Distribution of a Discrete Geometric Average Relative to the Terminal Price
Summary
The document derives the distribution of the logarithm of a discrete geometric average of stock prices divided by the terminal stock price under a geometric Brownian motion model. Taking logarithms turns the ratio into a linear combination of Brownian motion values and a deterministic drift term. Because that combination is normal, the log ratio is normally distributed.
The mean follows from the drift and the monitoring times; the variance is the variance of the Brownian combination scaled by volatility squared. The answer gives the covariance identity for Brownian motion, allowing the variance to be computed from the monitoring schedule. It does not carry out that final computation or simplify the result for equally spaced observations, so readers must apply the covariance identity themselves. The setup also uses a specific geometric Brownian motion model and discrete monitoring, so the conclusion need not hold unchanged under other price dynamics or sampling schemes.
Key ideas
- Taking logs expresses the average-to-terminal-price ratio as a linear combination of Brownian motion values.
- Under geometric Brownian motion, the log ratio is normally distributed.
- Its mean depends on the drift and the observation times.
- Its variance can be computed using the Brownian covariance identity for each pair of monitoring times.
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# Distribution of discrete Geometric average and Stock Price
# Distribution of discrete Geometric average and Stock Price
If we have $$S_t = S_0 e^{(r-\frac{1}{2} \sigma ^2) +\sigma W_t}$$ and a discrete geometric average of stock prices $$G_n = (\prod_{i=1}^{n} S_{t_i})^{\frac{1}{n}} $$ where the monitoring points are spaced equally as $t_i=i \frac{T}{N}$, where $0 \leq i \leq N$, i.e. $0 \leq t_i \leq T$
How can I find the distribution of $ln(\frac{G_n}{S_T})$, i.e. finding expected value and variance? I have tried rewriting $G_n$ using natural logs so that: $$ln(G_n) = \frac{1}{n} \left ( ln(S_{t_1})+ln(S_{t_2})+ln(S_{t_3})+...+ln(S_{t_N}) \right )$$ but I am unsure of how to combine it with $S_T$.
Note that since I am new to much of this notation there might be inconsistencies so please fix them.
## Answer by NN2 (score 1, accepted)
https://quant.stackexchange.com/a/75379
We have $$\begin{align} \ln \left(\frac{G_T}{S_T} \right)&= \frac{1}{n}\sum_{i=1}^n\ln(S_{t_i})-\ln(S_T)\\ &=\frac{1}{n}\sum_{i=1}^n \left(\ln(S_{t_i})-\ln(S_T)\right)\\ &=-\frac{1}{n}\sum_{i=1}^n \left((r-\frac{1}{2}\sigma^2)(T-t_i)+\sigma(W_T-W_{t_i})\right)\\ &=(r-\frac{1}{2}\sigma^2)\left(-T+\frac{1}{n}\sum_{i=1}^n t_i\right)+\sigma \left( -W_T+\frac{1}{n}\sum_{i=1}^n W_{t_i}\right)\\ &=\underbrace{(r-\frac{1}{2}\sigma^2)\left(-T+\frac{1}{n}\sum_{i=1}^n t_i\right)}_{=:\color{red}{\mu}}+\sigma \underbrace{\left( -W_T+\frac{1}{n}\sum_{i=1}^n W_{t_i}\right)}_{=:\mathcal{L}}\\ \end{align}$$
The term $\mathcal{L}$ follows the normal distribution $\mathcal{N}(0,\color{red}{\Sigma})$ as it is the sum of the $(W_{t_i})_{i=1,...,n}$. To determine $\Sigma$, it suffices to compute $$\begin{align} \Sigma &=\mathbb{E}\left(\left( -W_T+\frac{1}{n}\sum_{i=1}^n W_{t_i}\right)^2\right)\\ &=\mathbb{E}\left(W_T^2\right) -2\frac{1}{n}\mathbb{E}\left(W_T \sum_{i=1}^n W_{t_i}\right)+\frac{1}{n^2}\mathbb{E}\left(\left(\sum_{i=1}^n W_{t_i}\right)^2\right) \tag{1} \end{align}$$ I let you compute $\Sigma$ by using the property $\mathbb{E}(W_xW_y) = \min\{x,y \}$ (it is not difficult)
Finally, $$\ln \left(\frac{G_T}{S_T} \right) \sim \mathcal{N}(\color{red}{\mu}, \color{red}{\Sigma})$$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.