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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.

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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 $$ \frac{\sigma}{n}\sum_{i=1}^nW_{t_i}\,. $$

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  • $\begingroup$ So basically with this change of variables the expectation is simply the expectation of a lognormal ? $\endgroup$ 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.
    Nov 29, 2021 at 9:57

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