Original Black-Scholes paper assumptions -- "variance rate" - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2019-08-20T00:44:39Z https://quant.stackexchange.com/feeds/question/37374 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://quant.stackexchange.com/q/37374 1 Original Black-Scholes paper assumptions -- "variance rate" user217285 https://quant.stackexchange.com/users/30998 2017-12-14T22:56:56Z 2017-12-17T05:06:49Z <p>In the 5th page of Black and Scholes' original paper on option pricing formulas, they write the following assumption:</p> <blockquote> <p>b) The stock price follows a random walk in continuous time with a variance rate proportional to the square of the stock price. Thus the distribution of possible stock prices at the end of any finite interval is log- normal. The variance rate of the return on the stock is constant.</p> </blockquote> <p>I'm a bit confused about the first sentence and how it relates to the GBM assumption $$dS_t = S_t[ \mu dt + \sigma dW_t],$$ especially since the last sentence says that the "variance rate of return" (which I assume is $\sigma^2$) is constant over time. In particular, I don't see where $S_t^2$ comes into play.</p> <p>Also, I read somewhere that Merton cleaned up the derivation and put it in more mathematically rigorous language. Where can I find that paper?</p> <p>--</p> <p>Answer: if $dX_t = \mu(t,X_t,\dots) dt + \sigma(t,X_t, \dots) dW_t$, then $\mu$ is called the "drift rate" and $\sigma^2$ is called the variance rate. The connection to quadratic variation is that $d\langle X \rangle_t = \sigma(t,X_t,\dots)^2 dt$, i.e. the drift rate of the quadratic variation of $X_t$ is in fact the variance rate of $X_t$.</p>