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.

  • $\begingroup$ The question is strangely asked. I would interpret it as “first-order sensitivity (delta) to volatility”, i.e. vega, not $\frac{\partial \Delta}{\partial \sigma}$. $\endgroup$
    – siou0107
    Mar 23, 2020 at 16:32
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 24, 2020 at 10:21
  • $\begingroup$ How to obtain that ATMF implies $d_2=0$? $\endgroup$
    – Idonknow
    Mar 24, 2020 at 14:14
  • $\begingroup$ d_2 = 0 is not the ATMF strike. The answer by @siou0107 is the answer to your question. $\endgroup$ Mar 24, 2020 at 14:21
  • $\begingroup$ But he doesn't answer when the call is ATMF. $\endgroup$
    – Idonknow
    Mar 24, 2020 at 14:41

2 Answers 2


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.

  • $\begingroup$ What about at-the-money? $\endgroup$
    – Idonknow
    Mar 23, 2020 at 16:41
  • $\begingroup$ At the money it is fairly insensitive, as $d_2 \approx 0$ $\endgroup$
    – siou0107
    Mar 23, 2020 at 19:04
  • $\begingroup$ 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. $\endgroup$
    – Idonknow
    Mar 24, 2020 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 $$ \tag{3*} Vega \approx \sqrt{\frac{\tau}{2\pi}} $$
  • 1
    $\begingroup$ I think you are answering different question. $\endgroup$
    – Idonknow
    Mar 24, 2020 at 14:15
  • $\begingroup$ 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. $\endgroup$ Mar 24, 2020 at 15:12
  • $\begingroup$ I think you are calculating Vega. But what I want is how delta changes if volatility changes? $\endgroup$
    – Idonknow
    Mar 24, 2020 at 15:26
  • $\begingroup$ Funny enough: The result is the same, scaled by a factor of 2. $\endgroup$ Mar 24, 2020 at 18:52

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