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Suppose we have a geometric Brownian $S(t)$ which follows a lognormal process. Say $$ \begin{equation} dS_t = \mu S_t dt + \sigma S_tdW_t \end{equation} $$

My question is what is the distribution of $S(t+h)-S(t)$ where $h>0$?

I think this is a standard textbook question but I didn't find anything relevant to it yet. If it's duplicated question please refer me to the existed one. I'm working on it at the same time. Any help will be appreciated!

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    $\begingroup$ It'd be the difference of two log normal random variables, which isn't nearly as simple a concept as the ratio of two log normal random variables, which is again log normal. I have actually never seen the difference discussed in my studies, so I guess it's fairly uncommon in the Black Scholes framework at least. $\endgroup$
    – Slade
    Mar 16, 2019 at 3:20
  • $\begingroup$ It is going to be something very complicated and obscure, if indeed it is known at all. $\endgroup$
    – nbbo2
    Mar 16, 2019 at 3:37
  • $\begingroup$ I'd assume we are at t, right? $\endgroup$
    – James Spencer-Lavan
    Mar 16, 2019 at 7:10
  • $\begingroup$ Assuming we are at 0, intuitively it looks like a fat tailed lognormal shape. The movement of the stock from 0 to t effectively modifies the volatility in (t,t+h), almost like stoch vol. $\endgroup$
    – dm63
    Mar 16, 2019 at 9:08
  • $\begingroup$ @JamesSpencer-Lavan No, assume we're at t=0. $\endgroup$
    – Pandaaaaaaa
    Mar 17, 2019 at 20:05

1 Answer 1

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interesting question, as this problem is quite famous re stock prices I think. So I did some research on it and found this: https://stats.stackexchange.com/questions/238529/the-sum-of-independent-lognormal-random-variables-appears-lognormal

Seems to be very complex and the sum (and therefore the difference) of two lognormals is (generally) not lognormal. However in some cases it is approximated as lognormal.

Anyway, your question is more specific as S(t+h) and S(t) are supposed to be not independent, as it is the same stochastic process, but time shifted. So I guess, it goes more complex as it already is.

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – Pandaaaaaaa
    Apr 2, 2019 at 13:08

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