# Asian option IV less than vanilla option IV

I was wondering whether the following handwaving line of thought can be used to show that the IV of an Asian option is less than the IV of a vanilla option with the same strike and time to maturity:

For simplicity I'll take $$r=q=0$$. Furthermore I am going to assume (as pointed out in the comment by Kevin) that the asset $$S_u$$ is a diffusion (no jumps).

The price of an Asian option is $$E_0\left[ \left( \frac1T \int_0^T S_u\, du - K\right)_+ \right].$$ According to the intermediate/mean value property for integrals, there exists at least one $$t \in[0,T]$$ such that $$S_t = \frac1T \int_0^T S_u\, du.$$ Let $$t^*$$ be the first such $$t$$. It's clear that $$t^*$$ will be a random variable which is always less than or equal to $$T$$.

We can therefore write \begin{align} E_0\left[ \left( \frac1T \int_0^T S_u\, du - K\right)_+ \right] &= E_0 \left[ \left( S_{t^*} - K \right)_+ \right] \\ &\leq E_0 \left[ \left( S_T - K \right)_+ \right]. \end{align}

I think this is OK, but I still have some lingering doubts as $$t^*$$ is a random time.

Does anyone spot a blatant error above? Better yet, would someone be able to make make the 'proof' more rigorous (if it is correct) or point out where it is incorrect?

• why is $S_{t^*} \leq S_T$? Commented Jan 28, 2022 at 20:29
• @MainCom It's stated nowhere that $S_{t^*} \leq S_T$. What is stated is that $t^* \leq T$ under all scenarios, and hence $E_0 (S_{t^*}-K)_+ \leq E_0 (S_T-K)_+$.
– user34971
Commented Jan 28, 2022 at 21:36
• Doesn't it follow from the fact that the expectation of a call option increases with maturity? We show that the Asian option is equivalent to a Call option with some maturity t* that is smaller than T, so if you wanted to price the Asian option you ought to use an Implied Vol that's implied from a call option C(k, t*) rather than C(k, T). Commented Jan 29, 2022 at 10:16
• Intuitively, I can follow your ansatz. Obviously, the Asian feature is smoothing away the interim variation until $T$ and the Asian underlying behaves 'like' a shorter-term European underlying. But what about a world where $S_{t^*}$ is larger than $S_T$, on average? Is that ruled out by some other property, i.e. do we "only" need to assume no-calendar-arbitrage opportunity; or a monotoneously increasing total variation or some such? Commented Jan 31, 2022 at 11:56
• Nice question and ansatz, by the way. +1 Commented Jan 31, 2022 at 11:57

Your proof relies on the following claim: Let $$t^*$$ be a random variable that takes value in $$[0,T]$$, then $$E_0(S_{t^*} -K)_+ \leq E_0(S_T -K)_+$$ holds.

Counter example: Let $$t^*$$ be the last (or first, does not matter) time when $$S$$ achieves maximum on $$[0,T]$$. It is a random variable that takes value in $$[0,T]$$. However obviously $$E_0(S_{t^*} -K)_+ \geq E_0(S_T -K)_+$$ since $$S_{t^*} \geq S_T$$.

• The maximum of S has nothing to do with this. It's about application of the mean value theorem.
– user34971
Commented Jan 29, 2022 at 11:26
• I know what you mean. But how do you prove the last inequality in your proof? What I wrote down is about why something like this does not hold generally. Commented Jan 29, 2022 at 11:49
• i.e. you cannot deduce the expectation is smaller just because $t^* < T$. Commented Jan 29, 2022 at 11:51
– user34971
Commented Jan 29, 2022 at 19:18
• They are different. As you already have mentioned, here t is a random variable. I have already given you a counter example why this is WRONG. Commented Jan 30, 2022 at 1:23

Undeleting and editing my own answer, not to bump the question up again, but to try to close / settle it as I think there are some interesting subtleties in it.

I'll also briefly address @MainCom 's counterexample to show that in fact it isn't a counterexample.

As I'd like to use the mean value theorem I'll assume that the asset process is continuous on $$[0,T]$$. For example a local stochastic volatility model without asset price jumps will satisfy this condition.

For simplicity I've set the risk-free rate and dividend yield to zero.

We will be interested in vanilla options $$$$C\left(S_t,t,K,T\right) := E_t \left[ \left(S_T - K\right)_+ \right],$$$$ and Asian options $$$$C\left(A_t,t,K,T\right) := E_t \left[ \left(A_T - K\right)_+ \right],$$$$ with $$$$A_t := E_t \left[ \frac1T \int_0^T S_u \, du \right] .$$$$ Let $$BS\left(S_t,t,K,T,I_S (K)\right)$$ denote the Black-Scholes (BS) price of a vanilla option with implied volatility (IV) $$I_S (K)$$, and $$BS\left(A_t,t,K,T,I_A (K)\right)$$ the BS price of an Asian option with IV $$I_A (K)$$. These IVs are defined by \begin{align} BS\left(S_t,t,K,T,I_S(K)\right) &:= C\left(S_t,t,K,T\right), \\ BS\left(A_t,t,K,T,I_A(K)\right) &:= C \left(A_t,t,K,T\right). \end{align}

Lastly, recall also the mean-value theorem for integrals: Let $$f:[a,b]\rightarrow \mathbb{R}$$ be a continuous function. Then there exists at least one $$x\in[a,b]]$$ such that $$f(x) = \frac{1}{b-a} \int_a^b f(u)\, du.$$

Proposition:

An upper bound on the price $$BS\left(A_t,t,K,T,I_A (K)\right)$$ of an Asian option is $$$$BS\left(A_t,t,K,T,I_A(K)\right) \leq \lambda \, BS\left(S_t,t,\lambda^{-1}K' ,T,I_S(\lambda^{-1}K')\right),$$$$ with $$\lambda = \frac{T-t}{T}$$ and $$K' = K - \frac1T \int_0^t S_u \, du$$.

Proof:

We can write \begin{align*} BS\left(A_t,t,K,T,I_A (K)\right) &:= E_t \left[ \left(A_T - K\right)_+ \right] \\ &= E_t \left[ \left(\frac1T \int_0^T S_u \, du - K \right)_+ \right] \\ &= E_t \left[ \left(\frac{\lambda}{T-t} \int_t^T S_u \, du - K' \right)_+ \right] \end{align*} with $$\lambda = \frac{T-t}{T}$$ and $$K' = K - \frac1T \int_0^t S_u \, du$$. According to the mean-value theorem, for each path of the asset, there exists at least one $$\tau\in [t,T]$$ such that $$S_\tau = \frac{1}{T-t} \int_t^T S_u du.$$ Let $$\tau^*$$ be the first such $$\tau$$. It is clear that each $$\tau \in [t,T]$$ is a random variable, and in particular $$\tau^* \in [t,T]$$ is a random variable. The problem of determining the price of an Asian option can then be re-cast in the following form: $$BS\left(A_t,t,K,T,I_A (K)\right) = \lambda E_t \left[ \left(S_{\tau^*} - \lambda^{-1}K' \right)_+ \right].$$

Denote by $$q(r)$$ the distribution of $$\tau^*$$. Then \begin{align*} BS\left(A_t,t,K,T,I_A (K)\right) &= \lambda \int_t^T E_t\left[ \left(S_{\tau^*} - \lambda^{-1} K' \right)_+ | \tau^* = r \right] q(r)\, dr \end{align*}

Now, \begin{align*} E_t\left[ \left(S_{\tau^*} - \lambda^{-1} K' \right)_+ | \tau^* = r \right] &= E_t\left[ \left(E_{\tau^*}(S_T) - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\ &\leq E_t\left[E_{\tau^*} \left(S_T - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\ &= E_t\left[\left(S_T - \lambda^{-1} K' \right)_+ | \tau^* = r \right] \\ &=E_t\left[\left(S_T - \lambda^{-1} K' \right)_+\right] \end{align*} where the inequality follows from Jensen's inequality, and clearly $$S_T$$ is independent of $$\tau^*$$.

Hence, \begin{align*} BS\left(A_t,t,K,T,I_A (K)\right) &\leq \lambda \int_t^T E_t\left[ \left(S_T - \lambda^{-1} K' \right)_+ \right] q(r)\, dr\\ &= \lambda \, E_t\left[ \left(S_T - \lambda^{-1} K' \right)_+ \right] \\ &= \lambda \, BS\left(S_t,t,\lambda^{-1}K' ,T,I_S (\lambda^{-1}K')\right). \end{align*}

Corollary: The IV of a freshly minted Asian option is bounded above by the IV of a vanilla option with the same strike and time to maturity.

Proof: For a freshly minted Asian option $$t=0$$ and thus $$\lambda = 1$$ and $$K=K'$$.

As for MainCom's counterexample: Even though it is true that calendar arbitrage cannot in general be applied to random times, it is not a counterexample since the maximum of an asset on $$[0,T]$$ can't be written as an integral with integral limits equal to $$0$$ and $$T$$. This means that the mean value theorem can't even be applied to MainCom's example rendering the whole random time argument non-applicable.

Note that the bounds are in line with the more straightforward derivation given for instance here. However, I thought applying mean value theorem in this context is interesting as well.

Afterthought: The whole derivation can be shortened to \begin{align*} BS\left(A_t,t,K,T,I_A (K)\right) &= \lambda E_t \left[ \left(S_{\tau^*} - \lambda^{-1}K' \right)_+ \right] \\ & = \lambda E_t \left[ \left(E_{\tau^*}(S_T) - \lambda^{-1}K' \right)_+ \right] \\ &\leq \lambda E_t \left[ E_{\tau^*} \left(S_T - \lambda^{-1}K' \right)_+ \right] \\ &= \lambda E_t \left[ \left(S_T - \lambda^{-1}K' \right)_+ \right]. \end{align*}

• Still, you miss the core of the problem. Your claim the inequality follows from the absence of calendar arbitrage is WRONG because here $\tau^*$ is a random variable. To prove it you need to go back to the definition of your process $S$ and make use of property of it. Commented Feb 1, 2022 at 1:40
• @MainCom Why the need to capitalize the word 'wrong' twice now in your comments? Is something wrong?
– user34971
Commented Feb 1, 2022 at 7:31
• Because I want to emphasize it is a fundamental error which you still believe is true. Commented Feb 1, 2022 at 7:41
• I suspect you are the only one thus far who has spotted a fundamental error. That's good.
– user34971
Commented Feb 1, 2022 at 7:50
• If you are unwilling to accept that, you don't need to post the question at all... Commented Feb 1, 2022 at 8:03