I am trying to solve this problem

Consider the following one-dim. stochastic process
$$dX_t = b_t dt + \sigma_t dW_t$$ where $W$ is a one-dim. Brownian motion. The above SDE is well-defined. Consider a smooth and bounded function $g$, and put $$ Z_t := \exp(\int_0^t g(s,X_t)ds).$$ Calculate the stochastic differential $dZ$.

My answer:
Put $Y_t = \int_0^t g(s,X_t)ds$ . Then, it follow that $Z_t=e^{Y_t}$ , and from Ito formula, I have
$$dZ_t = Z_t(dY_t + \frac{1}{2}(dY_t)^2).$$ Thus, I want to know the stochastic differential $dY$.
If I can say that $$dY_t=g(t,X_t)dt$$ then, $$dZ_t = Z_t \bigl(g(t,X_t)dt + \frac{1}{2}(g(t,X_t)dt)^2 \bigl)$$ $$\Leftrightarrow dZ_t = Z_t g(t,X_t)dt .$$ follows.

My question: I am not sure if I can say that $$dY_t=g(t,X_t)dt.$$ I suspect that my answer is too simple to be true. Where did I make a mistake?

  • $\begingroup$ I am also not so sure you can say $dY_t=g(t,X_t)dt.$. In order to get Y differential I would define $f(t,x) =\int_0^t g(s,x)ds$ and then compute $f(t,X_t)$ through Ito Lemma. $\endgroup$
    – loxol
    May 25, 2019 at 14:23
  • $\begingroup$ @loxol I also guess that I need Ito formula to calculate $ dY $. <br> However, I have no idea how I can compute $ f(t,x) = \int_0^t g(s,x)ds $ through Ito formula. What is $ \frac{\partial}{\partial t} \int_0^t g(s,x)ds $ and $ \frac{\partial}{\partial x} \int_0^t g(s,x)ds $? $\endgroup$
    – David Khan
    May 26, 2019 at 6:05

1 Answer 1


In your notation, $$ dY_t = g(t,X_t) dt + \int_0^t dg(s,X_t) ds $$ where $$ dg(s, X_t) = \partial_{X_t} g(s,X_t) dX_t + \frac{1}{2} \partial^2_{X_t} g(s,X_t) (dX_t)^2 $$ The rest seems ok.

  • $\begingroup$ Can you elaborate why the first equality holds? Does it use Ito lemmma as loxol pointed out? Seemingly, the second equation used Ito lemma. But why isn't partial derivative w.r.t. $t$ included? $X_t$ is a function of $t$, isn't it? $\endgroup$
    – David Khan
    May 26, 2019 at 6:13
  • $\begingroup$ Well, if you write $X_t$ in the integrand, and not $X_s$, then if you differentiate with respect to $t$ you will also need to differentiate $X_t$. It's like normal calculus, except since $X_t$ is stochastic you take the Ito differential. $\endgroup$
    – user34971
    May 26, 2019 at 8:07
  • $\begingroup$ I understand it, thanks. Let me ask one more thing on calculation. If I put $dX_t = b_tdt + \sigma_t W_t$ into $dg(s,X_t)$, then I have something like $\int_0^t (\partial_{X_t}g(s,X_t)b_tdt) dt $ and so on. Can I say that this is equivalent to $b_tdt\int_0^t (\partial_{X_t}g(s,X_t)) dt$? If this is true, then by Ito rule $\bigl((dt)^2=0\bigl)$, I hope I could get relatively neat result on $dZ$. $\endgroup$
    – David Khan
    May 26, 2019 at 9:44
  • $\begingroup$ You don't have $(dt)^2$ because the integral is over $ds$ from $0$ to $t$, and not over $dt$. $\endgroup$
    – user34971
    May 26, 2019 at 11:55
  • $\begingroup$ Sorry for my abbreviation. I'm not sure if my intention is understood. What I wanted to say is this: $dY_t = g(t,X_t)dt + \int_0^t(\partial_{X_t}g(s,X_t) dX_t + \frac{1}{2} \partial^2_{X_t} g(s,X_t)(dX_t)^2)ds$ = $...+\int_0^t(\partial_{X_t} g(s,X_t)(dX_t)ds+... $= $...+\int_0^t(\partial_{X_t} g(s,X_t) (b_tdt+\sigma_t dW_t)ds+...$= $...+\int_0^t\partial_{X_t} g(s,X_t) b_tdtds+... $ holds. And, $(dY_t)^2=...+(\int_0^t\partial_{X_t} g(s,X_t) b_tdtds)^2+... $ follows. $\endgroup$
    – David Khan
    May 27, 2019 at 9:32

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