# arbitrage free volatility surface

Why is calendar spread arbitrage equivalent to $$\partial_t \omega(k,t) \geq 0, \forall k \in \Bbb{R}$$ where $$\omega(k,t) = \sigma^2(k,t) t$$ and $$\sigma(k,t)$$ represents the Black-Scholes implied volatility smile at $$t$$.

What is the motivation for this definition? Thanks for your help in advance :)

• Can you define terms? Dec 12, 2018 at 22:54
• Sorry i forgot. $\omega (k,t)= \sigma(k,t)_{BS}^2t$, so $\omega(k,t)$ is the implied variance of the Black scholes model times $t$. I try to understand the definition of calendar spread arbitrage of Gatheral. And I'm asking if there is another access. I was thinking about that the option prices for fix $k$ are increasing functions. But I'm not able to write down the connection.
– P.G.
Dec 12, 2018 at 23:07

You'll find here that in terms of European option prices, the absence of calendar arbitrage writes $$\frac{\tilde{C}(k\, F(0,t_2),t_2)}{F(0,t_2)} \geq \frac{\tilde{C}(k \, F(0,t_1),t_1)}{F(0,t_1)}, \forall k \in \Bbb{R}, \forall \, 0 < t_1 < t_2 \tag{1}$$ where $$\tilde{C}(K,t)$$ denotes the undiscounted European call price for strike $$K$$ and time to maturity $$t$$ and $$F(0,t)$$ the underlying forward price for delivery at $$t$$ as seen of $$0$$.

Suppose you would like to translate this inequality in terms of implied volatility i.e. by working in a Black-Scholes world. In that setting it is well known that $$\frac{\tilde{C}(k \, F(0,t),t)}{F(0,t)} =: \mathcal{C}(k,w) = N(d_+(k,w)) - k N(d_-(k,w))$$ with $$d_{\pm}(k,w) = -\frac{\ln(k)}{\sqrt{w}} \pm \frac{1}{2}\sqrt{w}$$ where we have let $$w = w(k,t) = \sigma^2(k,t) t$$.

Then inequality $$(1)$$ can be rewritten as $$\mathcal{C}(k, w(k,t_2)) \geq \mathcal{C}(k, w(k,t_1)) \tag{2}, \, \forall k \in \Bbb{R}, \forall 0 < t_1 < t_2$$ which is verified iff $$\forall k \in \Bbb{R}$$ $$\frac{\partial \mathcal{C}}{\partial t}(k,w(k,t)) \geq 0, \,\, \forall t\in \Bbb{R}^+$$ So that this translates to $$\frac{\partial \mathcal{C}}{\partial w}(k, w(k,t)) \frac{\partial w}{\partial t}(k,t) \geq 0$$ where the first term is positive (see link with BS vega) hence the conclusion.

• And is it right, that I can write the call price as $C(K,t2)=\mathbb{E}((S_{t_2}-K)^+)$ or do I need a conditional expactation?
– P.G.
Dec 14, 2018 at 15:09
• Hi @P.G. that would be the undiacounted call price indeed. The expectation is implicitly conditional on the information you have at $t=0$ i.e. the filtration $\mathcal{F}_0$. Dec 14, 2018 at 15:23
• So is this one right: $C(K,t_2)=\mathbb{E}((S_{t_2}-K)^+ | \mathcal{F}_0) =$(with the tower property) $\mathbb{E}( \mathbb{E} (S_{t_2}-K)^+ | \mathcal{F}_1) | \mathcal{F}_0)$ (with Jensen) $\geq \mathbb{E} ( \mathbb{E} (S_{t_2}-K | \mathcal{F}_1)^+ | \mathcal{F}_0)$ (since $S_t$ is a martingal) = $\mathbb{E}(S_{t_1}-K)^+ | \mathbb{F}_0)=(S_{t_1}-K)^+$. So I have $C(K,t_2)-(S_{t_1}-K)^+ \geq 0 \Rightarrow C(K,t_2)-(S_{t_1}-K)^+ +x>0$ with $x=C(K,t_1)-C(K,t_2)$ if $C(K,t_1)>C(K,t_2)$. But then there is a arbitrage opportunity, so $C(K,t)$ has to be a non decreasing function in $t$.
– P.G.
Dec 14, 2018 at 19:13
• And am I right, that these are the undiscounted values for an option price?
– P.G.
Dec 15, 2018 at 15:09
• In your first comment: $(S_t)$ is not a martingale except if there are no funding costs and no dividends, this is why i used the forward line in the answer linked in my post. Also the last line is incorrect since $\Bbb{E}_0[(S_{t_1}-K)^+] = C(K,t_1)$ not $(S_{t_1}-K)^+$ which is a random variable. Finally, yes, these are undiscounted call prices as mentioned in my previous comment (and the original answer). Dec 17, 2018 at 11:20

The simple explanation is that in the absence of calendar spread arbitrage, we should observe monotonic option prices with respect to maturity. And option prices are monotonic with respect to increase in volatility.

Let $$(X_t)_{t \geq 0}$$ be a martingale, $$L>0$$ and $$0\leq t_1, t_2$$, then we have $$E[(X_{t_{2}} - L)^{+}] \geq E[(X_{t_{1}}-L)^{+}]$$

for any $$i = 1,2$$, let $$c_{i}$$ be options with strikes $$k_i$$ and expirations $$t_i$$. If we assume the two options have the same moneyness ($$k1/F_{t_{1}} =k2/F_{t_{2}} = \alpha^{k}$$, then the process defined by $$x_t = s_t/F_t$$ for all $$t\geq 0$$ is a martingale and

$$c_2/k_2 = \alpha^{-k} E[(X_{t_{2}} - \alpha^{k})^{+}] \geq \alpha^{-k} E[(X_{t_{1}} - \alpha^{k})^{+}] = c_1/k_1$$

so keeping the moneyness constant, option prices are non-decreasing in time to expiration. So, for fixed $$k$$, the function $$w(k,.)$$ must be non-decreasing.

Assuming I understood your question correctly: in the asence of rates, dividends etc. Black & Scholes' formula for a call $$C(K,T)$$ is a function of $$V(K,T) = \sigma(K,T)^2*T$$ as you can easily check (I mean that sigma and T only affect BS through V). It follows that $$\frac{dC}{dT} = \frac{dC}{dV} * \frac{dV}{dT}$$and $$\frac{dC}{dV} > 0$$ (as you may also easily check) so $$\frac{dC}{dT} > 0$$ (as required for no arb, see for instance the beginning of this lectures ) is equivalent to $$\frac{dV}{dT} > 0$$.

• Thank you Jojo Tang for formatting the equations :) Dec 13, 2018 at 23:43

This condition is incorrect in presence of dividend and repo. The correct condition in general is that variance has to be monotonically increasing along the direction of the forward line.

This follows from this reasoning: if maturity is $$T$$ and strike is $$K$$ and spot at time $$t$$ is $$S_t$$ then assuming that you have a dividend d at a date $$T_d$$ then $$S_{T_d}^+ = S_{T_d}^- -d$$

From then you can write the relationship between calls payoffs maturing at $$T_d^+$$ and $$T_d^-$$ as

$$max(0,S_{T_d}^+ - K^+) = max(0, S_{T_d}^- - K^-)$$

Provided that

$$K_{T_d}^+ = K_{T_d}^- -d$$

Which is the definition of the forward line through a dividend

Now that relation between terminal payoffs is established it implies that

$$\sigma(K^+,T+) = \sigma(K^-,T^-)$$

Which is the continuity relation. You can derive for yourself the impact of repo by the same reasoning.