I was told that the Vega of an European option always increases when its time to expiry increases (all else equal). I found this confusing and potentially wrong, but there doesn't seem to be relevant sources online about this. Let's take an ATM option for simplicity, its Vega is: $S\sqrt(\tau)N'(d1)$, which is just $1 \over \sqrt(2\pi)$$S\sqrt(\tau)e^{-(r+{\sigma^2\over2})^2\tau\over2}$. Now as $\tau$ increases to a large range, I found that Vega certainly decreases as we have a $-\tau$ term in the exponent. However in small ranges of $\tau$ for example between 0 and 1, Vega does increases as $\tau$ increases. Am I mistaken here?

I also would like to see the relationship of European option price with respect to volatility and plotted a graph where the $y$ axis is the option price computed from Black-Scholes, and the $x$ axis is $\sigma$ (also holding everything else equal and use ATM options for simplicity). The graph surprisingly looks like a straight line. But from the formula above, the local slope of this line should just be Vega at different values of $\sigma$, and thus should be decreasing, so theoretically the line should be concave. Am I mistaken here?


1 Answer 1


To keep notations uncluttered, consider that $r=q=0$ in what follows, while focusing on the particular case of an ATM option i.e. $K=S$ (otherwise use the same reasoning with $K=F(0,T)=Se^{(r-q)T}$ i.e. an ATMF option, the conclusion won't change that much).

In your first question, you're looking for the sign of the derivative of Vega with respect to the residual maturity $\tau$. The closed-form expression of this Greek is given here under the name Veta. In our particular configuration we get $$ \text{Veta} = - S \phi(d_1) \sqrt{\tau} \left[ \frac{1 + d_1d_2}{2\tau} \right] $$ with $d_1 = \frac{1}{2}\sigma\sqrt{\tau}$ and $d_2 = -d_1$ such that $$ \text{Veta} = - S \phi(d_1) \frac{1}{2\sqrt{\tau}} \left[ 1 - \frac{1}{4}\sigma^2\tau \right] $$ since the prefactor has always the same sign, we see that a sign change occurs at $\tau = \frac{4}{\sigma^2}$. For typical levels of volatility -- say 20% -- this gives $\tau \approx 100Y$. This is easily confirmed by a quick graph (I've used $\sigma=0.2$, $r=q=0$, $S=K=1$). Since most traded instruments generally expire long before that time, this justifies the common belief of considering that $\text{Vega}$ is 'always' increasing.

enter image description here

In the second question, you're looking at $C = f(\sigma)$ all the other parameters being fixed where $C$ is the price of an ATM call. You're observation that $C$ is a linear function of $\sigma$ is due to the fact that this well known approximation holds quite well in practice for small $\tau$. Of course if you want the exact result you should again work out the closed-form expression of the second derivative of $C$ with respect to $\sigma$ (Volga or Vomma) and look at when it is significant vs. negligible.

  • 1
    $\begingroup$ It is surprising and fascinating to me that a relationship that I am completely used to seeing in every case I have looked at, actually is not always true but reverses itself for maturities longer than 100 years. It shows the limitations of financial intuition based on experience. Thanks for the insight. $\endgroup$
    – nbbo2
    Jan 26, 2018 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.