# Time integral of geometric brownian motion

Suppose $S_t$ is a geometric brownian motion. Then how to understand its time integral, i.e., $Y_t=\int_0^{t}S_udu$?

Is $Y_t$ still a stochastic process?

How to compute the expectation of $Y_t$?

How to compute the expectation of $\left[a-Y_t\right]^+$?

Here $[x]^+=\max\{x,0\}$.

• Search for "Asian option" in the site's search bar, there is plenty of Q&A addressing these issues. – Daneel Olivaw Oct 30 '17 at 14:39
• Daneel Olivaw is right. That being said, yes $Y_t$ is a stochastic process. The expectation operator being linear computing the expectation should be straightforward (just swap integral and expectation operators). For the non-linear payout you won't have a closed form result because the distribution of $Y_t$ is not analytically tractable, there are other options though (PDE, Monte Carlo). – Quantuple Oct 30 '17 at 16:07
• The two are right. If you look at my answer to your original question (I suppose the two are connected) then you I really think you need the Asian option for Levy processes. – Ric Oct 30 '17 at 18:17
• Thnks. Could you elabrate more why Asian option for Levy processes? – Zhiyuan Wang Oct 31 '17 at 0:55