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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?

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    $\begingroup$ why is $S_{t^*} \leq S_T$? $\endgroup$
    – MainCom
    Jan 28, 2022 at 20:29
  • $\begingroup$ @MainCom It's stated nowhere that $S_{t^*} \leq S_T$. What is stated is that $t^* \leq T$ under all scenarios, and hence $E_0 (S_{t^*}-K)_+ \leq E_0 (S_T-K)_+$. $\endgroup$
    – user34971
    Jan 28, 2022 at 21:36
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    $\begingroup$ 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). $\endgroup$
    – Oscar
    Jan 29, 2022 at 10:16
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    $\begingroup$ 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? $\endgroup$ Jan 31, 2022 at 11:56
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    $\begingroup$ Nice question and ansatz, by the way. +1 $\endgroup$ Jan 31, 2022 at 11:57

2 Answers 2

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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 $S_{t^*} \geq S_T$.

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  • $\begingroup$ The maximum of S has nothing to do with this. It's about application of the mean value theorem. $\endgroup$
    – user34971
    Jan 29, 2022 at 11:26
  • $\begingroup$ 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. $\endgroup$
    – MainCom
    Jan 29, 2022 at 11:49
  • $\begingroup$ i.e. you cannot deduce the expectation is smaller just because $t^* < T$. $\endgroup$
    – MainCom
    Jan 29, 2022 at 11:51
  • $\begingroup$ See this thread: quant.stackexchange.com/questions/69568/… $\endgroup$
    – user34971
    Jan 29, 2022 at 19:18
  • $\begingroup$ 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. $\endgroup$
    – MainCom
    Jan 30, 2022 at 1:23
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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 \begin{equation} C\left(S_t,t,K,T\right) := E_t \left[ \left(S_T - K\right)_+ \right], \end{equation} and Asian options \begin{equation} C\left(A_t,t,K,T\right) := E_t \left[ \left(A_T - K\right)_+ \right], \end{equation} with \begin{equation} A_t := E_t \left[ \frac1T \int_0^T S_u \, du \right] . \end{equation} 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 \begin{equation} 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), \end{equation} 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*}

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  • $\begingroup$ 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. $\endgroup$
    – MainCom
    Feb 1, 2022 at 1:40
  • $\begingroup$ @MainCom Why the need to capitalize the word 'wrong' twice now in your comments? Is something wrong? $\endgroup$
    – user34971
    Feb 1, 2022 at 7:31
  • $\begingroup$ Because I want to emphasize it is a fundamental error which you still believe is true. $\endgroup$
    – MainCom
    Feb 1, 2022 at 7:41
  • $\begingroup$ I suspect you are the only one thus far who has spotted a fundamental error. That's good. $\endgroup$
    – user34971
    Feb 1, 2022 at 7:50
  • $\begingroup$ If you are unwilling to accept that, you don't need to post the question at all... $\endgroup$
    – MainCom
    Feb 1, 2022 at 8:03

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