How Volatility Changes European Call Delta and At-the-Money Vega
Summary
The note examines the volatility sensitivity of a European call’s delta under the Black-Scholes model. Differentiating delta with respect to volatility yields a term whose sign depends on negative d2, since the normal density and volatility are positive. Under the stated convention, an in-the-money call has negative delta sensitivity to volatility, while an out-of-the-money call has positive sensitivity. This describes how delta moves; it does not show that delta converges to one-half as volatility rises.
For an at-the-money call, the response separately derives a formula for vega, the option price’s sensitivity to volatility, and gives a square-root-of-maturity approximation under practical bounds on implied volatility and time to expiry. Vega is larger for longer maturities in the discussion. The analysis assumes Black-Scholes and defines at-the-money using spot equal to the discounted strike; the approximation has stated bounds and is not a general result for all maturities or volatility levels.
Key ideas
- Under Black-Scholes, the sign of call delta’s volatility sensitivity depends on negative d2.
- The note states that volatility reduces delta for in-the-money calls and increases it for out-of-the-money calls.
- At-the-money vega increases with the square root of time to maturity in the given approximation.
- The stated vega approximation assumes bounded implied volatility and time to expiry.
- The delta-sensitivity result and the at-the-money vega calculation concern different option sensitivities.
Tags
Full text
# What is the delta of an at-the-money European call option with respect to volatility?
# 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.
## Answer by siou0107 (score 2, accepted)
https://quant.stackexchange.com/a/51781
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.
## Answer by Kermittfrog (score 1)
https://quant.stackexchange.com/a/51788
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 $$ \tag{3*} Vega \approx \sqrt{\frac{\tau}{2\pi}} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.