# why is the delta of a short call option negative? [closed]

Why is the delta of a short call option negative? In Black-Scholes-Merton equation the delta of a call option is always a probability function therefore it does not imply such a consequence. How do I interpret this fact from a mathematical/quantitative point of view?

Edit: My bad. I thought a long/short call refers to a call with long/short maturity time. Please disregard this question.

## closed as off-topic by Quantuple, Richard, LocalVolatility, Luigi Ballabio, vanguard2kFeb 7 '17 at 13:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Basic financial questions are off-topic as they are assumed to be common knowledge for those studying or working in the field of quantitative finance." – Quantuple, Richard, LocalVolatility, vanguard2k
If this question can be reworded to fit the rules in the help center, please edit the question.

• For a given S,T,E,r and $\sigma^2$ the Delta is a real number, not a "probability function" AFAIK. – Alex C Jan 20 '17 at 2:09
• @AlexC What I meant is that it is always positive and always between 0 and 1. – Ethan Jan 20 '17 at 2:14
• Well, when you are long the call it is between 0 and 1, when you are short the call it takes the opposite sign, and therefore is between -1 and 0. That is the convention regarding short positions. – Alex C Jan 20 '17 at 2:29
• Is it clear why the delta of a long option is positive and (!) what this means? If yes, then it should be clear why the delta of the short position is negative ... – Richard Jan 20 '17 at 9:12
• I'm voting to close this question because the OP stated it comes from a misunderstanding. – Luigi Ballabio Jan 25 '17 at 9:15

The delta of any position, $\Delta_P$, is the number of units you hold, $N$, multiplied by the delta of each unit, $\Delta$

$$\Delta_P = N\times \Delta$$

You are correct that for a call option you have $0\leq \Delta \leq 1$. If you are short a call option, then you have a negative position (that's what being short means) so $N<0$ and therefore $\Delta_P < 0$.

• This is spot on. Just to be very pedantic though, under BS assumptions, we should have $0 \leq \Delta \leq e^{-q\tau}$ with $q$ the continuously compounded div yield and $\tau$ the time to maturity (because $\Delta = e^{-q\tau}\Phi(d_1)$) – Quantuple Jan 20 '17 at 13:07
• Yes, thanks for the refinement! We might also note that this applies to non-equity instruments too, assuming you interpret $q$ appropriately (the foreign interest rate for a currency option, storage rate for a commodity option or yield for a treasury option). – Chris Taylor Jan 20 '17 at 14:16

A couple of different ways to look at it

• Delta is the change in price/value of the option per unit change in price of the underlying
• For a long call option - any +ve change in the price of underlying, regardless of ATM/OTM/ITM, can only increase the value of the option. Hence, the delta of a long call option is always positive
• A short call position is the mirror/counterparty of the long call position. So if the long increases in value - the short can only decrease in value. Hence the short call position always has a negative delta
• Another way to look at this would be in terms of replicating a stock with options
• Long a Put, and Short a call replicates a Short position in the Stock. Now a short position in the stock has a delta of -1. Long Put has a delta between -1 and 0. Hence the short call needs to have a negative delta