A Straddle Bought for $5 Pays on a $2 Move, If You Sell It

Held to expiry the position needs the full five dollars and a two dollar move loses three, but the same move sold the next morning is worth 5.4405. Nothing is assumed to get there: the five dollar price pins the volatility at 35.5424 percent and the position's slope at exactly a tenth. That slope is why the gain is lopsided, and why one dollar down loses money while two dollars down gains six cents.

A share trades at 25. You buy the call struck at 25 and the put struck at 25, and the pair costs you five dollars. Six months to expiry, no interest, no dividends. What are you hoping for?

The answer that comes back most often is a move of more than five dollars in either direction, and it is exactly right about one thing and wrong about the timing. Five dollars is the break-even if you hold the position to the last day. Nothing in the contract obliges you to. Sell the pair the morning after a two dollar jump and it is worth 5.4405, a gain of forty-four cents on a position whose expiry break-even is still three dollars away.

The position has two clocks

At expiry the call and the put cannot both be worth something, and whichever one pays gives $|S(T) - 25|$ . That is the V-shaped graph, and against a premium of five it puts the two break-evens at 20 and at 30. Two dollars of movement pays two dollars and loses three, which is the whole content of the reflex.

Before expiry the position is worth strictly more than that V at every share price, because there is still time in which something could happen. What you own on day one is the smooth curve, and the smooth curve is what a buyer would pay you for. Those are the two clocks: the payoff clock, which only strikes once, and the resale clock, which runs continuously.

Fig. 1 — The same two dollar move read on two clocks. It lands at 2.00 on the V and at 5.4405 on the curve above it.

The five dollar price tells you the volatility

The problem never states a volatility, and it does not need to, because the price contains it. At the money with a zero rate the straddle has an exact closed form with no cumulative normal left in it.

The at-the-money straddle, exactly

With $S = X$ and $r = 0$ , put-call parity gives $c + p = 2c - S + X = 2c$ , and $d_1 = -d_2 = \tfrac{1}{2}\sigma\sqrt{\tau}$ , so the two cumulative normals collapse into one error function: $\;c + p = 2S\,\mathrm{erf}\!\bigl(\sigma\sqrt{\tau}/(2\sqrt{2})\bigr)$ .

Put in $S = 25$ and $\tau = 1/2$ . The argument of the error function becomes $\sigma/4$ , and the whole thing reduces to a single equation with a single unknown:

50\,\mathrm{erf}(\sigma/4) = 5 \quad\Longrightarrow\quad \sigma = 4\,\mathrm{erf}^{-1}\!\left(\tfrac{1}{10}\right) = 35.5424\%

(1)

Every number in the rest of this article descends from that one, so nothing is assumed. The five dollars also pins two quantities exactly, without rounding. Since $50(2\Phi(d_1) - 1) = 5$ , the standardised moneyness satisfies $\Phi(d_1) = 0.55$ , and the slope of the position is

\frac{\partial(c+p)}{\partial S} = 2\Phi(d_1) - 1 = 0.10

(2)

A tenth, exactly. That number is the reason the pre-expiry gains are lopsided, and it is easy to miss because at expiry the position is perfectly symmetric.

What a two dollar jump is actually worth

Move the share to 27 with nothing else changed and reprice. The pair is worth 5.4405, up forty-four cents or 8.8 percent. Where the forty-four cents comes from is worth writing out, because the split explains the whole position:

\underbrace{0.10 \times 2}_{20\text{c of slope}} \;+\; \underbrace{\tfrac{1}{2}\times 0.12599 \times 2^2}_{25.20\text{c of curvature}} \;-\; \underbrace{1.15\text{c}}_{\text{higher order}} \;=\; 44.05\text{c}

(3)

The curvature term is the larger of the two, and it is the one blind to direction, since $(\Delta S)^2$ does not care about sign. That is what “long volatility” means in this position: most of what you own is the squared term. The slope term is small and signed, and it is the entire source of the asymmetry below.

The half of the answer that needs no move at all

There is a second route to profit here, and in this instance it is the bigger one. Leave the share exactly where it is and raise the volatility by a tenth of itself, from 35.5424 percent to 39.0966. Equation (1) reprices immediately:

50\,\mathrm{erf}\!\left(1.1 \times \mathrm{erf}^{-1}\!\left(\tfrac{1}{10}\right)\right) = 5.4970

(4)

Forty-nine and a half cents, against forty-four for the two dollar jump. So the honest answer to the question is a large move or a rise in volatility, and neither requires the five dollars. This is also where a common figure goes astray. A value of 5.50 circulates for the two dollar jump; 5.4970 is the volatility leg and it is what rounds to 5.50, while the jump gives 5.4405. Quoting 5.50 for the jump overstates it by about six cents and attaches the number to the wrong mechanism.

The gain is not symmetric before expiry

Equation (2) has a consequence that is easy to state and easy to forget. The position has a positive slope of a tenth, so a downward move fights the slope while an upward move is helped by it. Repricing the four small moves gives numbers that are nothing like symmetric.

Fig. 2 — Same size of move, very different results. One dollar down loses money outright, which the symmetric expiry payoff gives no hint of.

Two dollars down is worth 5.0602, a gain of six cents rather than forty-four. One dollar down is worth 4.9641, which is a loss of about four cents. At two dollars the curvature term finally overwhelms the slope term and the position is up; at one dollar it does not, since $\;\tfrac{1}{2}(0.12599)(1)^2 = 6.3\text{c}$ against ten cents of slope working the other way. Any statement that a straddle profits from a move “either way” belongs to expiry, where it is exact, and not to the resale clock.

What the argument needs to be true

Three conditions, and none of them is free. The first is a buyer. The whole argument turns on selling the position, so the pre-expiry curve is a price only where a market exists to quote it. Nothing stops you offering the pair tomorrow, and nothing promises a fill at 5.4405.

The second is time. With the share and the volatility both standing still, the position loses about two cents a trading day at the start: 4.9802 after one day, 4.9003 after five, and 4.5664 after a month. The forty-four cents from a jump is therefore worth roughly three weeks of standing still, which is generous but finite. A view that the move will happen “eventually” is not the same view as this trade.

The third is that the volatility itself holds up. Equation (4) cuts both ways: a fall of a tenth in volatility takes the pair to 4.5022, a loss of 49.78 cents, which is more than the two dollar jump earns. Buying a straddle after a price already known to be about to move is how the five dollars gets expensive, which is a statement about the market and not about the arithmetic.

One structural fact makes the whole thing possible: the smooth curve lies strictly above the V at every share price, so there is always something to sell. The two curves only meet in the limit of no time left, and it is at that limit, and only there, that the five dollar answer becomes the whole answer.

Sources and further reading

Wikipedia: Straddle for the payoff, and Put–call parity for the step that turns the pair into twice a call.
Wikipedia: Error function and Black–Scholes model, which together give the closed form in equation (1).
Louis Bachelier, Théorie de la spéculation (1900), the first treatment of an option value as a function of a diffusing price.

The expiry break-even was checked by evaluating the payoff at 10,001 prices from 20.000 to 30.000, and every price strictly inside that band loses money. The jumped value was reproduced on a recombining lattice with no normal distribution in its code path, agreeing to a tenth of a percent at 500, 2,000 and 8,000 steps, and by a 400,000-path simulation agreeing to a quarter of a percent. The pre-expiry value was verified to sit strictly above the payoff at all 10,001 of those prices.