I am looking to derive the call price of an asian option of the form $$\max\{A_T - K, 0\}$$ with $$A_T = \left(\prod_{i=1}^nS_{t_i}\right)^\frac{1}{n}$$ which has price under $\mathbb{Q}$ $$e^{-rT}[S_0e^{\mu_nT} \Phi(d_n) - K \Phi(d_n - \sigma_n \sqrt{T})]$$ where $\Phi$ is a standard normal and \begin{align} \mu_n & = (r-0.5\sigma^2) \frac{n+1}{2n}+0.5\sigma_n^2\\ \sigma_n^2 & = \frac{\sigma^2(n+1)(2n+1)}{6n^2}\\ d_n & = \frac{\ln (S_0/K) + (\mu_n +0.5\sigma_n^2)T}{\sigma_n \sqrt{T}} \end{align} I have tried splitting the expectation : \begin{alignat*}{2}\Pi_0&=e^{-rT}\mathbb{E}^\mathbb{Q}[\max\{A_T-K, 0\}|\mathcal{F}_0]\\ & = e^{-rT} \mathbb{E}^\mathbb{Q}[(A_T-K) \mathbb{1}_{A_T > K}|\mathcal{F}_0]\\ &= e^{-rT}\left( \mathbb{E}^\mathbb{Q}[A_T \mathbb{1}_{A_T > K}|\mathcal{F}_0]- \mathbb{E}^{\mathbb{Q}}[K \mathbb{1}_{A_T > K}|\mathcal{F}_0]\right) \end{alignat*} \begin{split} \mathbb{E}^\mathbb{Q}[A_T\mathbb{1}_{A_T > K}|\mathcal{F}_0]&=\int_K^\infty A f(A) dA\\ &=\int_{K}^\infty A \frac{1}{A \sigma_n \sqrt{2 \pi T}}\exp\left\{-\frac{(\log(A) -\left(\log(S_0) + \left((r - 0.5\sigma^2)\frac{n+1}{2n}T \right) \right)^2}{2 \sigma_n^2T}\right\}dA \end{split} and \begin{split} \mathbb{E}^{\mathbb{Q}}[K \mathbb{1}_{A_T > K}|\mathcal{F}_0] & = K \mathbb{E}^\mathbb{Q}[\mathbb{1}_{A_T > K}|\mathcal{F_0}]\\ &= K\mathbb{Q}(A_T > K)\\ &= K\mathbb{Q}\left(\exp\left\{\log(S_0) + \frac{n+1}{2}(r-0.5\sigma^2) \Delta_t + \sum_{i=1}^n \sigma_i Z_{n-i+1}\right\} > K\right) \end{split} with $Z_i \sim \mathcal{N}(0,1)$ and $\sigma_i=\frac{i\sigma}{n}\sqrt{\Delta_t}$ ($\Delta_t$ being the time between fixing periods). I think I'm on the right track but it seems like the calculations will be tedious. Is there a simpler way to do this ? I have so far proven the equality in the last part and derived the mean and variance which appear in the pdf.


1 Answer 1


Hint (too long for a comment): If you write $$ S_t=S_0e^{rt+\sigma W_t-\frac{\sigma^2 t}{2}} $$ then $$ A_T=\left(\prod_{i=1}^nS_{t_i}\right)^\frac{1}{n}=S_0\exp\left(\frac{r}{n}\sum_{i=1}^nt_i+\frac{\sigma}{n}\sum_{i=1}^nW_{t_i}-\frac{\sigma^2}{2n}\sum_{i=1}^nt_i\right)\,. $$ This can be thought of a lognormal variable: $$ A_T=S_0e^{\alpha+\beta Y-\beta^2/2} $$ where $Y$ is standard normal,

$$ \alpha-\frac{\beta^2}{2}:=\frac{r}{n}\sum_{i=1}^nt_i-\frac{\sigma^2}{2n}\sum_{i=1}^nt_i $$ and $\beta^2$ is the variance of $$ \frac{\sigma}{n}\sum_{i=1}^nW_{t_i}\,. $$

  • $\begingroup$ So basically with this change of variables the expectation is simply the expectation of a lognormal ? $\endgroup$ Commented Nov 28, 2021 at 17:25
  • $\begingroup$ @SimonCello94 : that's what the expectation is - even if you don't change the variables. $\endgroup$
    – Kurt G.
    Commented Nov 29, 2021 at 9:57

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.