# If S(t) is geometric Brownian motion, what is the distribution of S(t+h)-S(t)?

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

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!

• 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. – Slade Mar 16 '19 at 3:20
• It is going to be something very complicated and obscure, if indeed it is known at all. – noob2 Mar 16 '19 at 3:37
• I'd assume we are at t, right? – James Spencer-Lavan Mar 16 '19 at 7:10
• 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. – dm63 Mar 16 '19 at 9:08
• @JamesSpencer-Lavan No, assume we're at t=0. – Pandaaaaaaa Mar 17 '19 at 20:05