Calculating the expected value and time-average of an SDE

Question

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.)

Answer 1

$E[dV]$ 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.

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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.