Solving Difference Equations

Stochastic Foundations for Finance

Lorenzo Naranjo

Fall 2024

Simple Linear Difference Equations

Introduction

• In finance and economics, we are usually interested in describing the evolution of economic quantities over time.
• For example, we might want to describe how a stock price or an interest rate evolves as time goes by.
• In these notes, time can only take discrete values.
  • For example, we might assume that t = 0, 1, 2, 3, \ldots
  • In some cases, we might even consider t = \ldots, -2, -1, 0, 1, 2, \ldots
• We will then index the variable of interest by t.
  • For example, the stock price at time t is given by S_{t}.

A Simple Difference Equation

• Consider the evolution of y_{t} given by y_{t} = a + b y_{t-1} \tag{1} and assume that y_{0} is given.
• We can iterate to find future values of y_{t}, \begin{aligned} y_{1} & = a + b y_{0}, \\ y_{2} & = a + b y_{1} = a + b (a + b y_{0}) = a + b a + b^{2} y_{0}, \\ y_{3} & = a + b y_{2} = a + b (a + b a + b^{2} y_{0}) = a + b a + b^{2} a + b^{3} y_{0}. \end{aligned}
• We can see that there is a pattern such that y_{t} = a (b^{0} + b^{1} + b^{2} + \ldots + b^{t-1}) + b^{t} y_{0}.

A Useful Result

• Consider the sum S = 1 + b + b^{2} + \ldots + b^{t - 1}, and assume that b \neq 1.
• Multiplying S by b yields b S = b + b^{2} + b^{3} \ldots + b^{t}.
• Subtracting the second from the first expression allows us to solve for b S = \frac{1 - b^{t}}{1 - b}.
• If b = 1, it’s easy to see that S = t, and we can show that \lim_{b \rightarrow 1} S = t.

Solving the Difference Equation

• We can now see that y_{t} = a \left(\frac{1 - b^{t}}{1 - b}\right) + b^{t} y_{0}, is a solution of y_{t} = a + b y_{t-1} given y_{0}.
• If \left\lvert b \right\rvert < 1, we know that \lim_{t \rightarrow \infty} b^{t} = 0.
  • If 0 < b < 1 the convergence is monotonic but if -1 < b < 0 the convergence is alternating.
  • In finance applications we typically assume that 0 < b < 1 but some interesting economic models use -1 < b < 0.
• Denote by \bar{y} the limit of y_{t} as t goes to infinity, so that \bar{y} = \lim_{t \rightarrow \infty} y_{t} = \frac{a}{1 - b}.

Expressing the Solution as Deviations from Steady-State

• Note that \bar{y} solves (1) since \bar{y} = a + b \bar{y}.
  • Therefore \bar{y} is a particular solution.
• We can re-express the general solution of (1) as y_{t} = \bar{y} + b^{t} (y_{0} - \bar{y}).
  • The model defined by (1) implies that y_{t} starts at y_{0} and converges exponentially to its steady-state value \bar{y}.

Figure 1: The figure shows solutions to y_{t} = a + b y_{t-1} for different values of y_{0} and b > 0. In the figure, the dashed line represents the steady-state solution given by \bar{y} = a / (1 - b). The parameters a is chosen so that \bar{y} = 1.

Mean-Reversion

• The model defined by (1) exhibits mean-reversion.
  • The values of y_{t} converge to their long-run mean \bar{y}.
• Note that we could write it as \Delta y_{t} = (1 - b) (\bar{y} - y_{t - 1}), \tag{2} where \Delta y_{t} = y_{t} - y_{t - 1}.
  • Assume 0 < b < 1.
  • In this case, when y_{t-1} is above \bar{y}, we have that \Delta y_{t} < 0, implying that y_{t} < y_{t - 1}.
  • When y_{t-1} is below \bar{y}, the opposite happens and y_{t} > y_{t - 1}.

Figure 2: The figure shows the solution to y_{t} = a + b y_{t-1} when b = -0.90 and \bar{y} = a / (1 - b) = 1. The solution oscillates around its steady-state value until convergence.

Modeling Equity Prices

Discounting Dividends

• The price of a stock at time t should be equal to the present value of its price next period plus any dividend paid during the period, i.e., p_{t} = \frac{p_{t+1} + d_{t+1}}{1 + r}. \tag{3}
  • We assume here that the dividend is paid at the end of the period, meaning p_{t} is the price of the stock after it goes ex-dividend.
• We can iterate forward to find p_{t} = \frac{d_{t+1}}{1 + r} + \frac{d_{t+2}}{(1 + r)^{2}} + \ldots + \frac{d_{t+n}}{(1 + r)^{n}} + \frac{p_{t+n}}{(1 + r)^{n}}. \tag{4}

A Note on Bubbles

• Taking limits in (5) we get p_{t} = \sum_{n = 1}^{\infty} \frac{d_{t+n}}{(1 + r)^{n}} + \lim_{n \rightarrow \infty} \frac{p_{t+n}}{(1 + r)^{n}}. \tag{5}
• The first component in (5) is the fundamental value of the asset which is the present value of future dividends.
• The second component is a bubble!
  • If this term is not zero, the price of the asset can be anything.
  • To pin down a unique price for the stock, we assume \lim_{n \rightarrow \infty} \frac{p_{t+n}}{(1 + r)^{n}} = 0.
• In economics, this is known as the transversality condition.

The Dividend-Discount Model (DDM)

• Imposing the transversality condition, we must have that p_{t} = \sum_{n = 1}^{\infty} \frac{d_{t+n}}{(1 + r)^{n}}. \tag{6}
  • Equation (6) determines the fundamental value of the stock.
  • The fundamental value of the asset at any time is the present value of future dividends.
• In practice, it is quite hard to forecast dividends, so academics and practitioners add assumptions to model how dividends evolve over time.

Constant Dividend Growth

• The Gordon growth model assumes that dividends grow at a constant rate g < r, i.e., d_{t+1} = (1 + g) d_{t}.
• According to (6), the price of the stock next period is p_{t+1} = \sum_{n = 1}^{\infty} \frac{d_{t+1+n}}{(1 + r)^{n}} = (1 + g) \sum_{n = 1}^{\infty} \frac{d_{t+n}}{(1 + r)^{n}} = (1 + g) p_{t}.
• Thus, we can re-write (3) as p_{t} = \frac{(1 + g) p_{t} + d_{t+1}}{1 + r}. \tag{7}

The Gordon-Growth Model

• Equation (7) implies p_{t} = \frac{d_{t+1}}{r - g} = \frac{1 + g}{r - g} d_{t}. \tag{8}
• Equation (8) ties the price of the asset to fundamentals.
  • It is widely used by practitioners in corporate finance, investments and real estate.
• Thus, constant dividend growth implies that price-dividend ratios are constant over time, i.e., \frac{p_{t}}{d_{t}} = \frac{1 + g}{r - g}.

Present Value of a Growing Annuity

• We can use (4) and (8) to derive the present value of a growing annuity, i.e., PV = \frac{d_{t+1}}{1 + r} + \frac{d_{t+1}(1 + g)}{(1 + r)^{2}} + \ldots + \frac{d_{t+1}(1 + g)^{n-1}}{(1 + r)^{n}}.
• Note that \begin{aligned} PV & = p_{t} - \frac{p_{t+n}}{(1 + r)^{n}} = p_{t} - \frac{(1 + g)^{n} p_{t}}{(1 + r)^{n}} \\ & = p_{t} \left(1 - \frac{(1 + g)^{n}}{(1 + r)^{n}}\right) = \frac{d_{t+1}}{r - g} \left(1 - \frac{(1 + g)^{n}}{(1 + r)^{n}}\right). \end{aligned} \tag{9}

Application: Two-Stage DDM

• Amco last year paid a dividend of $0.20 per share. This dividend is expected to grow at 15% per year for the next three years, then at 10% per year until year 6, after which it is expected to grow at 5% in perpetuity. What is the stock’s value if your required rate of return is 10%?

  Time	0	1	2	3	4	5	6
  Dividend	0.2000	0.2300	0.2645	0.3042	0.3346	0.3681	0.4049
  Terminal Value						8.0971
  Cash Flows		0.2300	0.2645	0.3042	0.3346	8.4652

  • The price of the stock is the present value of the cash flows which is $6.14.

Where Does Growth Come From?

• Dividends are paid from earnings, so assume that a fraction 1 - \varphi of earnings is paid as dividends and the rest reinvested in projects generating a return on equity k > r.
• Earnings next period are thus equal to current earnings plus reinvested earnings times the return of equity, i.e., e_{t+1} = e_{t} + \varphi e_{t} k = (1 + \varphi k) e_{t}, implying that g = \varphi k.
• In this simple model, the dividend growth rate is equal to reinvested earnings times the return of equity.

Positive NPV Projects

• In which projects should the company invest?
• The price-to-future-earnings ratio is given by \frac{p_{t}}{e_{t+1}} = \frac{1 - \varphi}{r - \varphi k}.
• We can show that \frac{d}{d\varphi} \frac{p_{t}}{e_{t+1}} > 0 \Longleftrightarrow k > r.
  • The stock price increases if the company reinvests in projects with positive NPV, i.e., projects in which the IRR is greater than the cost of equity.
  • Of course, the company cannot reinvest everything forever, otherwise it would be a bubble.

Systems of Linear Difference Equations

Systems of Linear Difference Equations

• The previous model is an example of a system of difference equations.
• For example, we could model the evolution of x_{t} and y_{t} as \begin{pmatrix} x_{t} \\ y_{t} \end{pmatrix} = \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix} + \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \begin{pmatrix} x_{t-1} \\ y_{t-1} \end{pmatrix}. \tag{10}
• Systems such as (10) can explode, mean-revert or oscillate depending on the values of b_{ij}.
• Nevertheless, it is always possible to iterate forward to solve for future values of x_{t} and y_{t}

Solving the System

• It is convenient to write down (10) in more compact form as \mathbf{z}_{t} = \mathbf{a} + \mathbf{B} z_{t-1}, \tag{11} where \mathbf{z}_{t} = \begin{pmatrix} x_{t} \\ y_{t} \end{pmatrix}, \mathbf{a} = \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix}, and \mathbf{B} = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}.
• We can iterate again to find \mathbf{z}_{t+n} = \mathbf{a} (\mathbf{I} + \mathbf{B} + \mathbf{B}^{2} + \ldots + \mathbf{B}^{n-1}) + \mathbf{B}^{n} z_{0}, \tag{12} where \mathbf{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} is the identity matrix.
• We can write (12) more compactly as \mathbf{z}_{t+n} = \mathbf{a} (\mathbf{I} - \mathbf{B}^{n})(\mathbf{I} - \mathbf{B})^{-1} + \mathbf{B}^{n} z_{0}. \tag{13}

Note on Solutions

• With modern computers it is usually easier to solve the system directly by iterating equation (11).
• Equation (13) will converge to a steady-state solution if \lim_{n \rightarrow \infty} \mathbf{B}^{n} = 0, in which case \bar{\mathbf{z}} = \mathbf{a} (\mathbf{I - \mathbf{B}})^{-1}.
• This will happen whenever the absolute value of the eigenvalues of \mathbf{B} are strictly less than one.

Figure 3: The figure shows the solutions to \begin{pmatrix} x_{t} \\ y_{t} \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0.25 \end{pmatrix} + \begin{pmatrix} 0.85 & -0.3 \\ 0.4 & 0.9 \end{pmatrix} \begin{pmatrix} x_{t-1} \\ y_{t-1} \end{pmatrix}, given x_{0} = y_{0} = 0.

Second Order Difference Equations

• A second order difference equation is defined as x_{t} = a + b_{1} x_{t-1} + b_{2} x_{t-2}, \tag{14} where x_{0} and x_{1} are given.
  • This means you can only start computing x_{t} from t > 1 onwards.
• If we define y_{t} = x_{t-1}, we can write (14) as \begin{pmatrix} x_{t} \\ y_{t} \end{pmatrix} = \begin{pmatrix} a \\ 0 \end{pmatrix} + \begin{pmatrix} b_{1} & b_{2} \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_{t-1} \\ y_{t-1} \end{pmatrix}, \tag{15} where y_{1} = x_{0} and x_{1} is given.
  • The eigenvalues of \begin{pmatrix} b_{1} & b_{2} \\ 1 & 0 \end{pmatrix} can tell you whether x_{t} will converge to a steady-state value, oscillate, or diverge to infinity as t \rightarrow \infty.

Figure 4: The figure shows the solutions to \begin{pmatrix} x_{t} \\ y_{t} \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0 \end{pmatrix} + \begin{pmatrix} 0.5 & -0.8 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} x_{t-1} \\ y_{t-1} \end{pmatrix}, given x_{0} = y_{0} = 0.

The Impulse Response

Adding a Forcing Term

• We can add a non-autonomous component to a first-order difference equation y_{t} = b y_{t-1} + w_{t}. \tag{16}
  • We will now start at t = -1.
• To solve it we proceed as before, \begin{aligned} y_{0} & = b y_{-1} + w_{0}, \\ y_{1} & = b y_{0} + w_{1} = b (b y_{-1} + w_{0}) + w_{1} = b^{2} y_{-1} + b w_{0} + w_{1}, \\ y_{2} & = b y_{1} + w_{2} = b (b^{2} y_{-1} + b w_{0} + w_{1}) + w_{2} = b^{3} y_{-1} + b^{2} w_{0} + b w_{1} + w_{2}, \\ y_{t} & = b^{t} y_{-1} + (b^{t} w_{0} + b^{t-1} w_{1} + \ldots + b^{t-t} w_{t}) = b^{t} y_{-1} + \sum_{i = 0}^{t} b^{t-i} w_{i}. \end{aligned}

Interpretation

• In the next chapter, w_{t} will represent random shocks to the system transforming the process for y_{t} into a stochastic process.
• For the moment, though, we assume that w_{t} is deterministic and represents exogenous inputs to the model.

Impulse Response

• The impulse response represents the solution to (16) if w_{t} = \begin{cases} 1 \quad \text{if $t = 0$}, \\ 0 \quad \text{if $t > 0$}, \end{cases} assuming that the system is in steady-state, i.e., y_{-1} = 0.
• In this scenario, the solution to (16) is y_{t} = b^{t}, which is the impulse response of the system.

Optional Material

A Second Take on the Gordon Growth Model

• The Gordon growth model can be written as \begin{aligned} p_{t+1} & = (1 + r) p_{t} - (1 + g) d_{t}, \\ d_{t+1} & = (1 + g) d_{t}. \end{aligned}
• In matrix form \begin{pmatrix} p_{t+1} \\ d_{t+1} \end{pmatrix} = \begin{pmatrix} 1 + r & -(1 + g) \\ 0 & 1 + g \end{pmatrix} \begin{pmatrix} p_{t} \\ d_{t} \end{pmatrix}.
• We can iterate again to find \begin{pmatrix} p_{t+n} \\ d_{t+n} \end{pmatrix} = \begin{pmatrix} 1 + r & -(1 + g) \\ 0 & 1 + g \end{pmatrix}^{n} \begin{pmatrix} p_{t} \\ d_{t} \end{pmatrix}. \tag{17}

Explicit Solution

• It is possible to get an explicit solution for the matrix exponential in (17), \begin{pmatrix} p_{t+n} \\ d_{t+n} \end{pmatrix} = \begin{pmatrix} (1 + r)^{n} & \dfrac{1 + g}{r - g}((1 + g)^{n} - (1 + r)^{n}) \\ 0 & (1 + g)^{n} \end{pmatrix} \begin{pmatrix} p_{t} \\ d_{t} \end{pmatrix}, implying \begin{aligned} p_{t+n} & = (1 + r)^{n} p_{t} + \dfrac{1 + g}{r - g}((1 + g)^{n} - (1 + r)^{n}) d_{t}, \\ d_{t+n} & = (1 + g)^{n} d_{t}. \end{aligned} \tag{18}
• The second equation above is not a surprise, but the first equation tells how the future stock price relates to the current price.

No Bubbles

• Equation (18) implies \frac{p_{t+n}}{(1 + r)^{n}} = p_{t} - \dfrac{1 + g}{r - g}\left(1 - \frac{(1 + g)^{n}}{(1 + r)^{n}}\right) d_{t}. \\
• If we apply the transversality condition to the expression above we get, 0 = \lim_{n \rightarrow \infty} \frac{p_{t+n}}{(1 + r)^{n}} = p_{t} - \dfrac{1 + g}{r - g} d_{t}, as long as g < r.
• The transversality condition is necessary to relate the stock price to fundamentals.
  • Otherwise the stock price can be anything regardless of the dividends paid by the stock.