Some models use jumps as a way to explain volatility skew. I understand that if jumps exist, then you are "mishedged" as you no longer can continuously hedge. Options have a gamma component and being short an option means you may lose more on the option as you will be longer/shorter deltas than your replicating portfolio during a jump.

However, that in itself should not explain skew correct? All options have some gamma and ATM options have the most gamma. So what makes the wing options have the most IV? I would imagine a 2SD move during market close lead to a greater loss on a 50 delta option than a 1 delta option.



4 Answers 4


Jumps are an attempt to solve a math mistake in Modern Portfolio Theory. In the 19502-70s, economists were working on solving the variance-mean tradeoff. Furthermore, they needed to do so with punchcard computing. That radically restricted the set of computable, potential solutions. Both the normal distribution and the log-normal distribution are tractable with punchcard computing.

There are two problems, however. The first is one that most economists are unaware of. In 1958, a mathematician by the name of John White proved that there is no solution to an equation of the form $w_{t+1}=Rw_t+\epsilon_{t+1},R>1$ in Frequentist statistics. Of course, you would not invest money if $R\le{1}$. We will come back to this. The second is that Mandelbrot, beginning in 1963, started publishing articles that returns had heavy tails and could not be from a distribution with a variance. In other words, there is no variance-mean tradeoff because the first central moment does not exist.

Going back to the sixties and seventies with its heavy-tailed discussions and the results of the Fama-MacBeth work excluding the CAPM from empirical science, there was sort of a choice to be made. Embrace distributions without a mean and for which there was no undergirding of math for economists to work in, or decide for reality that there is a mean and just add jumps to try and cover the large shifts. That math was easily tractable.

That was an unfortunate choice. What makes it unfortunate is that the distribution of returns is $$R_{total}=R_G\times{\Pr(G)}+R_M\times{\Pr(M)}+0\times{\Pr(B)}+R_D\times{\Pr(D)}-R_L,$$ where $G$ denotes a going concern, $M$ denotes mergers, $B$ denotes bankruptcy, $D$ denotes dividends and $L$ denotes the lost return from liquidity costs. The distribution of $R_G$ is $$\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}\frac{\sigma}{\sigma^2+(R_G-\mu)^2}.$$

Going back to White, from earlier, his proof was that the sampling distribution of the slope estimator was the Cauchy distribution. Models like Black-Scholes are built on either Ito or Stratonovich calculus. Both assume that all parameters are known. As such, it is a model built on parameters. If you didn't know them, then you cannot build models on them. You would have built them on sufficient statistics instead. As sufficient statistics are independent of the parameter, you wouldn't reference the parameter.

So models like Black-Scholes are valid if the parameters are known but invalid, as per White and a later generalizing article by Sen, if the parameters are unknown. There cannot exist a Frequentist solution to models like the CAPM or Black-Scholes as it is known to be impossible unless you use non-mean and non-variance based tools.

That opens up the possibility of a Bayesian solution, except the Bayesian solution ends up having no mean or variance because the results do not come out the same. That should serve as a deep warning as well.

All Bayesian estimators are admissible estimators. Frequentist estimators are admissible only in two cases. The first is that the Bayesian and the Frequentist solution are the same for every sample. The second is that the Frequentist solution matches the Bayesian solution at the limit. That is why $\bar{x}$ is an admissible solution to estimate $\mu$ for the normal distribution but $\frac{\sum(\sin(x_i))}{n-33}$ is not.

Although Frequentist estimators do not have to be admissible and sometimes are not, when they are not, there should be an investigation. It could be the model was derived incorrectly.

The skew in the volatility is an artifact of the algorithm. Consider, instead, for returns rather than option prices as it is simpler to discuss, where returns are $R=\frac{FV}{PV}-1,$ an algorithm such as $$\Pr(\sigma|X,\mu)=\int_{-1}^\infty\frac{\prod_{i=1}^I\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}\frac{\sigma}{\sigma^2+(x_i-\mu)^2}\times{1}}{\int_{-1}^\infty\int_0^\infty\prod_{i=1}^I\left[\frac{\pi}{2}+\tan\left(\frac{\mu}{\sigma}\right)\right]^{-1}\frac{\sigma}{\sigma^2+(x_i-\mu)^2}\times{1}\mathrm{d}\mu\mathrm{d}\sigma}\mathrm{d}\mu,\forall\sigma\in\Re^{++}.$$

There will be a small natural amount of skew because the distribution should converge to the standard deviation ratio distribution but it will be small. It is related to the Snecdor's F distribution.

Note that I multiplied by $1$ and really should not have. Good priors exist for this but I didn't want to impose a prior on it as you should use your own.

The volatility skew is an artifact of the tool used to measure it and the fact that there is a non-existence theorem surrounding Ito models.


Actually, I do not think it's true. Jumps, when added to the Black-Scholes (BS) dynamics, do modify the volatility surface. However, the volatility skew may get inverted: the implied BS volatility may be higher when the strike is closer to the current value $S(0)$ of the underlying asset $S$.

Consider an idealized example: $$ \log(S(t+dt) / S(t)) ={\rm[normal\ variable\ with\ infinitesimally\ small\ volatility]} \pm 0.1 * {\rm Poisson}(3 * dt). $$

The second term is the jumps. Consider a one touch binary option $C$ with strike $K$ and expiration $dt$. This option pays \$1 if the underlying price $S(t)$ touches the strike. Consider portfolio $P = (1/dt $ units of $C)$. Then, if strike $K$ falls outside $[S(0)e^{-0.1}, S(0)e^{+0.1}]$, the true price of portfolio $P$ converges to 0 as $dt$ converges to 0. On the other hand, if strike $K\in [S(0)e^{-0.1}, S(0)e^{+0.1}]$, the true price of portfolio $P$ stays around $3 * \$1$ as $dt$ converges to 0.

The implied volatility in the Black-Scholes model has to compensate for this phenomenon (as $dt$ converges to 0). When $K$ moves from $0$ into $[S(0)e^{-0.1}, S(0)e^{+0.1}]$, the implied volatility jumps from 0 to a positive value. Likewise, when $K$ is leaving $[S(0)e^{-0.1}, S(0)e^{+0.1}]$ on the way to $+\infty$, the implied volatility drops from a positive value to 0.


Jumps do not imply fat tails. See the simulation in R. Note that the excess kurtosis of [normal variable + jump] is negative.

> set.seed(1)
> Normal_Variable <- rnorm(1e8)
> kurtosis(Normal_Variable)
[1] -0.000628316
> Jump <- 2 * ((runif(1e8) < 0.5) * 2 - 1)
> kurtosis(Normal_Variable + Jump)
[1] -1.280009

The argument given in the OP doesn’t convince me. Yes, the dollar gamma of ATM options is the largest. But so is the Vega. Therefore the amount of implied volatility increase necessary to compensate is not clear.

For me, the intuition is simply that jumps-> the distribution of log returns has fat tails relative to a normal distribution. And then it is a mechanical fact that the BS IV shape has a smile with OTM option vol > ATM option vol.

[edit]. However, it is shown by @stans that the presence of jumps is not sufficient to provide fat tails. In practice , a jump in the market is often accompanied by an increase in implied volatility, which would indeed increase the relative value of otm options. So it depends what type of jump is specified.

  • $\begingroup$ Jumps do not necessarily imply fat tails. See my answer. Also please check out the illustration in R displayed in my second answer. Note that the excess kurtosis of [normal variable + jump] is negative. $\endgroup$
    – stans
    Commented Oct 22, 2019 at 11:14
  • $\begingroup$ Is that true if normalized to the std dev ? Ie kurtosis / std dev ratio ? Thx $\endgroup$
    – dm63
    Commented Oct 22, 2019 at 12:21
  • $\begingroup$ I don't know, have not checked. But that is not the point. Excess kurtosis is an indicator of fat tails, not the ratio that you mentioned. Excess kurtosis already accounts for standard deviation. It is defined in terms of the [standardized random variable] = [original random variable] / Stand.Dev[original random variable]... So my message here: "fat tails" are not synonymous to "jumps". $\endgroup$
    – stans
    Commented Oct 22, 2019 at 12:37
  • $\begingroup$ Ok I see it. If you add a negative kurtosis jump such as a Bernoulli +/-1 coin flip, indeed this does not produce fat tails. $\endgroup$
    – dm63
    Commented Oct 22, 2019 at 17:30

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