### Why do some people claim the delta of an ATM call option is 0.5?

• 9 years ago

I am looking for a mathematical proof in terms of differentiating the BS equation to calculate Delta and then prove it that ATM delta is equal to 0.5. I have seen many books quoting delta of ATM call option is 0.5, with explanations like the probability of finishing in the money is 0.5, but I am looking for a mathematical proof. Since your original statement to prove is false, I've changed the question title to asking about the claim rather than asking for a proof. Can someone explain me why $\frac{(r+\sigma^2/2)*t}{\sigma *\sqrt{t}}$ is equal to zero? I see r and sigma as positive nos , so am I correct in thinking that mathematically a delta of call option is not exactly equal to 0.5? Are you asking when [(r+sigma^2/2)*(T-t)]/[sigma * sqrt(T-t)] is equal to zero? If so, please see my answer below thanks got it, so basically at maturity only the delta of ATM call option is equal to 0.5 and not during the life of option. Well, as Tal Fishman and Richard pointed out, it depends on what you mean by an ATM call option. In my answer I assumed K=S(t), and then the delta is exactly equal to 0.5 only at maturity. Delta is not defined at maturity when S=K I agree to @FKaria, Delta at maturity does not make too much sense, especially if the option is ATM. I have a call (or a put) with strike $100$, the stock price is $100$ and time to maturity is zero. This is probably not the set-up that people talk about when they talk about options with $0.5$ Delta.

• 9 years ago

Your question is not really well formulated since you do not specify at which time the delta is equal to 0.5. What you claim is in fact only true for an ATM call option at the time of maturity.

In the Black-Scholes model the price of a call option on the asset S with with strike price $K$ and time of maturity $T$ equals

$$c(t,S(t),K,T) = S(t)\Phi\left(\frac{\ln\frac{S(t)}{K} + \left(r+\frac{\sigma^2}{2} \right)\tau}{\sigma \sqrt{\tau}} \right) - Ke^{-r \tau}\Phi\left(\frac{\ln\frac{S(t)}{K} + \left(r-\frac{\sigma^2}{2} \right)\tau}{\sigma \sqrt{\tau}} \right)$$

where $r$ is the risk-free rate, $\sigma$ the volatility and $\tau = T-t$. The "delta" in the Black-Scholes model is

$$\Delta(t,S(t),K,T) = \frac{\partial c}{\partial S}(t,S(t),K,T) = \Phi\left(\frac{\ln\frac{S(t)}{K} + \left(r+\frac{\sigma^2}{2} \right)\tau}{\sigma \sqrt{\tau}} \right)$$

In the case of an at the money call option we have $K=S(t)$ which means that we get

$$\ln\frac{S(t)}{K} = \ln(1) = 0$$

and we are left with

$$\Delta(t,S(t),S(t),T) = \Phi\left(\frac{\left(r+\frac{\sigma^2}{2} \right)\tau}{\sigma \sqrt{\tau}} \right)$$

This expression equals $0.5$ when $\tau = 0$ that is when $t=T$. This is because $\Phi(x)=0.5$ if and only if $x=0$.

Hope this helps you understand. Otherwise, do not hesitate to ask again! The Delta of a call is not defined when $t=T$ and $S_T=K$, there is a spike in the payoff function. We can say that Delta will be 0.5 every time that $\log(F_T/K) = -0.5v*T$ (where $F_T$ is the forward at time $T$ conditioned by $S_t$). So necessarily \$F_T