Can delta of an option be greater than 1? Please illustrate it with an example.

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    $\begingroup$ No, the delta of a call (put) option can not be greater (lower) than 1 (-1). A positive gamma increases the delta value of call options towards 1 and decreases the delta value of put options towards -1 when these options get in the money. However, gamma does not remain constant and gets close to zero as delta reaches the bounds above. $\endgroup$ Oct 17, 2019 at 7:35
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    $\begingroup$ +1. Intuitively, being long an option exposes you to the performance of one share of an underlying asset. Since the Delta represents the number of shares to hold to replicate that performance (REM: it needs to be rebalanced continuously for this to work), it can never go above 1 in absolute value. This can easily be seen at expiry: if you are long an option which is ITM you replicate the perf with 1 share of the underlying. If it's OTM you replicate it with zero share. Before maturity, it's just a matter of how much ITM/OTM you are. $\endgroup$
    – Quantuple
    Oct 17, 2019 at 10:25
  • $\begingroup$ Unless you wanted to tinker with the payouts, a classic vanilla option is constrained between +/-1. You have to buy multiple options to give you an aggregate >1. Else you would need non-linear payouts, e.g. an option that paid the square of (strike-spot). Else if you get VERY cute, a European cash or nothing can go >1 delta if interest rates go VERY negative. N(d2) is constrained; but e(-rt) can lever this up if r is sufficiently negative. It back in the real world, short answer is " no" $\endgroup$
    – demully
    Oct 17, 2019 at 10:56

2 Answers 2


Only constrained to be <1 in the simplified Black-Scholes setting with zero cost of carry on the underlying. In the more realistic and common setting where the cost of carry of the underlying is higher than the discounting rate, then it is entirely possible for a call to have a delta > 1.

This is the case because your future costs are proportional to the spot value and hence require an additional hedge in terms of spot.

Consider for example a one-year zero-strike call in a situation where rates are 0 and the financing cost for the underlying is an annual rate of 10% of spot (USD 100) paid continuously. Then clearly as the option seller this is initially expected to cost you $10 in financing over the year but will actually cost you more in USD terms if the spot doubles today. This risk needs to be hedged, and calls for you to buy 110% delta in total.

The delta of a call is not $N(d_1)$ but $e^{qT}N(d_1)$ where q is the excess of the cost of carry over the risk-free rate, i.e. financing spread minus dividend yield.

  • $\begingroup$ Thanks for your input. Even under Black Scholes assumptions if we change convention of paying premium, then delta could be greater than 1. For example: if we change the premium paying currency in long put EURUSD from USD to EUR, it is likely that delta would be greater than 1. $\endgroup$
    – Ussu
    Oct 17, 2019 at 16:51
  • $\begingroup$ Theoretically, the premium included example is correct but practically, EURUSD is generally Delta premium excluded in the market. $\endgroup$
    – AKdemy
    Apr 25, 2021 at 21:26
  • $\begingroup$ I like using the basket option with weights for each underlying as an example. If you have a weight >1 then your Delta will also be >1. Say the weight for 1 component is 2. Then the Delta will max out at 2 when deep ITM. $\endgroup$
    – Matt
    Jul 23 at 20:39

The delta of a European call is: $$ \Delta(call)=N(d_1) $$ where $N$ is the cumulative probability function which return value between 0 and 1. Therefore, for a traditional option, your $\Delta$ cannot be greater than 1.

  • $\begingroup$ Hi Hydraxize, surely you meant the delta of a European call? $\endgroup$
    – Quantuple
    Oct 17, 2019 at 10:23
  • $\begingroup$ Yes, European (exercised at maturity), I've updated the answer. $\endgroup$
    – Hydraxize
    Oct 17, 2019 at 10:33

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