# Continuous Geometric Asian Options

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.

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 $$\ln(S_t) =\ln(S_0)+\left(r-q-\frac{1}{2}\sigma^2\right)t+ \sigma W_t$$.

Thus,

\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.

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