# 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.

• 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. – Tal Fishman Jun 21 '12 at 15:02
• 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? – ladz Jun 21 '12 at 16:25
• 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 – DoubleTrouble Jun 21 '12 at 16:29
• 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. – ladz Jun 21 '12 at 16:32
• 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. – Ric Jun 22 '12 at 9:18

## 5 Answers

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 $F_T<K$ always, otherwise the Delta is greater than 0.5 – FKaria Jun 21 '12 at 19:35

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.

• Richard, I started writing essentially the same answer when yours came up, so I decided I'd clean up yours instead. – Tal Fishman Jun 21 '12 at 14:55
• Thank you Tal, I appreciate your cleaning. I think it is worth mentioning that Delta is not excactly 1/2 ... as we are doing strict mathematics here. – Ric Jun 21 '12 at 15:00
• Forwards cannot be deduced from Put-Call parity. The ATM forward of S is F_T = S_0*exp(rT) so I assume that K=S_0. Regardless, the forward price has to be less than the strike to obtain a Delta of 0.5 – FKaria Jun 21 '12 at 19:41
• FKaria , of course it makes sense to imply a forward price from put-call-parity. Look at wikipedia and rearrange terms just like e.g. Bloomberg does. – Ric Jun 22 '12 at 7:20
• @Richard, You're right, it can be done but only for ATM strikes, which is the case we're talking about. – FKaria Jun 22 '12 at 17:58

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.

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...

• First: could you please use latex. The second part of your question is a pracitioners approach. Maybe you could go into more detail? Without model $- dc(T)/dk$ doesn't make to much sense mathematically- but it probably does as a trade. – Ric Jan 17 '14 at 13:17
• I’m just saying that the delta of a call is similar to negative 1 times differentiating the payoff at maturity of a vanilla call with respect to strike.. this is because a call pays max(ST -k,0), which is a indicator function. Now dc/dk (using central? finite differences) is actually a bull call spread. – rajn Mar 23 '19 at 20:32

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 $\sigma \to \infty$.

• Clearly the best answer to the OP's question. – Robino Dec 10 '14 at 18:47