Twelve Dollars for the Finish, Seven for the Average, and the Root Three Between Them

An option settling on the mean of a share's closes is strictly cheaper than one settling on the closing price, and the reason is convex order rather than any pricing model: for a martingale share every intermediate price is a forecast of the last one, so the average is dominated at every strike, for calls and for puts. Quantitatively the time average of a Brownian path carries variance T/3 against T, a swing ratio of 1/sqrt(3) = 0.57735, which turns 11.9235 into 6.9013 at a 30 percent volatility. A finite grid of 252 fixings sits at 0.33532 rather than 1/3, which accounts for most of the gap to the 6.918 measured by simulation on the true arithmetic average.

Two options on the same share, same strike, same expiry. The first settles on the closing price on the last day. The second settles on the arithmetic mean of every daily close over the option's life. Which one is worth more?

Almost everybody says the second. It is path dependent, it takes a whole year of data to settle, and it has a name that sounds exotic. Complexity feels like it should cost. It does not. The averaged contract is strictly cheaper, and the reason has nothing to do with which pricing model you write down.

The answer arrives before any model does

Suppose only that the share is a martingale, which is what pricing with zero rates and no dividend gives you. Write $S_T$ for the closing price and $\bar S = \tfrac1n\sum_{i=1}^{n} S_{t_i}$ for the average. Each intermediate price is a forecast of the final one,

S_{t_i} = \mathbb{E}\!\left[\,S_T \mid \mathcal{F}_{t_i}\right],

(1)

and a forecast is a smoothed version of what it forecasts. Jensen's inequality, applied conditionally, turns that sentence into an ordering: for any convex $\varphi$ ,

\mathbb{E}\big[\varphi(S_{t_i})\big] = \mathbb{E}\Big[\varphi\big(\mathbb{E}[S_T \mid \mathcal{F}_{t_i}]\big)\Big] \le \mathbb{E}\big[\varphi(S_T)\big].

(2)

Averaging cannot undo that. A convex function of an average is at most the average of the convex function, so the same inequality survives the sum in $\bar S$ .

Convex order

A random variable $A$ is below $B$ in the convex order when $\mathbb{E}[\varphi(A)] \le \mathbb{E}[\varphi(B)]$ for every convex $\varphi$ . Taking $\varphi(x)=x$ and $\varphi(x)=-x$ forces the two means to agree, so convex order compares spread at fixed mean and nothing else.

Call and put payoffs are both convex, so the conclusion is much stronger than the one the question asks for. The averaged contract is worth no more than the plain one at every strike, for calls and for puts, under any law the share cares to have. I checked it by brute force on all 256 paths of a symmetric five-dollar walk, at every integer strike from 70 to 130, and no combination reversed it.

The equal means in the definition matter more than they look. With zero rates $\mathbb{E}[\bar S] = \mathbb{E}[S_T] = S_0$ , so both contracts sit at the money at the same strike and the comparison is honest. Restore a positive rate and that breaks, which is a caveat I come back to at the end.

Fig. 1 — One path. The finish lands at 114.6 and pays 14.6 on a strike of 100. The average of the same path lands at 99.6 and pays nothing.

How much calmer an average is

Convex order says cheaper. It does not say by how much, and for that the model has to come back. Let $W$ be a standard Brownian motion, the driver of the log price, and compare the endpoint $W_T$ with the time average. The covariance kernel $\operatorname{Cov}(W_s, W_t) = \min(s,t)$ does all the work:

\operatorname{Var}\!\left(\int_0^1 \! W_t\,dt\right) = \int_0^1\!\!\int_0^1 \min(s,t)\,ds\,dt = 2\int_0^1\!\!\int_0^t s\,ds\,dt = 2\int_0^1 \frac{t^2}{2}\,dt = \frac13.

(3)

The endpoint has variance $1$ over the same window. So the average carries a third of the variance and, since a swing is a standard deviation,

\frac{\text{swing of the average}}{\text{swing of the endpoint}} = \frac{1}{\sqrt3} = 0.57735,

(4)

the 58 percent that gets quoted. Nothing in (3) is specific to a year: the same calculation over $[0,T]$ gives $T/3$ against $T$ , so the ratio is the same at every maturity.

A real contract averages a finite number of closes

No contract integrates. It sums 252 fixings, or 12, or 4. On a uniform grid $t_i = i/n$ the same kernel gives $\sum_{i,j} \min(i,j) = n(n+1)(2n+1)/6$ , and dividing by $n^3$ leaves

\operatorname{Var}\!\left(\frac1n\sum_{i=1}^n W_{i/n}\right) = \frac{(n+1)(2n+1)}{6n^2} = \frac13 + \frac{3n+1}{6n^2}.

(5)

The correction term is positive for every $n$ , so a sampled average is always a little more volatile than the continuous one and the discrete contract is always a little more expensive than the idealised one. At $n=1$ the formula returns $1$ , as it has to, because averaging one observation is not averaging. At $n=2$ it returns $5/8$ , which you can confirm by hand from $\operatorname{Var}\big(\tfrac12(W_{1/2}+W_1)\big) = \tfrac14\left(\tfrac12 + 1 + 2\cdot\tfrac12\right)$ . At 252 fixings it returns $0.33532$ , six tenths of a percent above the limit.

Fig. 2 — Equation (5) approaches one third from above, never from below. Twelve monthly fixings already get within 13 percent of the limit.

Turning a swing into a price

At the money with zero rates the standard European call price collapses to something with no cumulative normal in it at all. Both $d_1$ and $d_2$ reduce to $\pm\sigma\sqrt{T}/2$ , so

C = S\left[2\Phi\!\left(\tfrac{\sigma\sqrt T}{2}\right) - 1\right] = S\,\operatorname{erf}\!\left(\frac{\sigma\sqrt T}{2\sqrt2}\right).

(6)

With $S=100$ , $\sigma = 30\%$ and one year that is $11.9235$ . Feed it the reduced swing $\sigma/\sqrt3 = 17.32\%$ instead and it returns $6.9013$ . Since $\operatorname{erf}$ is increasing, which is vega being positive, the direction never needed the two numbers.

One honest warning about that substitution. It is exact for the continuous geometric average, whose logarithm is normal, and only an approximation for the arithmetic average a real contract settles on. A sum of lognormals has no elementary law, so the arithmetic case has to be simulated: 400 000 paths over 252 daily closes price it at $6.918 \pm 0.018$ . Put the discrete variance $0.33532$ from (5) into (6) rather than $1/3$ and the closed form moves to $6.9217$ , inside that error bar. So the visible gap is mostly the finite grid, not the choice of mean.

The price ratio $6.92/11.92 = 0.579$ lands suspiciously close to $0.5774$ . That is a coincidence of these parameters, not an identity. Equation (6) is only nearly linear in $\sigma$ at this level, and at a 90 percent swing the two ratios part company completely.

Three places the discount shrinks or disappears

The averaging window has to be the whole life of the option. Average over only the last stretch of length $h$ and split the average into the level reached at $T-h$ plus the fresh wandering inside the window, which are independent:

\operatorname{Var}\!\left(\frac1h\int_{T-h}^{T} \! W_t\,dt\right) = (T-h) + \frac{h}{3} = T - \frac{2h}{3}.

(7)

A one-month window on a one-year option leaves 94.4 percent of the variance, a swing of $29.16\%$ and a price of $11.59$ against $11.92$ . Almost the whole discount lives in the early observations, because they are the ones that disagree with the finish.

Positive rates spoil the comparison rather than the mathematics. The average of a share drifting upward has mean $S_0(e^{rT}-1)/(rT)$ , which sits below the share's own forward $S_0e^{rT}$ . At $r=5\%$ those are $102.54$ and $105.13$ , so a call struck at the share's forward is out of the money against the average and part of the price gap has become moneyness. Convex order needs equal means and no longer applies. This is why the clean version of the problem pins the rate at zero.

Convexity is a hypothesis, not decoration. A digital, a capped call, anything whose payoff stops bending upward is outside (2), and reducing dispersion can raise its value instead. Nothing above licenses the sentence "averaging always makes an option cheaper" once the payoff is allowed to be non-convex.

Sources and further reading

The contract itself, and the absence of an elementary arithmetic formula — Asian option
The $\min(s,t)$ kernel behind (3) and (5) — Wiener process
The ordering in (2), stated for its own sake — Stochastic ordering
The inequality that produces it — Jensen's inequality
The formula in (6) — Black-Scholes model

Everything above was checked before publication on three channels that share no step. The identities in (3) and (5) symbolically, including the limit and the exact values at $n = 2, 3, 4, 12$ and $252$ . The prices by 400 000 simulated paths of 252 daily closes, which reproduce $11.94 \pm 0.03$ for the plain call and a measured log-variance ratio of $0.3348$ against the predicted $0.33532$ . And the ordering itself by exhaustive enumeration of a 256-path walk that never mentions a lognormal, at 61 strikes, for calls and puts.