When an At-the-Money Call Has 50 Delta in Black–Scholes

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


# 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 by DoubleTrouble (score 16, accepted)

https://quant.stackexchange.com/a/3651

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!

## Answer by Richi Wa (score 14)

https://quant.stackexchange.com/a/3652

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.

## Answer by joelhoro (score 9)

https://quant.stackexchange.com/a/3669

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

- all the paths that finished below the strike in the original MC will probably still finish below in the bumped MC. So for these the payoff of the option is unchanged.

- for those paths which ended in the money, given that your original spot is 1% higher, that means that the simulated spot is also 1% higher, but that 1% if obviously larger than 1% of spot because the simulated spot is ITM. So for these the payoff is increased by more than 1% of spot.

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.

## Answer by AFK (score 4)

https://quant.stackexchange.com/a/9864

First of all, even though you can read it in many places, $\Delta_{Call} = \Phi(d_+)$ is not the probability of finishing ITM. For two reasons:

- it is computed under the risk neutral probability. Not the historic one.

- even under the risk neutral probability $Q(S_T>K) = \Phi(d_-)$ not $d_+$.

In the BS model under the historic probabilty $P(S_T>K) = P(S_te^{(\mu-\sigma^2/2)(T-t)+\sigma(W_T-W_t)} > K) = \Phi\left(\frac{1}{\sigma\sqrt{T-t}}(\log(S_t/K) + (\mu-\sigma^2/2)(T-t))\right)$. Basically you get $\Phi(d_-)$ but with the real drift $\mu$ instead of the risk-free rate.

So what does the Delta really looks like? As was explained: $$ \Delta_{Call} = \Phi(d_+) = \Phi\left(\frac{1}{\sigma\sqrt{T-t}}(\log(S_te^{r(T-t)}/K) + \sigma^2(T-t)/2))\right) $$ So if you are ATMF i.e. $S_te^{r(T-t)} = K$ then the log term cancels out: $$ \Delta_{Call}^{ATMF} = \Phi\left(\frac{1}{2}\sigma\sqrt{T-t}\right) $$ In particular, the volatility term is positive so $\Delta_{Call}^{ATMF} > 0.5$. We can then make a first order approximation: $$ \Delta_{Call}^{ATMF} = \Phi\left(\frac{1}{2}\sigma\sqrt{T-t}\right) \approx \Phi(0) + \Phi'(0)\frac{1}{2}\sigma\sqrt{T-t} \approx 0.5 + 0.2\sigma \sqrt{T-t}. $$ since $\Phi'(0) = \frac{1}{\sqrt{2\pi}} \approx 0.4$. So you have a nice simple approximation for the Delta of an ATMF Call.

PS: You should ask yourself what happens when $\sigma \to \infty$.

## Answer by turtle_in_mind (score 2)

https://quant.stackexchange.com/a/9843

i have a really interesting answer.. clearly delta of an ATM call is approximately 0.5 in BS setting since we can do a taylor series expansion of N(d1) = 1/2 + 1/root(2pi) *d1 which goes to 1/2 at maturity..assuming r=q=0

but above is a model dependent result...it assumes log normal distribution in stock price..

a model independent result.... at maturity...we know delta is either 1 or 0 depending on whether the stock is ITM or OTM.. But delta = - dc(T)/dk...which is a digital...and this can be replicated by 1/2dk of portfolio of call spreads...at maturity...the call spread value is 1/2...