Case study: the perpetual American put

A put that never expires

A put gives the right to sell a stock at the strike $K$. Exercised when the stock price is $S$, it pays $K - S$; it is worth exercising only when $S < K$. An American put may be exercised at any time (case study 5.11.4), and a perpetual one has no expiry date at all.

The inputs are those of case study 2.4.6: strike $K = 100$, interest rate $r = 0.03$ (continuous) and volatility $\sigma = 0.2$, both yearly decimals. The stock starts at $S = 100$ and pays no dividends.

With no expiry, nothing changes as time passes except the stock price. A holder facing the price 80 next year faces the same problem as a holder facing 80 today. So the value is a function of the price alone, $V(S)$.

The equation

For an option whose value $V(S, t)$ depends on the price and the time, the Black–Scholes equation holds wherever the holder keeps the option:

This case study only states it. Case study 6.3.4 states the full equation and checks it on the Black–Scholes formula; the Stochastic Calculus course derives it.

Here $V$ does not depend on $t$, so the first term is 0 (case study 6.3.4 notes this too). The partial derivatives (lesson 4.1.2) become ordinary ones:

That is the equidimensional equation of lesson 6.2.2, with $x = S$, $a = \frac{1}{2}\sigma^2$, $b = r$ and $c = -r$.

Its terms all have the same units. $\sigma^2$ and $r$ are per year and $V$ is in dollars; $V'$ is in dollars per dollar of $S$, and $V''$ in dollars per dollar squared. So each term is in dollars per year.

Trying $V = S^p$

Lesson 6.2.2's trial $V = S^p$ turns the equation into one for $p$:

Multiply by $\frac{2}{\sigma^2}$ and write $\beta = \frac{2r}{\sigma^2}$, a pure number with no units (beta). The equation becomes

So $p = 1$ or $p = -\beta$, and the general solution is $V = AS + BS^{-\beta}$. For the inputs above, $\beta = \frac{2 \times 0.03}{0.04} = 1.5$.

The solution $V = S$ is the stock itself: a share solves the equation. The other, $S^{-1.5}$, falls as the price rises, as a put's value should.

The boundary conditions

Two conditions fix $A$ and $B$, one at each end of the range where the holder keeps the put.

Far away. As $S \to \infty$, the stock is ever less likely to fall below the strike, and the put is worth ever less: $V \to 0$. The term $AS$ grows, so $A = 0$, as in lesson 6.2.2's condition at infinity. That leaves $V = BS^{-\beta}$.

At the exercise price. The holder exercises when the price falls to some level $S^*$, the exercise boundary, and receives $K - S^*$ there. So $V(S^*) = K - S^*$, which gives $B = (K - S^*)(S^*)^{\beta}$ and

Below $S^*$ the put is exercised at once, so $V(S) = K - S$ there.

Which boundary?

The two conditions give a value for every choice of $S^*$, so the boundary is part of the answer: this is a free-boundary problem. The holder chooses the $S^*$ that makes the put worth most. With the stock at 100:

Exercising early, at 80, gives up too much of the chance of a deeper fall. Waiting for 40 risks never getting there. The best is 60.

Calculus finds it exactly (lesson 2.4.1). Only $(K - S^*)(S^*)^{\beta}$ depends on $S^*$. Its derivative is $(S^*)^{\beta - 1}\big[\beta(K - S^*) - S^*\big]$, which is 0 when

The best boundary does not depend on today's price $S$. So the rule is the same wherever the stock is: wait until it first falls to 60, then exercise.

Smooth pasting

At $S^* = 60$ the slope of the value is $V'(S^*) = -\beta\,\frac{K - S^*}{S^*} = -1.5 \times \frac{40}{60} = -1$. That is the slope of the payoff $K - S$. The value curve meets the payoff line without a corner.

This is called smooth pasting: at the best boundary, value and payoff agree in level and in slope. Where no formula is available, it is the extra condition that pins the boundary down. Any other boundary breaks it, and the figure shows what goes wrong.

Exercising at 50 fails in a telling way. Just above 50 its value is below the payoff: at $S = 55$ it gives 43.3392, while exercising there pays 45. A holder following that rule would do better to exercise early, so the rule cannot be best.

The value and the exercise rule

For the shared inputs, $V(S) = 40\left(\frac{S}{60}\right)^{-1.5}$ above 60:

At $S = 100$ the put is worth 18.5903, though exercising now would pay nothing. The factor $\left(\frac{100}{60}\right)^{-1.5} = 0.464758$ has a reading: it is the value today of $1 paid the first time the stock falls to 60. So the put is worth 40 such dollars.

Below the boundary, use the payoff, not the formula. At $S = 55$ the formula gives $40\left(\frac{55}{60}\right)^{-1.5} = 45.5767$, but the holder exercises at once and gets 45. The formula only holds where holding is best, for $S \ge 60$.

Against the tree of case study 5.11.4

Case study 5.11.4 priced American puts on a binomial tree. A finite expiry is worth at most as much as no expiry, since the perpetual holder can copy any strategy of the finite one. So the tree's prices should rise towards 18.5903 as $T$ grows.

With 2,000 steps the tree gives:

The first three match case study 5.11.4's table, 14.74, 17.66 and 18.28. The approach is slow: 50 years still leaves 31 cents. And a fixed number of steps gets coarser as $T$ grows, since $\Delta t = \frac{T}{n}$ is a tenth of a year at $T = 200$.

A finer tree with 40,000 steps, too slow for this page, gives 18.5599 at $T = 100$ and 18.5889 at $T = 200$: within 0.0014 of the formula.

The tree's exercise boundary falls in the same way. With half a year left it was about 81 (case study 5.11.4). A finer tree puts it near 64 with 10 years left and near 61 with 50, closing in on 60.

Solving the boundary-value problem numerically

Once $S^*$ is known, the value on $S \ge 60$ solves a boundary-value problem, and lesson 6.2.2's solve_bvp can check it. A grid cannot reach infinity, so a desk stops it at a large price $L$ and sets $V(L) = 0$ there.

That far condition is not quite right. The true value at $L$ is small but positive, and setting it to 0 drags the whole curve down.

At $L = 200$ the value is $2.49 too low, and the slope at the boundary is no longer $-1$. Putting the true value $V(1{,}000) = 0.5879$ at the far end instead recovers 18.5903. Case study 6.5.4 makes the same choice of a far boundary for its grid.

# The shared put of case study 6.2.4: K = 100, r = 0.03, sigma = 0.2, so beta = 1.5


def expect_value_error(call, why):
    try:
        call()
    except ValueError:
        return
    raise AssertionError(f"expected ValueError: {why}")


def test_shared_put():
    """Without dividends the put at S = 100 is worth 18.5903, with boundary 60"""
    value, boundary = perpetual_put(100, 100, 0.03, 0.2)
    expect_close(boundary, 60.0, what="boundary", rel=1e-9)
    expect_close(value, 18.590320061795605, what="value at S = 100", rel=1e-9)


def test_below_boundary():
    """Below the boundary the put is worth its payoff: 50 at S = 50"""
    value, boundary = perpetual_put(50, 100, 0.03, 0.2)
    expect_close(value, 50.0, what="value at S = 50", rel=1e-12)
    expect_close(boundary, 60.0, what="boundary", rel=1e-9)


def test_at_boundary():
    """At the boundary the value equals the payoff: 40 at S = 60"""
    value, _ = perpetual_put(60, 100, 0.03, 0.2)
    expect_close(value, 40.0, what="value at S = 60", rel=1e-9)


def test_dividend():
    """A 2% dividend yield gives p = −1: boundary 50 and value 25 at S = 100"""
    value, boundary = perpetual_put(100, 100, 0.03, 0.2, q=0.02)
    expect_close(boundary, 50.0, what="boundary with q = 0.02", rel=1e-9)
    expect_close(value, 25.0, what="value with q = 0.02", rel=1e-9)


def test_higher_rate():
    """At r = 5% the boundary rises to 71.43 and the value falls to 12.32"""
    value, boundary = perpetual_put(100, 100, 0.05, 0.2)
    expect_close(boundary, 71.42857142857142, what="boundary at r = 0.05", rel=1e-9)
    expect_close(value, 12.320032867762636, what="value at r = 0.05", rel=1e-9)


def test_other_inputs():
    """S = 120, K = 110, r = 5%, σ = 30%, q = 1%: boundary 55 and value 25.2083"""
    value, boundary = perpetual_put(120, 110, 0.05, 0.3, q=0.01)
    expect_close(boundary, 55.0, what="boundary", rel=1e-9)
    expect_close(value, 25.208333333333336, what="value", rel=1e-9)


def test_smooth_pasting():
    """Just above the boundary the value falls with slope −1, like the payoff"""
    _, boundary = perpetual_put(80, 90, 0.04, 0.25, q=0.01)
    h = 1e-6 * boundary
    slope = (perpetual_put(boundary + h, 90, 0.04, 0.25, q=0.01)[0] - (90 - boundary)) / h
    expect_close(slope, -1.0, what="slope just above the boundary", rel=1e-4)


def test_never_below_payoff():
    """The value is never below the payoff, from S = 10 to S = 300"""
    for S in [10, 30, 47, 48, 55, 70, 90, 100, 150, 300]:
        value, _ = perpetual_put(S, 90, 0.04, 0.25, q=0.01)
        assert value >= max(90 - S, 0) - 1e-12, f"value {value} at S = {S} is below the payoff {max(90 - S, 0)}"


def test_bad_inputs():
    """Non-positive S, K, σ or r, or a negative q, raise ValueError"""
    expect_value_error(lambda: perpetual_put(0, 100, 0.03, 0.2), "S must be positive")
    expect_value_error(lambda: perpetual_put(100, -5, 0.03, 0.2), "K must be positive")
    expect_value_error(lambda: perpetual_put(100, 100, 0.03, 0), "sigma must be positive")
    expect_value_error(lambda: perpetual_put(100, 100, 0.0, 0.2), "with r = 0 there is no exercise boundary")
    expect_value_error(lambda: perpetual_put(100, 100, -0.01, 0.2), "with r < 0 there is no exercise boundary")
    expect_value_error(lambda: perpetual_put(100, 100, 0.03, 0.2, q=-0.01), "q cannot be negative")