Could someone please suggest with detailed steps and/or a reference,

1) How to convert the below discrete time summation to continuous time form and write it as an integral?

2) Any methods to solve it?

$$ \sum_{t=0}^{T}\left[\left(K-X_{t}\right)\right]\left(Y_{t}-Y_{t+1}\right) $$

Here, $K$ is a constant. $X$ is a geometric brownian motion. $Y$ is another geometric brownian motion.

Please let me know if anything is not clear.

Posted initially on Math Forum with no response; hence posting here. Please let me know if I should delete the posting on the other forum.


  • $\begingroup$ you mean at the limit ? $\endgroup$ Sep 2, 2016 at 13:15
  • $\begingroup$ @MJ73550 Yes, as I understand it, that is one way ... If there are other methods, please do let know. $\endgroup$
    – texmex
    Sep 2, 2016 at 13:24

1 Answer 1


By definition

let $t^n_k=\frac{k}{n}T$, $$I^n = \sum_{k=0}^{n-1} (K-X_{t_k})(Y_{t_k}-Y_{t_{k+1}}) \to_{n\to\infty} I = \int_0^T (X_t-K)dY_t $$

  • $\begingroup$ Thanks for your answer. How would this change, in terms of notation and how to express it, if we just had $Y_{t}$ instead of $Y_{t}-Y_{t+1}$? Any ways to solve this? Also, how about introducing time discounting into this? Please let me know if anything is not clear and if I should modify the question and add these additional questions as sub questions ... $\endgroup$
    – texmex
    Sep 4, 2016 at 2:14
  • $\begingroup$ with just $Y_t$ it wont converge anymore, what do you mean by "solving this" ? $\endgroup$ Sep 5, 2016 at 7:49
  • $\begingroup$ Could you please elaborate more on the lack of convergence? I meant solving as in solving the integral to get a simpler expression if possible. Also, how would we write it in this case? Would this be correct? $$\int_0^T (X(t)-K)Y(t)dt $$ $\endgroup$
    – texmex
    Sep 11, 2016 at 10:54

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