While I was recently testing a delta-hedging on a few products, I got a P&L result of 20% for some of them.

First, I thought that the implementation was incorrect. But I couldn't find an outright error.

After a deeper look in my (back)test-data, I found that delta-hedging might not be enough even for simple products like 'vanilla calls'.

Here's an example (with payoff = (S(T) - K )+): (T = 20 days, K =50.5, (hypothesis : Black-Scholes model))


day 0 = 50 ;

day 1 = 50.5 ;

day 2 = 50.3 ;

... ;

day 15 = 52.7 ;

day 16 = 49.3 ;

day 17 = 37.5 ;

day 18 = 36.4 ;

day 19 = 36.8;

day 20 = 37.7.

I rebalance every day and because of the huge price fall on day 17, and because my delta was close to 1, I lost a lot of money (call price is almost 0, while my porfolio is about -15), causing a big P&L.

So, my questions are :

1/ Is it correct that : "delta-hedging is ok for continuous data (and continuous heding). Price-jumps are not hedged with delta-hedging"..?

2/ Assuming that we can't hedge more than once a day, and having recurrent (there are a few) prices falls in my data, what do you suggest for a hedging ?

Thanks a lot,

  • $\begingroup$ Delta hedging is inherently statistical - even in absence of jumps, there are paths where standard deviation is exactly matching the implied volatility yet you make or lose money. example - you buy an ATM option but the underlying realizes smaller moves in a single direction but then realizes large moves when the option is OTM and your gamma is much lower. $\endgroup$ Commented Mar 28, 2017 at 0:25
  • $\begingroup$ @StudentInFinance, The answer of Daneel Olivaw concerning the fact that jumps are not hedgeable is perfectly correct. However, note that if you used real market data (and notably real IVs) to your theoretical replication error: $.5\Gamma(S_t,\sigma)S_t^2( (dS_t/S_t)^2 - \sigma^2 dt)$ also adds the mark-to-market error (out-of-model risk) due to the fact that IV changes as well i.e. $ \nu (\sigma_{t+dt}-\sigma)$ where $\nu$ is the BS vega and $\sigma_{t+dt}$ the new implied volatility for your option ($\sigma$ = the IV you've used to hedge). $\endgroup$
    – Quantuple
    Commented Mar 28, 2017 at 8:08
  • $\begingroup$ Thus, rather than simple BS hedging, investment banks incorporate some rules to capture the spot/IV dynamics (so called sticky rules, or stickiness assumptions). $\endgroup$
    – Quantuple
    Commented Mar 28, 2017 at 8:10

1 Answer 1


Regarding your 1st question, jumps are indeed unhedgeable. From a theoretical point of view, you might want to look at Merton's "Option pricing when underlying stock returns are discontinuous", the original paper that adapted Black-Scholes framework to include jumps. If you look at page 7, just after equation $(9)$:

Unfortunately, in the presence of the jump process, $dq$, the return on the [hedging] portfolio [...] will not be riskless. Moreover, inspection of [equation] $(7c)$ shows that there does not exist a set of portfolio weights [...] that will eliminate the 'jump' risk [...].


Note: the return on the portfolio is a 'pure' jump process because the continuous parts of the stock and the option price movements have been 'hedged' out.

So, as regards to question 2, you can for example increase the frequency of your hedging portfolio reallocation to get closer to purely continuous trading; but even there, if you assume a Merton framework, you will still be exposed to the jump risk.

Edit: jump risk is unhedgeable because there isn't a tradeable asset allowing to hedge against it. For example, Heston's original stochastic volatility model was incomplete (as is Merton's) because the volatility risk couldn't be hedged out. However, if you include a volatility derivative in your model, then volatility risk becomes hedgeable. If you had at your disposal a tradeable asset allowing you to hedge against jump risk, then the 'model' would be complete and you could hedge out the risk.

Edit 2: I am writing down some additional thoughts related to your comment "[...] how banks hedge their (for instance) long term structured products [...]?" @StudentInFinance.

Going back to Merton's article, consider equation $(10)$:

$$ \frac{dP}{P} = \left(\alpha_P - \lambda k_P \right)dt + dq $$

$\frac{dP}{P}$ is the return of the hedging portfolio (stock, option and zero-coupon bond), $\alpha_P$ its instantaneous return, $\lambda$ the mean number of jumps per unit of times, $k_P$ the expected percentage change in the portfolio value if a jump occurs and $q$ the Poisson process modelling jumps.

The key feature is that Merton argues that the random jump component of the portfolio is uncorrelated to the market. Emphasis mine:

The total change in the stock price is posited to be the composition of two types of changes: (1) the 'normal' vibrations in price, [...]. (2) The 'abnormal' vibrations in price are due to the arrival of important new information about the stock that has more than a marginal effect on price. Usually, such information will be specific to the firm or possibly its industry.


[...] the stock price dynamics were described as the result of two components: the continuous part [...] and the jump part which is a reflection of important new information that has an instantaneous, non-marginal impact on the stock. If the latter information is usually firm [...] specific, then it may have little impact on stocks in general (i.e. the 'market'). [...]

If the source of the jumps is such information, then the jump component of the stock's return will represent 'non-systematic' risk, i.e. the jump component will be uncorrelated with the market.

Now, consider that some investment bank has a position in 2 different options written on 2 different stock prices, $S_t^{(1)}$ and $S_t^{(2)}$, with different jump risk sources. The return of each hedging portfolio will be:

$$ \begin{align} & R_1 \equiv \frac{dP_1}{P_1} = \left(\alpha_{P,1} - \lambda_1 k_{P,1} \right)dt + dq_1 \\[12pt] & R_2 \equiv \frac{dP_2}{P_2} = \left(\alpha_{P,2} - \lambda_2 k_{P,2} \right)dt + dq_2 \end{align} $$


$$ \mathbb{C}\text{ov}\left[R_1,R_2\right] = \mathbb{C}\text{ov}[dq_1,dq_2] $$

From Merton's comments we can posit that:

$$ \mathbb{C}\text{ov}[dq_1,dq_2]=0 $$

Hence, letting $\sigma_i^2 \equiv \mathbb{V}\text{ar}[R_i]$, $i \in \{1,2\}$, the bank's return variance of these 2 portfolios is $-$ if held in proportions $w_1, w_2$ of its total book, such that $w_1, w_2>0$ and $w_1+w_2=1$:

$$ \mathbb{V}\text{ar}\left[w_1R_1+w_2R_2 \right] = w_1^2\sigma_1^2 + w_2^2\sigma_2^2 < w_1\sigma_1^2 + w_2\sigma_2^2$$

You see there is some "diversification effect" kicking in: the book's risk is lower than the weighted risk of each individual portfolio. Non-systematic risk is lowered by including additional (uncorrelated) jump risks in your portfolio, hence based on this interpretation banks, by writing options on multiple stock underlying, are "naturally" hedging out part of their jump risk.

  • 1
    $\begingroup$ Thank you a lot for your answer. If I understood well, if we are under Merton's hypothesis, the best we can do is increase the hedging frequency and hope that no jump occurs. But, then how banks hedge their (for instance) long term structured products, that sometimes last more than 10 years ? High-frequency hedging would be too expensive because of the transaction costs ? But on the other hand, not doing so could be very risky. Is the solution a tradeoff between these two aspects ? $\endgroup$
    – user25844
    Commented Mar 27, 2017 at 20:19
  • $\begingroup$ @StudentInFinance I am extending my answer, giving some thoughts related to your last comment. $\endgroup$ Commented Mar 28, 2017 at 9:35
  • $\begingroup$ Thank you for this well detailed answer, does this mean that continuous delta hedging of derivatives for which the underlying is quoted once a day (Libor) is not very consistent as jumps are always present? $\endgroup$
    – Xman
    Commented Sep 2, 2022 at 16:23

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