# Intuitively speaking, why do at the money options have no volga/convexity?

I was wondering if someone could give me an intuitive explanation as to why the vega of at the money options doesn't increase with volatility. I've seen some mathematical explanations showing the derivative of vega with respect to vol when strike=fwd price to be 0, but I didn't follow intuitively why that would be.

Edit: Volga: change in vega/change in vol

• Can you please also define volga/convexity? Not everyone knows it. Jan 7, 2016 at 2:27
• Because Vega is maximum ATM? Jan 7, 2016 at 8:06

Suppose S = 100, and K = 100. Imagine it is 1 second prior to expiry. And only two outcomes are possible UP or Down. Sup = 101 and Sdow = 99. Your call will pay either 1 or 0 with 50% probability. Thus price of option is 0.50

Now, imagine same situation but with higher vol. Sup = 102 and Sdow = 98. Same proba, thus price of call is 1.

Repeat with higher vol: Sup = 104, Sd = 96. Call is 2.

As you can see it increases linearly with vol, cause the down leg always gives you zero. And the payout on the upside is linearly increasing with vol. (clearly not a mathematically sophisticated answer, but at least paints the picture)

Consider the graph of price vs implied volatility of an at-the-money call option. At 0 volatility, the price is zero, as with zero vol the spot remains constant and finishes at the strike for zero payoff. For low volatilities, there is a famous approximation that call value is about $0.4 S \sigma \sqrt T.$ That gives the graph initially increasing linearly with a slope $0.4 S \sqrt T.$ But for very high vol, there is a no-arbitrage limit that the price of the call cannot be higher than the price of the underlying. So the graph must asymptote to being flat at $S$ as volatility goes to infinity. The slope of this graph is vega. So vega is positive (about $0.4 S \sqrt T$) for low vol but decreases to $0$ in the limit as vol goes to infinity.

This is probably the mathematical explanation you are referring to... LINK

I tend to think of this problem semi-mathematically with remembering the whole formula. Say the time value of the option comes from sigma*sqrt(T). So ATM option is approximately linear in volatility. Similar conclusion can be drawn in terms of the trend of time decay of an option, which is approximately a square root function.

Suppose you have two stocks, perfectly correlated, both initially at 100, but one has exactly twice the dollar standard deviation of the other. Then the payouts of a 100 call are exactly in a 2:1 ratio for every possible path. Notes: (1) I've assumed both spot and forward price are 100 (2) this doesn't strictly hold if the distribution is not Normal, but I think the intuition is ok. For an ATM call option, the vega is given by \begin{align*} \frac{\partial C}{\partial \sigma} &= SN'(d_1)\sqrt{T}\\ &=\frac{1}{\sqrt{2\pi}}S\,e^{-\frac{d_1^2}{2}}\sqrt{T}\\ &=\frac{1}{\sqrt{2\pi}}S\,e^{-\frac{\sigma^2}{8}T}\sqrt{T}. \end{align*} Then, the volga is given by \begin{align*} \frac{\partial^2 C}{\partial \sigma^2} &=-\frac{1}{4\sqrt{2\pi}}S\,e^{-\frac{\sigma^2}{8}T} \sigma\, T^{3/2}, \end{align*} which is negative. That is, the vega does not increase with the volatility. Depending on the magnitude ofS$,$\sigma$, and$T$, the volga does not have to be insignificant. For example, for$S=100$,$\sigma=0.35$, and$T=5\$, \begin{align*} \frac{\partial^2 C}{\partial \sigma^2} \approx -36.15, \end{align*} which does not appear small.