Let $Y_{t}$ be

\begin{equation} Y_{t}=\int_{\Omega} g(X_{u}) du \end{equation}

where $g(.)$ is a deterministic function and $\Omega=[t_{0},t]$ continuos partition of $\mathbb{R}$. Furthermore let $X$ be an Ito process \begin{equation} X_{u}= X_{0}+\int_{0}^{u}\mu(s)ds+\int_{0}^{u} \sigma(s) dW_{s}^{\mathbb{P}} \end{equation} for som well behaved $\mu$ and $\sigma$ and $(W_{s}^{\mathbb{P}})_{0\leq s}$ is standard brownian motion under objective probability measure $\mathbb{P}$.

What is differential of $Y_{t}$?

\begin{equation} dY_{t}=? \end{equation}

  • $\begingroup$ Where is the t in the rhs of your equation ? $\endgroup$ – MJ73550 Oct 9 '16 at 8:09
  • $\begingroup$ @Bob I can relatively easy derive the expression for $g(X_{t})=X_{t}$ namely\begin{equation} dY_{t}=-X_{t}dt+ \int_{\Omega} dX_{u} du\end{equation}. This is done by assuming that the mapping of probability is a continuous function and then using classic calculus (derivative under integral) $\endgroup$ – Lost in Oct 9 '16 at 11:22
  • $\begingroup$ @MJ73550 $t \in \Omega$ in other words $t\in [t_{0}, t_{1}]$ or if your prefer the integral can be written as \begin{equation} \int_{t_{0}}^{t_{1}} g(X(u)) du\end{equation} where $u$ is a dummy. $\endgroup$ – Lost in Oct 9 '16 at 11:28
  • 1
    $\begingroup$ @Lost in your certainly mean $\Omega=[t_0,t]$ otherwise there is no point on writing the LHS as $Y(t)$ right? Now is the integrand $g(X_u)$ or $g(u,X_u)$? $\endgroup$ – Quantuple Oct 9 '16 at 13:05
  • 1
    $\begingroup$ in your answer, there is still no dependency on $t$ on your $\int_{t_0}^{t_1}g(X_u)du$, so for the moment $dY_t =0$... $\endgroup$ – MJ73550 Oct 10 '16 at 9:57

Under some probability space $(\Omega,\mathcal{F},\Bbb{P})$ equipped with the (augmentation of the) natural filtration ${\bf{F}}=(\mathcal{F}_t)_{t \geq 0}$ of a $\mathbb{P}$-Wiener process $(W_t)_{t\geq 0}$, consider the Itô process $$ X_t = X_0 + \int_0^t \mu(s) ds + \int_0^t \sigma(s) dW_s \tag{1} $$

for some sufficiently well-behaved functions $\mu$ and $\sigma$, such that the stochastic integration can be defined in the Itô sense.

Define the integral $$Y_t = \int_0^t X_u du $$

From $(1)$ it follows that \begin{align} Y_t &= \int_0^t \left( X_0 + \int_0^u \mu(s) ds + \int_0^u \sigma(s) dW_s \right) du \\ &= X_0 t + \int_0^t \int_0^u \mu(s) ds du + \int_0^t \int_0^u \sigma(s) dW_s du \end{align} Using (stochastic) Fubini theorem one can permute the integration order and write \begin{align} Y_t &= X_0 t + \int_0^t \int_s^t \mu(s) du ds + \int_0^t \int_s^t \sigma(s) du dW_s \\ &= X_0 t + \int_0^t (t-s) \mu(s) ds + \int_0^t (t-s) \sigma(s) dW_s \\ &= \left(X_0 + \int_0^t \mu(s) ds + \int_0^t \sigma(s) dW_s\right) t - \int_0^t s \mu(s) ds - \int_0^t s \sigma(s) dW_s \\ &= X_t t - \underbrace{\int_0^t s \mu(s) ds}_{\text{classic integral}} - \underbrace{\int_0^t s \sigma(s) dW_s}_{\text{Itô integral}} \\ \end{align} And one can now appeal to the usual "differential" definition (whether from standard calculus or Itô calculus) to write: \begin{align} dY_t &= \underbrace{X_t dt + t dX_t + 0}_{d(X_t t)\,\,\,\text{Itô's lemma}} - t \mu(t) dt -t \sigma(t) dW_t \\ &= X_t dt + t dX_t - t \underbrace{(\mu(t) dt + \sigma(t) dW_t)}_{dX_t} \\ &= X_t dt \end{align} Now as mentioned in the comments, because any smooth function $g(X_t)$ will also be an Itô process, you can repeat the reasoning with $\tilde{X}_t := g(X_t)$ to get, for your particular problem, $$ dY_t = \tilde{X}_t dt = g(X_t) dt $$

[Remark] Should $X_u = X(u) \to X(t,u)$ with an additional, explicit dependence on $t$ things can get more complicated. See this related question on math SE.

[Edit] Just saw that this was discussed here as well.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.