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.
Tell me more
×
Quantitative Finance Stack Exchange is a question and answer site for
finance professionals and academics. It's 100% free, no registration required.
|
|
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! |
|||
|
|
Given that the mathematical proofs have already been given above, let me stress the intuitive aspects of it. If you use a normal model, then you will find that the delta of an ATM option is equal to 50%, and at the same time, the probability of ending ITM (in the money) is also 50%. Now, with a lognormal model, there is a difference between the probability and the delta. The reason is actually very simple. Imagine you run a Montecarlo to figure the delta of an ATM call option. Say you've got around half the paths ending above the strike, and half below. Then clearly, if you were to rerun the Montecarlo, but starting from a slightly higher spot (say 1% because you want to calculate the delta so you 'bump' the spot up), then roughly speaking
If you combine the 2 points above, the price impact of bumping spot up by 1% is going to be 50% x 0 + 50% x (something > 1% ), so the delta is going to be higher than the probability of ending ITM. You can even see that the 'something' is itself very tied to that actual value of the call option. Obviously this relationship works irrespective of whether the option is ATM or not. |
|||
|
|
|
If you look at the BS formula as you find it e.g. in wikipedia straight forward differentiation of the call price gives the call's Delta. You find the formula for the Delta on the wikipedia page under "The Greeks". $\Delta=\Phi(d_1)$ where $\Phi$ is the standard normal cdf and $d_1$ is given by $$d_1 = \frac{\ln(\frac{S}K)+(r+\frac{\sigma^2}2)(T-t)}{\sigma \sqrt{T-t}}$$ where I assume that all parameters are clear. You also find it on wikipedia. If $d_1=0$ then $\Phi(d_1)=\frac12$ per the definition of the normal cdf. When people refer to ATM options having 50 delta they usually mean ATMF, or at the money forward, given by $$S=Ke^{-r(T-t)}$$ Note that sometimes forward prices are derived from put-call parity. Then the forward price can be different) Thus, when the stock is ATMF, $\ln(\frac{S}K)+r(T-t) = 0$, but the terms with sigma remain. In this case $d_1$ is very small but not exactly zero, and $\Delta$ is close to 1/2. |
|||||||||||
|
