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I want to show that the assumption that stock prices follow geometric Brownian motion is consistent with the weak efficient market hypothesis. I first want to show that the stock price defined by the following SDE is a martingale. \begin{equation} d S_t=\mu S_tdt+\sigma S_td B_t\label{jhblyd} \end{equation} By ITO's formula, \begin{equation*} d\ln {S_t} = \left( {\frac{1}{{{S_t}}}\mu {S_t} + \frac{{ - 1}}{2}\frac{1}{{{S_t}^2}}{\sigma ^2}{S_t}^2} \right)dt + \frac{1}{{{S_t}}}\sigma {S_t}d{B_t} = \left( {\mu - \frac{1}{2}{\sigma ^2}} \right)dt + \sigma d{B_t} %\label{dlnst} \end{equation*} Integrate with respect to $t$, \begin{equation*} \begin{aligned} \ln {S_t} &= \ln S_0+\int_{0}^t\left(\mu-\frac{1}{2}\sigma^2\right)ds+\int_0^t\sigma dB_s\\ & = \ln S_0+\left(\mu-\frac{1}{2}\sigma^2\right)t+\sigma B_t. \end{aligned} \end{equation*} Then, \begin{equation} S_t=S_0e^{(\mu-\frac{1}{2}\sigma^2)t+\sigma B_t}. \end{equation} $S_t$ is a martingale if and only if $\mu=0$.

So when the drift rate is not equal to zero, is that a violation of the weak efficient market hypothesis?

Yifei Zhang
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  • These formulas are better known than the weak efficient market hypothesis. What exactly is that? – Kurt G. Mar 30 '23 at 03:38
  • The weak efficient market hypothesis refers to the fact that the price of a stock has fully incorporated the information contained in the historical share price. Under this assumption, stock prices have martingale properties.@KurtG. – Yifei Zhang Mar 30 '23 at 03:42
  • If the converse does not hold the answer to your question is then obviously no. – Kurt G. Mar 30 '23 at 03:46
  • Yes, seen from my work, the stock price sequence is not a martingale (unless the drift rate is zero), but the textbook says it is a martingale, such as this problem https://math.stackexchange.com/q/1737119/907140. Puzzled. – Yifei Zhang Mar 30 '23 at 04:04
  • If you just want to know if it is a martingale (or more typically the discounted stock price) we neither need a textbook nor do we need a weak efficient market hypothesis. All we need to know is what filtration applies. That of the BM obviously. And correct. It is not a martingale for every $\mu.$ – Kurt G. Mar 30 '23 at 04:10
  • Could you please further explain the sentence that "filtration applies to BM"? What is "BM" stand for? Thanks for your patience. – Yifei Zhang Mar 30 '23 at 04:22
  • BM means the Brownian motion. I read a bit the other post and about EMH in the mean time. EMH deals with empirical research of real stock prices. In contrast, your equations are only the simplest stock price model of them all. This model is as you know a martingale if and only if $\mu=0$. That's all. A model cannot violate a statement about real markets. – Kurt G. Mar 30 '23 at 04:36
  • After your reminder, I realize that my previous measure of testing EMH was too simple. I'll take a closer look. Thank you very much. – Yifei Zhang Mar 30 '23 at 05:34

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