Prove that the first derivative of expectation is increasing Let us define
$$
X_t = x e ^{\mu t+\sigma B_t}, \quad x>0, \ t\in[0, T]
$$
where $\mu, \ \sigma$ are some constant values and $B_t$ is the standard Brownian motion.
I want to show that $v'(x)$ is increasing in $x$ parameter, for 
$$
v(x) = \mathbb{E}\left[e^{-\int_0^T \omega(X_u)du} g(X_T)\right],
$$
where $\omega$ and $g$ are continuous.
I calculated first derivative of the above equation:
\begin{equation}
\begin{aligned}
\frac{\partial}{\partial x}v(x) &= \frac{\partial}{\partial x} \mathbb{E}\left[e^{-\int_0^T \omega(X_u)du} g(X_T)\right] \\ &= 
\mathbb{E}\left[\frac{\partial}{\partial x} \left(e^{-\int_0^T \omega(xe^{\mu u+\sigma B_u})du} g(xe^{\mu T+\sigma B_T})\right)\right] \\ &= 
\mathbb{E}\left[e^{-\int_0^T \omega(xe^{\mu u+\sigma B_u})du} \left(-\int_0^T \omega'(xe^{\mu u+\sigma B_u})e^{\mu u+\sigma B_u} du \ g(xe^{\mu T+\sigma B_T}) \\ + g'(xe^{\mu T + \sigma B_T}) e^{\mu T + \sigma B_T} \right)\right].
\end{aligned}
\end{equation}
Based on this form I conclude that if $g$ is convex ($g'$ is increasing) and decreasing and $\omega$ is concave ($\omega'$ is decreasing) and decreasing, then $v'(x)$ is increasing. 
I want to ask is it correct or do we need some additional assumptions to prove that $v'(x)$ is increasing?
 A: I think we need some extra conditions on the functions $g$ and $\omega$ (e.g. $\mathcal{C}^1_b$) in order to justify the derivation under the integral sign.
In order to show this, let us simplify the problem by supposing $\omega\equiv 0$. We denote $X_t^x = xe^{\mu t+\sigma B_t}$ and assume $\sigma \geq 0$ and $x>0$. Hence, the problem is reduced to
\begin{equation}
v(x) = \mathbb{E}\left[g(X^x_T)\right]
\end{equation}
We remark that
\begin{equation}
\forall t \geq 0, \quad \frac{\partial X_t^x}{\partial x} = e^{\mu t+\sigma B_t} \quad a.s. \tag{1}
\end{equation} and
\begin{align}
\forall t \geq 0, \quad\frac{\partial g}{\partial x}(X_t^x) &=  g'(X_t^x)e^{\mu t+\sigma B_t} \quad a.s.
\end{align}
Plus, we have the domination of the derivative :
\begin{align}
\forall t \geq 0, \quad\big|\frac{\partial g}{\partial x}(X_t^x)|  \ &=|g'(X_t^x)|e^{\mu t+\sigma B_t} \\
&\leq \sup|g'(X_t^x)|e^{\mu t+\sigma B_t} \in L_1
\end{align}
Then by the Lebesgue theorem, 
\begin{align}
\frac{\partial v}{\partial x}(x) &= \mathbb{E}\left[\frac{\partial g}{\partial x}(X_T^x)\right] \\
&= \mathbb{E}\left[g'(X_T^x)e^{\mu T+\sigma B_T}\right] \\
&= \int g'(xe^{\mu T+\sigma \sqrt{T}u})e^{\mu T+ \sigma\sqrt{T} u}\phi(u)du\\
&= \frac{1}{x\sigma \sqrt{T}}\left[[g(xe^{\mu T+ \sigma\sqrt{T} u})\phi(u)]^{+\infty}_{-\infty} + \int g\left (xe^{\mu T+ \sigma\sqrt{T} u}\right)u \phi(u)du \right] \quad \text{Integration by parts} \tag{2}\\
&= \frac{1}{x\sigma T}\mathbb{E}\left[g(X_T^x)W_T\right]
\end{align}
If you suppose that $g$ is positive, then you can deduce than $v'$ is positive. Some remarks:


*

*$\phi$ is the p.d.f of a standard Gaussian variable.

*The derivative of $X_.^x$ w.r.t $x$ is called the tangent process, i.e. $Y_. = \frac{\partial X_.^x}{\partial x}$.

*In $(2)$, we have derived and integrate w.r.t $u$. In a sense, we have "derived" without explicitly saying the Brownian motion. This is properly introduced in the Malliavin Calculus.

*Here we supposed that $g\in \mathcal{C}^1_b$. This can be relaxed by only assuming polynomial growth.

*The results that you are looking for are used in option pricing. (We are basically computing the sensitivity of model parameters to a price function). For further readings, you can look at this paper.

