# What is the delta of an at-the-money European call option with respect to volatility?

Question: What is the delta of an at-the-money European call option with respect to volatility?

Note that $$\frac{\partial\Delta}{\partial\sigma} = N'(d_1) \frac{\partial d_1}{\partial\sigma} = N'(d_1) \frac{- d_2}{\sigma} = \frac{-N'(d_1)d_2}{\sigma}$$ where $$N(\cdot)$$ is the CDF of the standard normal distribution. I am not able to deduce anything from this equation.

This QFSE post states that higher volatility for in-the-money option will have lower delta whereas higher volatility for out-of-the-money options will have higher delta.

Based on this website, it seems that higher volatility will lead to $$\Delta = 0.5.$$ But I am not able to show this.

• The question is strangely asked. I would interpret it as “first-order sensitivity (delta) to volatility”, i.e. vega, not $\frac{\partial \Delta}{\partial \sigma}$. – siou0107 Mar 23 at 16:32
• As an aside: the "holy trinity" of strikes are d2=0, ATMF, d1=0, and the holy trinity of Greeks are Vega,Vanna, and Volga. Understand the behaviour of these three Greeks at those three strikes, and you understand 95% of the volatility smile. – ilovevolatility Mar 24 at 10:21
• How to obtain that ATMF implies $d_2=0$? – Idonknow Mar 24 at 14:14
• d_2 = 0 is not the ATMF strike. The answer by @siou0107 is the answer to your question. – ilovevolatility Mar 24 at 14:21
• But he doesn't answer when the call is ATMF. – Idonknow Mar 24 at 14:41

Based on your computation, you can observe that the $$N’$$ term is always positive, between 0 and 0.4. As $$\sigma$$ is always positive, you can focus on the $$-d_2$$ term. When $$d_2 > 0$$, i.e. call is ITM, delta has a negative sensitivity to volatility ; conversely for OTM call. That is in line with your remark.

• What about at-the-money? – Idonknow Mar 23 at 16:41
• At the money it is fairly insensitive, as $d_2 \approx 0$ – siou0107 Mar 23 at 19:04
• How do you obtain that $d_2\approx 0$ for at-the-money option? I have $d_2 = \left( \frac{r}{\sigma} - \frac{\sigma}{2}\right) \sqrt{\tau}$ only. – Idonknow Mar 24 at 8:40

In the following, I am assuming the BS73 model and I assume that "ATM" means

$$S = Xe^{-r\tau}$$ The pricing formula for a European call then becomes $$\tag{1} O\propto N\left(+\frac{1}{2}\sigma\sqrt{\tau}\right)-N\left(-\frac{1}{2}\sigma\sqrt{\tau}\right)$$ times some scaling factor which is irrelevant for our purpose. Clearly, $$Vega\equiv\frac{\partial O}{\partial \sigma}=\frac{1}{2}\sqrt{\tau} \cdot{} n\left(\frac{1}{2}\sigma\sqrt{\tau}\right)+\frac{1}{2}\sqrt{\tau} \cdot{} n\left(-\frac{1}{2}\sigma\sqrt{\tau}\right)$$ Leading us to $$\tag{2} \frac{\partial O}{\partial \sigma}=\sqrt{\tau}\frac{e^{-\frac{1}{2}\left(0.5\sigma\sqrt{\tau}\right)^2}}{\sqrt{2\pi}}$$ Thus:

• For longer maturities, the Vega is larger than for smaller maturities
• For all practical purposes (i.e. $$IV<75\%$$, $$\tau<1yr$$, you can approximate the ATM Vega to $$\tag{3*} Vega \approx \sqrt{\frac{\tau}{2\pi}}$$
• I think you are answering different question. – Idonknow Mar 24 at 14:15
• I interpreted your question " What is the delta of an at-the-money European call option with respect to volatility?" as the first derivative (hence Delta) w.r.t. Volatility. – Kermittfrog Mar 24 at 15:12
• I think you are calculating Vega. But what I want is how delta changes if volatility changes? – Idonknow Mar 24 at 15:26
• Funny enough: The result is the same, scaled by a factor of 2. – Kermittfrog Mar 24 at 18:52