Assume the risk-free bond $B_t$ and the stock $S_t$ follow the dynamics of the Black & Scholes model without dividends (with interest rate r, stock drift $\mu$ and volatility $\sigma$). Let $c(t; St;Gt;K)$ and $p(t; St;Gt;K)$ be the prices at time t of the (continuous) Geometric Asian call option and put option with strike $K$. Find a put-call parity relation for Geometric Asian options. In other terms, and an explicit expression for $c(t; St;Gt;K)-p(t; St;Gt;K)$.

So far, this is what I have: $G_T=\exp\{\frac{1}{T}\int_{0}^{T}\log S_udu\}\\ X_T=\frac{1}{T}\int_{0}^{T}\log S_udu\\ G_T=e^{X_T}$

Payoff functions are: $c_{fix}=(G_T-K)^+=(e^{X_T}-K)^+\\ p_{fix}=(K-G_T)^+=(K-e^{X_T})^+\\ c_{fix}-p_{fix}=G_{T}-K$

By risk neutral evaluation: $c_{fix}-p_{fix}=e^{-r(T-t)}E^{Q}[e^{X_T}-K]$.

Hoping to understand how to compute this without the standard normal variable.


1 Answer 1


Whether arithmetic or geometric averaging, you always get \begin{align*} \mathrm{AsianCall} - \mathrm{AsianPut} = e^{-rT} (\mathbb{E}[\bar{S}]-K). \end{align*}

So, let’s compute the expectation. You know that $\bar{S}=\exp\left( \frac{1}{T} \int_0^T \ln(S_t)\mathrm{d}t \right)$ where $\ln(S_t) =\ln(S_0)+\left(r-q-\frac{1}{2}\sigma^2\right)t+ \sigma W_t$.


\begin{align*} \ln(\bar{S}) &= \frac{1}{T} \int_0^T \ln(S_t)\mathrm{d}t \\ &= \frac{1}{T} \int_0^T \left( \ln(S_0)+\left(r-q-\frac{1}{2}\sigma^2\right)t + \sigma W_t \right) \mathrm{d}t \\ &= \frac{1}{T}\left( \ln(S_0)T + \frac{1}{2}\left(r-q-\frac{1}{2}\sigma^2\right)T^2+\sigma\sqrt{\frac{1}{3}T^3}Z \right) \\ &= \ln(S_0) + \frac{1}{2}\left(r-q-\frac{1}{2}\sigma^2\right)T+\frac{\sqrt{3}}{3}\sigma\sqrt{T}Z, \end{align*} using that $\int_0^T W_t\mathrm{d}t\sim N\left(0,\frac{1}{3}T^3\right)$ as shown here.

Consequently, \begin{align*} \ln(\bar{S}) \sim N\left( \ln(S_0) + \frac{1}{2}\left(r-q-\frac{1}{2}\sigma^2\right)T, \frac{1}{3}\sigma^2 T\right). \end{align*}

Hence, $\bar{S}$ is log-normally distributed and $\mathbb{E}[\bar{S}]=e^{m+\frac{1}{2}s^2}$, where $m$ and $s$ are the mean and standard deviation of $\ln(\bar{S})$ as computed above.

  • $\begingroup$ so the final answer would then be $e^{m+\frac{1}{2}s^2}-Ke^{-rT}$ right? $\endgroup$
    – Anon
    Commented Nov 3, 2019 at 12:38
  • 1
    $\begingroup$ @Anon It would be $e^{-rT}(e^{m+\frac{1}{2}s^2}-K)=e^{m+\frac{1}{2}s^2-rT}-Ke^{-rT}$, where $m=\ln(S_0) + \frac{1}{2}\left(r-q-\frac{1}{2}\sigma^2\right)T$ and $s^2=\frac{1}{3}\sigma^2 T$ $\endgroup$
    – Kevin
    Commented Nov 3, 2019 at 12:40
  • $\begingroup$ oh right, forgot to multiply $e^{-rT}$ to the first term! thank you! $\endgroup$
    – Anon
    Commented Nov 3, 2019 at 12:41
  • $\begingroup$ @Anon Happens to the best of us :) $\endgroup$
    – Kevin
    Commented Nov 3, 2019 at 12:42
  • $\begingroup$ Um, just have another question, this wouldn't lead to an answer with N(d) right? $\endgroup$
    – Anon
    Commented Nov 4, 2019 at 8:18

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