European option Vega with respect to expiry and implied volatility

Question

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?

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.

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.