Calculating the expected value and time-average of an SDE In the theory of real options, the following calculation is pretty standard.
Suppose an asset $R_t$ follows a GBM:
$$
 d R_t = R_t (\mu\, dt + \sigma\, dz)\ .
$$
We are interested in deriving a differential equation for a function $V(R)$ which represents the value of waiting to invest in said asset. The first thing ones does is apply Ito's lemma to obtain
$$
 d V = \left( \frac{1}{2} \sigma^2 R^2 V''+ \mu R V' \right) dt + \sigma R V' dz\ ,
$$
where the primes represent differentiation with respect to $R$. Then, from an arbitrage argument, one claims that the following must be satisfied:
$$
 \mathbb{E}[dV] = \rho V dt\ ,
$$
whence one derives a Cauchy-Euler equation for $V$.
We will just take the above condition on $\mathbb{E}[dV]$ as given. The crux of my question is the following: in derivations of the Cauchy-Euler equation, it is claimed that
$$
 \mathbb{E}[dV] = \frac{1}{2} \sigma^2 R^2 V''+ \mu R V'\ ,
$$
that is, the stochastic part of the Ito formula simply drops out. I would be grateful for a first-principles derivation of this claim. I can't even figure out how to proceed from the definition of the expected value
$$
 \mathbb{E}[X] = \int_\Omega X(\omega) dP(\omega)\ \implies ... ?
$$
I am willing to take the properties of the stochastic integral on faith; I would just like to know how to apply them here!
The reason that I would like to do so, and this is the second part of my question, is that I would like to be able to compute the time-average of $dV$, defined for a process $X$ as
$$
 \mathbb{T}[X] = \lim_{T \to \infty} \frac{1}{T} \int_0^T X(t) dt\ .
$$
I have a feeling that understanding how $\mathbb{E}[dV]$ is computed will help me understand how to proceed with $\mathbb{T}[X]$...but if anyone has any specific hints for this latter calculation, I would be more than grateful.
(I have a hunch that
$$
 \mathbb{T}[dV] = \frac{1}{2} \sigma^2 R^2 V''+ \left(\mu - \frac{\sigma^2}{2} \right) R V'\ ,
$$
but I do not know how to show this.)
 A: $E[dV]$ is not rigorous, so the argument is heuristic. The infinitesimal generator $Af$ is what you're referring to. For an Ito diffusion $dX_t=\mu(X_t)dt+\sigma(X_t)dW_t$, if $f$ is nice enough it can be proved that
$$Af(x):=\lim_{t \downarrow 0}\frac{E[f(X_t)|X_0=x]-f(x)}{t}=\mu(x)f_x(x)+\frac{1}{2}\sigma^2(x)f_{xx}(x)$$
For a reference see Oksendal.

The arbitrage argument you're looking for should go like the classic Black-Scholes one. Suppose we construct a self-financing portfolio $\Pi=V(X_t)-hX_t$ which is s.t.
$$d\Pi=dV-hdX$$
So by using Ito on $dV$
$$d\Pi=\bigg(\mu(X_t)(V_x(X_t)-h)+\frac{1}{2}\sigma(X_t)^2V_{xx}(X_t)\bigg)dt+\sigma(X_t)(V_x(X_t)-h)dW_t$$
We choose $h$ for perfect hedge (we make the random term vanish), so $h=V_x(X_t)$. By no arbitrage, our portfolio gains at the risk free rate $d\Pi=\rho \Pi dt$ so by simplifying
$$\rho V(X_t)dt=\bigg(\rho X_t V_x(X_t)+\frac{1}{2}\sigma(X_t)^2V_{xx}(X_t)\bigg)dt$$
This is where the heuristic $E[dV]=\rho Vdt$ is claimed (under the risk neutral measure we have $dX_t=\rho X_t dt + \sigma(X_t)dW_t$). So the Cauchy-Euler equation you get is
$$\rho V(x)=\rho x V_x(x)+\frac{1}{2}\sigma(x)^2V_{xx}(x)$$
By substituting $\sigma^2(x)=\sigma^2x^2$ you obtain the Black-Scholes case.
