Case study: present value of a continuous cash flow

A stream of cash as an integral

Say money arrives at a rate of $c(t)$ dollars per year at time $t$, measured in years. In a short time $\Delta t$, about $c(t)\,\Delta t$ dollars arrive. Discounted continuously at the rate $r$ (case study 1.4.6), they are worth $c(t)\,e^{-rt}\,\Delta t$ today.

Adding these pieces over $[0, T]$ is a Riemann sum (lesson 2.5.1). As $\Delta t \to 0$, it becomes an integral.

For a constant rate $c$, an antiderivative of $e^{-rt}$ is $-\frac{e^{-rt}}{r}$: divide by $k = -r$, as in lesson 2.5.2. The fundamental theorem (lesson 2.5.3) then gives a closed form.

For the toll road, $c = 10{,}000$, $r = 0.05$ and $T = 10$, so $PV$ = $78,693.87. Undiscounted, the same 10 years pay $100,000.

Units on a desk

The formula needs one unit of time throughout: $c$ in dollars per year, $r$ per year and $T$ in years. Then $rT$ is a pure number, as an exponent must be.

• Convert cash flows first. $850 a month is $c = 10{,}200$ dollars a year.
• Rates enter as decimals. Typing $r = 5$ for 5% gives $2,000, nonsense for ten years of $10,000.
• Time is in years. Typing $T = 120$ months with a yearly rate gives $199,504.25, almost the value of tolls forever.

Continuous against yearly payments

At the same discount rate, a schedule of separate payments is a Riemann sum of the discounted rate. So lesson 2.5.1 says which way it errs.

Ten payments of $10,000 at the year-ends are worth $\sum_{k=1}^{10} 10{,}000\,e^{-0.05k}$ today. That is the right sum of $10{,}000\,e^{-0.05t}$ with $\Delta t = 1$. The discounted rate falls, so the right sum is too small: $76,742.92, short by $1,950.95.

Payments at the start of each year form the left sum, which is too large: $80,677.61. The two schedules differ by $\Delta t\,\big(f(0) - f(10)\big) = 10{,}000\,(1 - e^{-0.5})$, or $3,934.69: the gap rule of lesson 2.5.1, with $f$ the discounted rate.

Monthly payments close most of the gap, and mid-year payments nearly all of it, because midpoint sums are accurate. That is why valuations often treat a year's cash as arriving at mid-year.

Same rate, same convention

A comparison is fair only when both sides use the same rate. A rate of 5% compounded once a year is not a continuous 5%.

Compounded yearly, 5% discounts year $k$ by $1.05^{-k} = e^{-0.04879k}$: a continuous rate of $\ln 1.05 = 4.879\%$ (case study 1.4.6). Valued that way, the year-end payments are worth $77,217.35, not $76,742.92. Convert both sides to one convention before you compare them.

A growing stream

If the tolls grow at a continuous rate $g$, then $c(t) = c_0\,e^{gt}$, where $c_0$ is the rate now. The discounted rate becomes $c_0\,e^{-(r - g)t}$, and the same antiderivative, with $r - g$ in place of $r$, gives the present value.

When the formula breaks

• $g = r$. The formula reads $\frac{0}{0}$. But then the discounted rate is the constant $c_0$, so the present value is $c_0 T$: $100,000 here.
• $g$ almost equal to $r$. Rates often come out of other sums. $0.15 - 0.1$ is stored as 0.04999999999999999, so a test g == r against $r = 0.05$ fails. The formula then divides two tiny numbers, $1 - e^{-x}$ loses its digits, and it returns $80,000 instead of $100,000.
• $T \to \infty$. If $r > g$, then $e^{-(r - g)T} \to 0$ and $PV \to \frac{c_0}{r - g}$: the growing perpetuity behind dividend-discount models. If $g \ge r$, the area grows without limit and there is no finite value; lesson 2.6.3 treats such integrals.

math.expm1(x) computes $e^x - 1$ accurately even for tiny $x$. Writing the top of the fraction as $-\text{expm1}\big(-(r - g)T\big)$ repairs the second case.

The perpetuity also shows how sensitive values are when $g$ is close to $r$. At $r = 5\%$ it is $500,000 with $g = 3\%$, $1,000,000 with 4% and $2,000,000 with 4.5%. A half-point change in a growth guess can double the answer.

Default intensity

Credit desks model a borrower's default with a hazard rate $\lambda$ (lambda), also called a default intensity. It is a rate per year: over a short time $\Delta t$, a borrower who has not yet defaulted defaults with probability about $\lambda\,\Delta t$.

With a constant $\lambda$, the probability of surviving to time $t$ is $e^{-\lambda t}$. The default time then has density $\lambda e^{-\lambda t}$: a rate of probability, per year. Integrating that rate over the years gives a total probability (lesson 2.5.3):

At $\lambda = 2\%$ a year, that is 9.52% within 5 years and 18.13% within 10. The shortcut $\lambda T$ gives 10% and 20%: close while $\lambda T$ is small, worse as it grows. Level 5 develops densities properly.

A stream that stops at default

Suppose the tolls stop for good if the operator defaults, with nothing recovered. Cash at time $t$ arrives only if there has been no default by then, which has probability $e^{-\lambda t}$. So each moment's discounted cash is weighted by it.

At $\lambda = 2\%$, the toll road's 10 years are worth $71,916.39, which is $6,777.48 less than without default risk.

import math

from scipy.integrate import quad


def test_constant_stream():
    """$10,000 a year for 10 years at 5% is worth about $78,693.87"""
    expect_close(pv_stream(10_000, 0.05, 10), 78693.86805747332, what="pv_stream(10000, 0.05, 10)")


def test_growing_stream():
    """Growing at 3%, it is worth about $90,634.62"""
    expect_close(pv_stream(10_000, 0.05, 10, g=0.03), 90634.62346100906, what="pv_stream(..., g=0.03)")


def test_risky_stream():
    """With a 2% hazard rate, it is worth about $71,916.39"""
    expect_close(pv_stream(10_000, 0.05, 10, hazard=0.02), 71916.38517265576, what="pv_stream(..., hazard=0.02)")


def test_matches_quad():
    """It matches numerical integration for a growing, risky stream"""
    expected = quad(lambda t: 2500 * math.exp(0.01 * t) * math.exp(-0.07 * t) * math.exp(-0.015 * t), 0, 25)[0]
    expect_close(pv_stream(2500, 0.07, 25, g=0.01, hazard=0.015), expected, what="pv_stream(2500, 0.07, 25, g=0.01, hazard=0.015)")


def test_growth_equals_rate():
    """When g equals r the answer is c × T"""
    expect_close(pv_stream(1000, 0.04, 8, g=0.04), 8000.0, what="pv_stream(1000, 0.04, 8, g=0.04)")


def test_growth_almost_equals_rate():
    """g = 0.15 − 0.1 is stored as 0.04999999999999999; at r = 0.05 the answer is still c × T"""
    expect_close(pv_stream(10_000, 0.05, 10, g=0.15 - 0.1), 100_000.0, what="pv_stream(10000, 0.05, 10, g=0.15 - 0.1)", rel=1e-9)


def test_bad_inputs():
    """A negative term or a negative hazard rate raises ValueError"""
    for args in [dict(T=-1), dict(T=10, hazard=-0.01)]:
        try:
            pv_stream(10_000, 0.05, **args)
        except ValueError:
            continue
        raise AssertionError(f"expected ValueError for {args}")


def test_default_probability():
    """A 2% hazard rate gives a 9.52% chance of default within 5 years"""
    expect_close(default_probability(0.02, 5), 0.09516258196404048, what="default_probability(0.02, 5)")