9.9 Why systematic option selling has negative skew, and how to size it

Motivation

A short strangle wins most weeks. The question is not whether it wins often but whether the rare losses are larger than the sum of the wins, and how much capital to commit given the answer.

The shape, qualitatively The P&L of a short option position held to expiry is $C_0e^{rT}-\Pi(S_T)$ with $\Pi\ge0$ convex. It is bounded above by the credit and unbounded below (for naked calls; bounded by the strike for puts). When most of the probability lies between the strikes, the P&L piles up at its maximum, and the convex payoff stretches the left tail further than the right: the distribution is negatively skewed, more so the fatter the tails of $S_T$ . There is no clean theorem here; the quantitative content comes from simulation.

9.9.1Simulated weekly strangles#

Simulation (m09_short_vol_sim.py) Daily log returns come from toolkit.data.synthetic_returns: GARCH(1,1) with Student-t(4) shocks and rare negative jumps. On one 20,000-day path (seed 5, annualised vol 16.5%, daily skewness −0.74), we sell 3,996 consecutive weekly (5-day) strangles, struck one implied standard deviation from spot, $K=S\exp(\pm\sigma_i\sqrt{5/252})$ , priced at $\sigma_i=m\times$ (20-day trailing realised vol). The markup $m$ stands in for the variance premium. P&L is in basis points of spot.

Markup $m$	Mean (bp)	Std dev (bp)	Skewness	Win rate	Worst trade (bp)
0.9	−21.0	125.6	−5.16	69.0%	−2,075
1.0	−11.5	120.3	−5.41	73.4%	−1,986
1.1	−2.7	115.4	−5.65	77.4%	−1,896
1.2	+5.4	110.9	−5.88	80.7%	−1,806
1.3	+12.7	106.8	−6.08	83.7%	−1,716

Three lessons:

Skewness is about −5 to −6 in every row. Negative skew is structural, not a symptom of mispricing.
At $m=1$ the seller loses. Pricing options at the trailing realised vol is not fair in a world with volatility clustering, fat tails and jumps: the trailing estimate lags spikes, and Black–Scholes underprices tail moves. Here the break-even markup is about 1.13. A variance premium has to be large enough to pay for the model's own error before it is profit.
The win rate is not the edge. At $m=1$ the strangle wins 73% of the time and loses money. Moving the strikes further out raises the win rate and worsens the skew: with $m=1.2$ , strikes at 0.5, 1, 1.5 and 2 standard deviations gave win rates of 70%, 81%, 90% and 94%, means of +19.7, +5.4, −3.2 and −5.5 bp, and skewness −3.3, −5.9, −8.8 and −11.6. Far-OTM options in this world are underpriced by a flat-vol Black–Scholes, which is the skew's job in a real market (9.5).

Relation to the VRP For a delta-hedged position, (9.11) says the expected P&L is roughly vega times $(\sigma_i-\sigma_r)$ , the premium in vol units. For the unhedged strangle, the premium shows up as the mean in the table: each 0.1 of markup adds about 8 bp per week. The skew comes from the convex payoff and fat tails, whatever the premium.

9.9.2Kelly sizing with negative skew#

Sizing: Kelly with skewed payoffs M03 derived the Kelly fraction $f^*=\arg\max_f E\ln(1+fX)$ for a per-trade return $X$ , and the Gaussian approximation $f^*\approx\mu/\sigma^2$ . For short vol that approximation is dangerous.

Proposition 9.11negative skew lowers Kelly

Let $X$ take finitely many values (for example, an empirical sample of trade returns), with $E X=\mu_1>0$ , $P(X<0)>0$ , raw moments $\mu_2=EX^2$ , $\mu_3=EX^3$ , and let $L=\max(-X)$ (the worst loss per unit staked). Then:

(a) $g(f)=E\ln(1+fX)$ is strictly concave on $[0,1/L)$ and has a unique maximiser $f^*\in(0,1/L)$ .

(b) If $\mu_3\le0$ , then $f^*<\mu_1/\mu_2<\mu_1/(\mu_2-\mu_1^2)=\mu/\sigma^2$ .

Proof

(a) $g''(f)=-E[X^2/(1+fX)^2]<0$ . $g'(0)=\mu_1>0$ , and since $P(X=-L)>0$ , the term $P(X=-L)\cdot(-L)/(1-fL)$ in $g'(f)$ tends to $-\infty$ as $f\uparrow1/L$ while the other terms stay bounded. By continuity and strict monotonicity of $g'$ there is a unique root. (b) The algebraic identity $\frac{X}{1+fX}=X-fX^2+\frac{f^2X^3}{1+fX}$ gives $g'(f)=\mu_1-f\mu_2+f^2E\left[\frac{X^3}{1+fX}\right]$ . Let $f_0=\mu_1/\mu_2$ . If $f_0\ge1/L$ , then $f^*<1/L\le f_0$ and we are done. Otherwise $1+f_0X>0$ , and pointwise $X^3/(1+f_0X)\le X^3$ : for $X>0$ the denominator exceeds 1; for $X<0$ it lies in $(0,1)$ and $X^3<0$ . Equality holds only at $X=0$ , and $P(X<0)>0$ , so $E[X^3/(1+f_0X)]<\mu_3\le0$ . Hence $g'(f_0)=f_0^2E[X^3/(1+f_0X)]<0$ , and by concavity $f^*<f_0$ . Finally $\mu_2>\mu_2-\mu_1^2$ .∎

The expansion $g(f)\approx f\mu_1-\tfrac12f^2\mu_2+\tfrac13f^3\mu_3$ shows the mechanism: the cubic term penalises leverage when the third moment is negative.

9.9.3Sizing in numbers#

Numbers from the simulation Treat the strangle as requiring margin of 10% of notional (an assumption for illustration; check the exchange's SPAN plus ELM for the real figure), so the return on margin is $X=\text{P\&L}/0.10$ . With $m=1.2$ , exact Kelly is $f^*=0.32$ of capital posted as margin per trade, against a Gaussian $\mu/\sigma^2=0.44$ . With $m=1.3$ , $f^*=0.52$ against 1.12. In the second case the Gaussian answer exceeds $1/L=0.58$ , so betting it would have been ruined by the worst week in the sample.

Practitioner rules that follow

Size from the worst loss you can imagine, not the sample. The sample's worst week (18% of notional here) is a lower bound on the true worst week. If you cap the stress loss at $\ell$ of capital, the margin fraction is $f\le\ell/L_{stress}$ .
Use fractional Kelly (M03). With estimation error in $\mu$ (the premium is small and noisy) and in the tail, overbetting is far more costly than underbetting.
Buy the wings. An iron condor has a lower mean than the strangle (it pays the premium on the wings) but its $L$ is known, so the Kelly constraint $f<1/L$ is computable.
Concentration in time matters. Weekly expiries mean 52 bets a year on one index, all sharing the same crash. These are not independent bets in the sense of M03.