Put-Call Parity for Continuous Geometric Asian Options
Summary
The document derives put-call parity for continuously monitored geometric-average call and put options under Black–Scholes dynamics. Since the difference between their terminal payoffs equals the geometric average minus the strike, the price difference is the discounted risk-neutral expectation of that amount. The remaining task is to find the expected geometric average.
Under risk-neutral dynamics, the average of the log stock price is normally distributed because it combines deterministic terms with the time integral of Brownian motion. The integral has a known normal distribution, yielding a lognormal geometric average. Its expectation follows from the mean and variance of the log average, giving an explicit route to the parity value. The derivation assumes the stated continuous averaging and model dynamics; the source’s initial setup specifies no dividends, while its calculation includes a dividend yield parameter.
Key ideas
- The call-minus-put payoff for a geometric Asian option is the geometric average minus the strike.
- Put-call parity follows by discounting the risk-neutral expectation of that payoff difference.
- The time average of log prices is normally distributed under the stated Black–Scholes dynamics.
- The geometric average is therefore lognormal, and its expectation can be computed from the log distribution’s mean and variance.
- The derivation relies on continuous averaging and the model assumptions used for the risk-neutral stock process.
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Full text
# Continuous Geometric Asian Options
# 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.
## Answer by Kevin (score 3, accepted)
https://quant.stackexchange.com/a/49501
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.