# Differential of integral of a stochastic process

Let $Y_{t}$ be

$$Y_{t}=\int_{\Omega} g(X_{u}) du$$

where $g(.)$ is a deterministic function and $\Omega=[t_{0},t]$ continuos partition of $\mathbb{R}$. Furthermore let $X$ be an Ito process $$X_{u}= X_{0}+\int_{0}^{u}\mu(s)ds+\int_{0}^{u} \sigma(s) dW_{s}^{\mathbb{P}}$$ 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 $Y_{t}$?

$$dY_{t}=?$$

• Where is the t in the rhs of your equation ? – MJ73550 Oct 9 '16 at 8:09
• @Bob I can relatively easy derive the expression for $g(X_{t})=X_{t}$ namely$$dY_{t}=-X_{t}dt+ \int_{\Omega} dX_{u} du$$. This is done by assuming that the mapping of probability is a continuous function and then using classic calculus (derivative under integral) – Lost in Oct 9 '16 at 11:22
• @MJ73550 $t \in \Omega$ in other words $t\in [t_{0}, t_{1}]$ or if your prefer the integral can be written as $$\int_{t_{0}}^{t_{1}} g(X(u)) du$$ where $u$ is a dummy. – Lost in Oct 9 '16 at 11:28
• @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 $g(u,X_u)$? – Quantuple Oct 9 '16 at 13:05
• 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 $dY_t =0$... – 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.