Why the delta of ATM call option is slightly larger than 0.5?

danielfang

New Member
I am learning the BS model from the Internet. While the fact that the delta of the ATM call option is approximately 0.5 looks intuitive for me, I have difficulty understanding why it is slightly larger than 0.5 in an intuitive way. From the formula, it is pretty clear that before maturity, d1 is larger than 0 so that N(d1) is larger than 0.5. But how to interpret this without the formula? I have tried searching this on Google, but those answers don't align well.
 

rajanS

Active Member
I am learning the BS model from the Internet. While the fact that the delta of the ATM call option is approximately 0.5 looks intuitive for me, I have difficulty understanding why it is slightly larger than 0.5 in an intuitive way. From the formula, it is pretty clear that before maturity, d1 is larger than 0 so that N(d1) is larger than 0.5. But how to interpret this without the formula? I have tried searching this on Google, but those answers don't align well.
Hmm, without a model. Here’s my interpretation. Assume European call, non dividend paying. At maturity, it’s either in the money or out of money. So delta is either 0 or 1. This has the same payoff of a digital call that pays 0 or 1. which as we know can be approximated by 1/2 of call spreads (long 1/2dk call struck at k-dk and short 1/2dk of same amount at k+dk). Using finite difference approximation). At maturity, if stock price ends in strike, u make half. At any time before maturity, it’s approximately the price of those spreads
 
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rajanS

Active Member
And since anytime before maturity, stock price is volatile, we can expect delta to be slightly higher than 0.5 to account for this fact.
 

danielfang

New Member
And since anytime before maturity, stock price is volatile, we can expect delta to be slightly higher than 0.5 to account for this fact.
Hmm...Can I say the stock price is more likely to grow than to drop because of the risk-free rate? Thus, before maturity, the chance of exercising the option at maturity is slightly larger than 0.5.

btw, I tried to understand how the BS model is derived from ground, but looks like the model involves stochastic calculus that I haven't learned before and will be taught during MFE studies. Am I right?
 

rajanS

Active Member
Hmm...Can I say the stock price is more likely to grow than to drop because of the risk-free rate? Thus, before maturity, the chance of exercising the option at maturity is slightly larger than 0.5.

btw, I tried to understand how the BS model is derived from ground, but looks like the model involves stochastic calculus that I haven't learned before and will be taught during MFE studies. Am I right?
Hopefully...
 

Daniel Duffy

C++ author, trainer
A possible cross is to solve for that value of S in the nonlinear equation

N(d_1) = 0.5

where d_1 = log(S/K) + all the rest.
 

danielfang

New Member
After reading the above paper, I guess I have figured out why it is slightly larger than 0.5 without calculating N(d1).
Given that the stock price follows a lognormal distribution, the expected value of the stock price over the current stock price E(St/S0) = exp(µt) where µ is the drift term of geometric Brownian motion. By risk-neural pricing, risk-free rate Rf is used instead of µ, so the E(St/S0) = exp(Rf*t). Before maturity, t is larger than 0, so the expected value is larger than 1. That being said, the expected value of the stock price at maturity with time-to-maturity = t is larger than the current stock price S0. Given that the call option is at the money, the current stock price equals the exercise price. Therefore, the stock price at maturity is expected to be slightly larger than the exercise price, which causes the probability of exercising the option to be slightly larger than 0.5. So the delta of the ATM call option prior to maturity is slightly larger than 0.5.

In simple words, the stock price is expected to grow at a risk-free rate like how forwards are priced but using continuously compounded interest rate instead. So the delta is > 0.5.

Correct me if I am wrong..
 
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danielfang

New Member
I just realized that the paper was written by the current director of Columbia MAFN. 😲
 
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