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For a pure equity process (with interest rate, dividend, etc., being zero) not necessarily the geometric Brownian motion, is the delta of a European call option always no higher than $1$? I am NOT asking for the Black-Scholes delta, but a model free general property of the European call delta. We can consider this question with and without the martingale property that the expected underlying price should be the current price.

I have also formulated this question a more formal fashion here as a calculus of variation or linear programming problem.

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    $\begingroup$ do you want a pure Black--Scholes model answer or a general model-free result? $\endgroup$ – Mark Joshi Dec 7 '15 at 3:50
  • $\begingroup$ @MarkJoshi: I want a general model-free result, as I did not mention Black-Scholes. Thank you for asking. I have put in a phrase to emphasize that point. Please check. $\endgroup$ – Hans Dec 7 '15 at 4:05
  • $\begingroup$ Could the down-voter please explain which part of the question is unclear, erroneous or uninteresting? $\endgroup$ – Hans Dec 7 '15 at 4:27
  • $\begingroup$ I doubt this is true but it will take me a bit to come up with a good example. I agree that this is a good question. $\endgroup$ – Mark Joshi Dec 7 '15 at 4:56
  • $\begingroup$ My feeling is that 1 should be the upper bound. In the end, if we can show that delta is a decreasing function of strike, in the “best" case scenario we have an option with strike zero whose delta must be one to maintain no arbitrage. $\endgroup$ – fni Dec 7 '15 at 7:13
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It is false. Here is an example. Let $$ dS_t = rS_t dt + f(S_0) S_t dW_t, $$ $$ dB_t = r dt. $$ The price is then the Black-Scholes price with volatility $f(S_0).$ The delta is the BS delta plus $$ f'(S_0) \times \operatorname{BS Vega}. $$ Picking $f$ appropriately, we can make the Delta as big as we like.

Note that the example is highly artificial in that volatility is a function of $S_0$ rather than $S_t.$

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    $\begingroup$ +1 Clever! It is indeed artificial though in that the volatility depends on the when one starts observing $S$. $\endgroup$ – Hans Dec 13 '15 at 22:14
  • $\begingroup$ Thanks. it is a very clever and parsimonious construction. $\endgroup$ – Gordon Dec 15 '15 at 14:19
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In any arbitrage free model, you can define the BS implied volatility $\sigma_{BS}(S;T,K)$ of the model by writing call prices as $$ C_{Mdl}(S;T,K) = C_{BS}(S;T,K;\sigma_{BS}(S;T,K)) $$ So the model's Delta is $$ \Delta_{Mdl}(S;T,K) = \partial_S C_{Mdl}(S;T,K) = \Delta_{BS} + Vega_{BS} \times \partial_S\sigma_{BS} $$ The second term is a corrective term that corresponds to the dynamic of the implied vol surface in your model. If $\partial_S\sigma_{BS}$ is positive (resp. negative) enough, the model delta of your calls (resp. puts) will be greater than 1 (resp. lower than -1).

The quantity $\partial_S\sigma_{BS}$ is sometimes called the backbone of the model. The regimes of volatility introduced by Derman can be seen as an out-of-model specification of this quantity.

PS: If you want a concrete example, I would suggest looking at a stochastic volatility model with very high/low correlation between spot and instantaneous volatility. In a Heston model, you have a semi-closed form for call prices so you should be able to compute the model delta somewhat explictly and show that is it not bounded by 1.

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  • $\begingroup$ Good suggestion for looking at the Heston model. However, are you claiming there exists a set of parameters of the Heston model that will produce delta greater than 1? Could you please prove it? $\endgroup$ – Hans Dec 9 '15 at 2:26
  • $\begingroup$ My guts tell me that the delta hedge ratio cannot be more than 1, but I do not yet have a good justification. $\endgroup$ – Gordon Dec 9 '15 at 15:19
  • $\begingroup$ @Gordon: I agree with your gut feeling that the delta can not be greater than $1$. I think I can now prove it with the maximum principle of the parabolic PDE. I will write it up later. $\endgroup$ – Hans Dec 9 '15 at 19:04
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    $\begingroup$ Looking forward to seeing your derivation. $\endgroup$ – Gordon Dec 9 '15 at 19:18
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    $\begingroup$ the delta is the probability of being in the money in the stock measure for a large class of models including Heston, so it must be less than 1 for Heston. See Chapter 4 of More Mathematical Finance. $\endgroup$ – Mark Joshi Dec 10 '15 at 23:38
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It seems fairly simple to demonstrate an example in discrete space where delta>1. Consider a 2 step binomial tree on a dividend free stock, with interest rates at zero. Let the initial stock price be 100, and each step on the tree have risk neutral p(up)=p(down)=0.5. Let the tree be as follows:
100-(101,99)-(200,2,99,99)
meaning that on the first step it goes up or down a dollar. On the second step the 101 goes to either 200 or 2, and the 99 stays at 99 for certain. On this tree, consider a 100 call. On the 101 node it is worth 50, and on the 99 node it is worth 0. Hence the delta on the initial node must be (50-0)/(101-99) = 25 !
Consider now a second tree with the same notation:
100-(149.5,50.5)-(200,99,99,2) This tree has the same terminal distribution as the first one, but the delta at the initial node is totally different (close to 0.5). So the delta is a function of the local dynamics and not just the terminal distribution. Sufficiently toxic local dynamics can exist that make it >1.

ps this was motivated by the below post about Heston. In the first tree above, implied vol goes ballistic if stock goes up and goes to zero if stock goes down.

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  • $\begingroup$ You are right! See my answer below expanding on it. However, the Heston model delta will NOT be greater than $1$. $\endgroup$ – Hans Dec 11 '15 at 23:39
  • $\begingroup$ understood! i'll be interested to see what constraints the process needs to have for the delta to be always <1 $\endgroup$ – dm63 Dec 11 '15 at 23:43
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@dm63 has demonstrated in his answer that without additional constraint, the delta in discrete setting can exceed $1$. Analogously, in the continuous time and stock price setting, assume the volatility of stock price below a positive threshold vanishes and is a positive constant above that threshold. Set the European call strike at the threshold. We see that the current European call price is zero when the current price is right below the threshold and positive right above the threshold. So the delta at the threshold is infinity.

With additional smoothness and dependency constraints on the volatility, however, the European call delta will be no greater than $1$. I shall supply such a condition and the associated proof.

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