black scholes - Why do some people claim the delta of an ATM call option is 0.5?

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.

Answer

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!