, the interest rates are prespecified, hence the aforementioned quantities are constants.

Two types of models: stochastic model

Stochastic model: The investment rates are allowed to vary with the use of probability/statistics methods. Pros:

allow us to have a range of answers to our problem
good for risk management (variance can be useful)

Cons:

calculation is more challenging
sometimes is not mathematically friendly
models can be wrong

Note that in a stochastic model, the aforementioned quantities are random variables.

Fixed rate model

In a fixed rate model

the investment rate follows a distribution function
it is determined right after the investment is made
the rate will be a constant throughout the period of the investment
the rate is applied as a compound rate
Note that it belongs to the class of stochastic models.

Fixed rate model: example A I

Suppose that an investment of \(5000\) is made and will be accumulate at the investment rate \(i_k\). The annual investment rate \(i\) in a fixed rate model follows a three-point distribution: \(\p(i=0.06)=0.2\), \(\p(i=0.08)=0.7\), and \(\p(i=0.10)=0.1\). Consider the following questions.

Find the expectation and variance of the accumulated investment value at the end of year 5.
Find the accumulated investment value at the mean rate of return.

Fixed rate model: example A II

Fixed rate model: example A III

Fixed rate model: example B I

Consider events \(A\) and \(A^c\) which denote the two states of an economy (e.g., good and bad) such that \(\p(A)=0.5\). In state \(A\), the rate of return is uniformly distributed on \(0.1\) and \(0.2\). In state \(A^c\), the rate of return is uniformly distributed on \(0.3\) and \(0.4\). Find the mean of \(S_2\).

Fixed rate model: example B II

Fixed rate model: example C

Suppose the rate of return \(i\) follows the distribution \[F(x)= \begin{cases} &0,~~~0\le x<0.2,\\ &x/0.3,~~~0.2\le x\le 0.3. \end{cases}\] Find the mean of \(V_1\).

Varying rate model

In a varying rate model,

each investment rate follows a distribution
it is determined for each period of investment
the rate of return for each period will be one realization from the distribution
the rates in different periods are independent of each other
the rates are applied as compound rates

Key difference: the rates in a fixed rate model is constant throughout the life of the investment while those in a varying rate model can be different in each period of investments.

Varying rate model: example

Assume that the rate of return for each period takes values \(0.02\), \(0.04\), and \(0.06\) with equal probabilities. What is the probability that \(S_n=1.02*1.06^{n-1}\)?

Varying rate model

Consider a time interval \([0,n]\) with subintervals \([0,1],\dots,[n-1,n]\). We fix the following notations:

\(P_t\): the investment made at time \(t\)
\(F_t\): the accumulated value of the investment right before time \(t\) (i.e., before any investment is made at time \(t\))

Then we have the recursive relation for \(t=1,\dots,n\): \[F_t=(1+i_t)(F_{t-1}+P_{t-1})\]

Varying rate model: a simple example

Suppose that \(F_0=0\), \(P_0=1\), \(P_2=2\) and \(\E[i_t]=0.1\). Find the mean of accumulated value at time 3.

Varying rate model: \(S_n\) and \(V_n\)

Moments of \(S_n\)

Note that \(\E[V_n]\neq \E[S_n]^{-1}\)

Moments of \(S_n\): example A

Suppose that each of \(i_1,\dots,i_n\) has mean \(\mu\) and variance \(\sigma^2\). Find the expectation and variance of \(S_n\).

Moments of \(S_n\): example B

In a varying rate model, let \(i_1,\dots,i_n\) have the following distribution: \[ i_t= \begin{cases} \mbox{0.04 with probability 0.25},\\ \mbox{0.06 with probability 0.60},\\ \mbox{0.08 with probability 0.15}. \end{cases} \] Calculate the mean and variance of \(S_n\) for \(n = 5, 10, 20\), and comment on the values.

We have

\[\E(S_5)=1.3256,\E(S_{10})=1.7573,\E(S_{20})=3.0883,\] and \[\var(S_5)=0.0012,\var(S_{10})=0.0045,\var(S_{20})=0.0266.\]

Moments of \(S_n\): example B

As \(n\) increases:

The expected accumulation increases. This is as we would expect since the longer the period of the accumulation, the greater the accumulation, and its expected value, must be.
The variance of the accumulation increases. That is, the longer the period into the future, the more uncertain we are about what the accumulation will be.

Moments of \(V_n\): example A

In a varying rate model, let \(i_1,\dots,i_n\) have the following distribution: \[ i_t= \begin{cases} \mbox{0.04 with probability 0.25},\\ \mbox{0.06 with probability 0.60},\\ \mbox{0.08 with probability 0.15}. \end{cases} \] Calculate the mean and variance of \(V_n\) for \(n = 5, 10, 20\), and comment on the values.

We have

\[\E(V_5)=0.7549,\E(V_{10})=0.5698,\E(V_{20})=0.3247,\] and \[\var(V_5)=0.00040,\var(V_{10})=0.00045,\var(V_{20})=0.00029.\]

Moments of \(V_n\): example A

As \(n\) increases:

The expected discounted value decreases. This is as we would expect.
The variance of the discounted value decreases. Does it mean that, the longer the period into the future, the less uncertain we are about what the discounted value will be?
Another way to measure variability is the coefficient of variation (cv): For a random variable \(X\), we have \[cv(X)=\frac{\sd(X)}{\E(X)}.\]
We have \(cv(V_5)=0.0264,cv(V_{10})=0.0373,cv(V_{20})=0.0528,\) which increase as \(n\) increases. Hence, we have more uncertainty in this perspective.

Moments of \(V_n\): example B

Assume that \(i_1,\dots,i_n\) are iid and \(i_t\sim U(0.05,0.09)\) for \(t=1,\dots,n\). Find the mean and variance of the present value of a unit payment in 20 periods.

Varying rate model: \(A_n\) and \(P_n\)

Recursive formula

Unlike \(S_n\) and \(V_n\), it is generally not easy to derive analytical expressions for the moments of \(A_n\) and \(P_n\).
For this, we can use recursive formulas of \(A_n\) and \(P_n\) to derive their moments.

Moments of \(A_n\): mean

Moments of \(A_n\): second moment

Since \[A_n^2=(1+i_n)^2(A_{n-1}+1)^2,\] and \(i_1,\dots,i_n\) are independent, we have \[\E[A_n^2]=\E[(i_n^2+2i_n+1)]\E[(A_{n-1}^2+2A_{n-1}+1)].\]

If \(i_1,\dots,i_n\) have the same mean \(\mu\) and variance \(\sigma^2\), we get a recursive relation \[\begin{align*} \E[A_n^2]=(1+2\mu+\mu^2+\sigma^2)(\E[A_{n-1}^2]+2\E[A_{n-1}]+1), \end{align*}\] where \(\E[A_{n-1}]\) is known.

Moments of \(A_n\): example

Suppose that in a varying rate model \(i_1,\dots,i_n\) follow a uniform distribution on \([0.02,0.06]\). Find the mean and variance of \(A_5\).

We first note that for \(t=1,\dots,n\), \[\mu=\E[i_t]=\frac{0.02+0.06}{2}=0.04,\] and \[\sigma^2=\var(i_t)=\frac{(0.06-0.02)^2}{12}\approx0.00013333.\]

The mean of \(A_5\) is \[\E[A_5]=\frac{(1+\mu)^5-1}{d}=\frac{1.04^5-1}{0.04/1.04}=5.63298.\]

Moments of \(A_n\): example

To get the variance of \(A_5\), we need to use the recursive formula \[\begin{align*} \E[A_n^2]&=(1+2\mu+\mu^2+\sigma^2)(\E[A_{n-1}^2]+2\E[A_{n-1}]+1)\\ &=1.08173333(\E[A_{n-1}^2]+2\E[A_{n-1}]+1). \end{align*}\] We first compute \[\E[A_1]=1.04,\E[A_2]=2.1216,\E[A_3]=3.24646,\] \[\E[A_4]=4.41632.\]

Moments of \(S_n\) and \(A_n\): example IV

Using the recursive relation, we get \[\E[A_1^2]=1.08173,\E[A_2^2]=4.50189,\E[A_3^2]=10.54158,\] \[\E[A_4^2]=19.50853,\E[A_5^2]=31.73933.\] Hence, the variance is \[\var(A_5)=31.73933-5.63298^2=0.0089172.\]

Moments for \(P_n\) I

We can use similar techniques to derive moments for \(P_n\) as well. We have \[\begin{align*} P_n&=\sum_{t=1}^{n}\prod_{k=1}^{t}(1+i_k)^{-1}\\ &=\prod_{k=1}^{1}(1+i_k)^{-1}+\prod_{k=1}^{2}(1+i_k)^{-1}+\dots+\prod_{k=1}^{n}(1+i_k)^{-1}\\ &=(1+i_1)^{-1}\left(1+(1+i_2)^{-1}+\dots+\prod_{k=2}^{n}(1+i_k)^{-1}\right)\\ &=(1+i_1)^{-1}(1+P_{n-1}^*), \end{align*}\] where \(P_{n-1}^*\) is the time-1 value of payments of 1 at times \(2,\dots,n\).

Moments for \(P_n\) II

Note that \(P_{n-1}^*\) and \(P_{n-1}\) have the same distribution for iid rates. Hence, \[\begin{align*} \E(P_n)&=\E((1+i_1)^{-1}(1+P_{n-1}^*))\\ &=\E((1+i_1)^{-1})(1+\E(P_{n-1}^*))\\ &=\E((1+i_1)^{-1})(1+\E(P_{n-1})), \end{align*}\] and \[\begin{align*} \E(P_n^2)&=\E((1+i_1)^{-2}(1+P_{n-1}^*)^2)\\ &=\E((1+i_1)^{-2})(1+2\E(P_{n-1})+\E(P_{n-1}^2)). \end{align*}\]

Infinite series of payments

Suppose that returns \(i_1,i_2,i_3\dots\) are iid. The time-0 value of payments of 1 at times \(0,1,2,\dots\) is denoted by \(c_{\infty}\). Find the first two moments of \(c_{\infty}\).

Infinite series of payments

Moments for \(P_n\): analytical solutions

Suppose \(i_1,\dots,i_n\) are iid. Denote by \(u_k\) the \(k\)th moment \(\E\left((1+i_t)^{-k}\right)\). Find the moments of \(P_n\).

We have \[\begin{align*} \E(P_n)&=\E\left(\sum_{t=1}^{n}\prod_{k=1}^{t}(1+i_k)^{-1}\right)\\ &=\sum_{t=1}^{n}\prod_{k=1}^{t}\E\left((1+i_k)^{-1}\right)\\ &=\sum_{t=1}^{n}u_1^t=\frac{u_1(1-u_1^n)}{1-u_1}. \end{align*}\]

Moments for \(P_n\): analytical solutions

Finding the second moment is more difficult. We look at the case when \(n=2\): We have \[\begin{align*} \E(P_2^2)&=\E\left(\left((1+i_1)^{-1}+(1+i_1)^{-1}(1+i_2)^{-1}\right)^2\right)\\ &=\E\left((1+i_1)^{-2}\right)+2\E\left((1+i_1)^{-2}\right)\E\left((1+i_2)^{-1}\right)\\ &~~~+\E\left((1+i_1)^{-2}\right)\E\left((1+i_2)^{-2}\right)\\ &=u_2+2u_1u_2+u_2^2. \end{align*}\]

Moments for \(P_n\): analytical solutions

Next, we find the second moment of \(P_n\). We have \[\begin{align*} \E(P_n^2)&=\E\left(\left(\sum_{t=1}^n\prod _{k=1}^t(1+i_k)^{-1}\right)^2\right)\\ &=\E\left(\left(\sum_{t=1}^nV_t\right)^2\right)\\ &=\E\left(\sum_{t=1}^nV_t^2+2\sum_{j< k}V_jV_k\right)\\ &=\sum_{t=1}^n\E\left(V_t^2\right)+2\sum_{j< k}\E\left(V_jV_k\right). \end{align*}\] We know \[\sum_{t=1}^n\E\left(V_t^2\right)=\sum_{t=1}^nu^t_2=\frac{u_2(1-u_2^n)}{1-u_2}.\]

Moments for \(P_n\): analytical solutions

The rest is to find the expectation of the cross term. For \(j<k\), we have

Moments for \(P_n\): analytical solutions

Then \[\begin{align*} \sum_{j<k}\E(V_jV_k)&=\sum_{j=1}^{n-1}\sum_{k=j+1}^n u_2^ju_1^{k-j}\\ &=\sum_{j=1}^{n-1}\left(\frac{u_2}{u_1}\right)^j\sum_{k=j+1}^n u_1^{k}\\ &=\sum_{j=1}^{n-1}\left(\frac{u_2}{u_1}\right)^j\frac{u_1^{j+1}-u_1^{n+1}}{1-u_1}\\ &=\frac{u_1}{1-u_1}\left(\sum_{j=1}^{n-1}u_2^j-u_1^n\sum_{j=1}^{n-1}\left(\frac{u_2}{u_1}\right)^j\right)\\ &=\frac{u_1u_2}{1-u_1}\left(\frac{1-u_2^{n-1}}{1-u_2}-u_1\frac{u_1^{n-1}-u_2^{n-1}}{u1-u2}\right). \end{align*}\]