Considering an American option in a Black-Scholes model, is there a relation between Vega and Gamma as it holds in the European case?

I am aware an exact relation would be difficult to find. But in practice is there a kind of approximate relation between these two quantities?


2 Answers 2


No, you should not expect such a relationship to hold in general. The reason is that American options have an "exercise barrier" which European options don't, and this results in different prices and greeks.

In the case of put options (with interest rate $r>0$) as the spot price falls, at some point it becomes optimal to exercise early and take the cash. Beyond this point, the option behaves like a short position in the stock, so its delta is exactly -1 and its gamma is zero. As the spot crosses through this barrier, the gamma jumps.

Vega, on the other hand, does not have such a discontinuity. The chart below shows the price of a European and American put with strike 100, three months to maturity, risk-free rate 15% and volatility 20%. The European prices and greeks are from the Black-Scholes model, and the American prices and greeks are from a binomial tree.

enter image description here

Note that when the spot price is sufficiently far from the early exercise barrier, it is unlikely to cross it, and the option behaves like a European. In this case you expect the standard relationship between gamma and vega,

$$\nu = \sigma\tau S^2 \Gamma$$

will approximately hold. The plot below shows how this relationship holds when the spot price is sufficiently far above the strike, but breaks down when it approaches the early exercise barrier.

enter image description here

  • 2
    $\begingroup$ Interesting diagrams, which I don't remember seeing before, thanks. $\endgroup$
    – nbbo2
    Commented Jul 18, 2017 at 15:27
  • $\begingroup$ Will you take a look at my question quant.stackexchange.com/q/75501/6686? Your Gamma plot seems to offers an answer in the positive. Does the inequality always hold for all time ? Does it also hold for calls with continuous dividends? If so, do you know of a proof? $\endgroup$
    – Hans
    Commented May 11, 2023 at 16:10

One way to think of American-exercise options is to break their value $V_A$ down into a value due to the european exercise, $V_E$, and a "premium" due to the possibility of early exercise, $V_P$

$$ V_A = V_E + V_P $$

Since the differentiation operator is linear, we then have similar terms for gamma and vega

$$ \Gamma_A = \Gamma_E + \Gamma_P $$

$$ \aleph_A = \aleph_E + \aleph_P $$

For the "european component" the standard relation holds

$$ \aleph_E = S^2 \sigma \tau \Gamma_E $$

To the extent that the exercise premium $V_P$ and its derivatives are small compared to $V_E$ and its derivatives, the standard european option relationship holds in approximation

$$ \begin{aligned} \aleph_A &= S^2 \sigma \tau \Gamma_E + \aleph_P \\ &= S^2 \sigma \tau (\Gamma_A-\Gamma_P) + \aleph_P \\ &= S^2 \sigma \tau \Gamma_A + \underbrace{(\aleph_P -S^2 \sigma \tau\Gamma_P)}_{\text{Small terms}} \end{aligned} $$ and similarly

$$ \Gamma_A = \frac{\aleph_A}{S^2 \sigma \tau} + \underbrace{\Gamma_P - \frac{\aleph_P}{S^2 \sigma \tau}}_{\text{Small terms}} $$

  • 1
    $\begingroup$ Is it true that $\Gamma_E\aleph_E = 1/\sigma$? I have always thought of the relationship as $\aleph_E = \sigma\tau S^2\Gamma_E$ where $\tau$ is the time to expiry. $\endgroup$ Commented Jul 18, 2017 at 14:03
  • $\begingroup$ I agree with Chris Taylor. $\endgroup$
    – nbbo2
    Commented Jul 18, 2017 at 14:15
  • $\begingroup$ Dumb mistake on my part -- thank you for pointing it out, Chris $\endgroup$
    – Brian B
    Commented Jul 18, 2017 at 15:33
  • 1
    $\begingroup$ Not sure why this answer of mine was accepted rather than Chris'. I like his better, so I urge everyone to read it. $\endgroup$
    – Brian B
    Commented Jul 19, 2017 at 15:44
  • $\begingroup$ What is the reason that the residual terms above should be small? $\endgroup$
    – Hans
    Commented May 12, 2023 at 13:34

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.