counterparty risk - Parametrizing the Radon Nikodym

The following is an excerpt from the dialogue in the book "Counterparty Risk and Funding - A Tale of Two Puzzles", regarding the statistics such as the volatility and correlation estimation under the real-world probability measure $P$ and the risk-neutral probability measure $Q$.

My question is then: What does "Parametrizing the Radon Nikodym" mean? Any ideas or elaborations?

Additional information: Consider a portfolio of $n$-assets. If there is only a single real-world probability measure $P$, then by a Girsanov transformation, we can move from $P$ to $Q$ with a single Radon-Nikodym derivative. However, if we assume that there is a real-world probability measure $P_s$ for each individual asset $s$, then we have a Radon-Nikodym derivative corresponding to asset $s$. Does the phrase "Parametrizing the Radon Nikodym", in the above excerpt, mean the Radon-Nikodym derivative parametrized by assets?

Answer

@Gordon, here is how I would interpret this excerpt.

Assume that the discussion starts with $\bf{SALVA}$ explaining to $\bf{SAGE}$ that, in the absence of arbitrage opportunities, it is always possible to identify an equivalent martingale measure (EMM) $\mathbb{Q}$ under which prices of financial instruments can be written as expectations, with no concern for investors' risk-aversion. $\bf{SALVA}$ goes on by explaining the concept of self-financing strategies, no arbitrage etc. but $\bf{SAGE}$ interrupts... he wants something more pragmatic.

$\bf{SAGE}$. This seems like a beautiful theory. But assume that the underlying asset follows a geometric Brownian motion in the real-world $$\frac{dS_t}{S_t} = \mu dt + \sigma dW_{t}^\mathbb{P}$$ How does that translate to your so-called risk-neutral world?

$\bf{SALVA}$. Well you would appeal to the Girsanov theorem to find out. It goes like this. Define $$\lambda = \frac{\mu - r}{\sigma}$$ as the Sharpe ratio of your underlying asset, which you could interpret as a return risk premium. Your SDE can then be re-written as $$\frac{dS_t}{S_t} = r dt + \sigma d(W_{t}^\mathbb{P} + \lambda t)$$ now defining the risk-neutral measure $\mathbb{Q}$ through its Radon-Nikodym derivative $$\left. \frac{d\mathbb{P}}{d\mathbb{Q}} \right\vert_{\mathcal{F}_t} = \mathcal{E}[{\color{blue}{-\lambda W_t^{\mathbb{P}}}}]$$ Gisanov theorem states that the process $$W_t^{\mathbb{P}} - \langle W^{\mathbb{P}}, {\color{blue}{-\lambda W^{\mathbb{P}}}} \rangle_t = W_t^{\mathbb{P}} + \lambda t$$ is a $\mathbb{Q}$-Brownian motion so that we can write: $$\frac{dS_t}{S_t} = r dt + \sigma dW_{t}^\mathbb{Q}$$

$\bf{SAGE}$. Wait, isn't your Girsanov theorem saying that instantaneous vols and correlations for diffusions are the same under the two measures?

$\bf{SALVA}$. [looking at Sage carefully] As usual, you pretend to be a jerk but you are quite smart. Yes, you would be right. But it's not that simple. Parametrizing the Radon-Nikodym...

... wait, let me give you a practical example. Consider a diffusion model where volatility is allowed to be stochastic. Under the real-world this model for instance reads $$\begin{gather} \frac{dS_t}{S_t} = \mu_t dt + \sqrt{v_t} dW_t^\mathbb{P} \\ dv_t = \alpha dt + \beta dB_t^\mathbb{P} \tag{1} \\ d\langle W^\mathbb{P}, B^\mathbb{P} \rangle_t = \rho dt \end{gather}$$ The instantaneous variance of your underlying asset (actually that of its log-returns) is $v_t$ with $v_t$ given by $(1)$. Now let's see how that changes when we move to the risk-neutral measure. I'll cut the details but under $\mathbb{Q}$, the price of any self-financing strategy, discounted at the risk-free rate, should be a martingale. We shall assume that the underlying pays no dividend, so that investing in it is a self financing strategy...

$\bf{SAGE}$. You're about to lose me in your beautiful theories once again.

$\bf{SALVA}$. Wait, all this boils down to saying that you should be able to write $$ \frac{dS_t}{S_t} = r dt + \text{}\ dW_t^{\mathbb{Q}} $$ as we did in the GBM case.

$\bf{SAGE}$. I guess I could live with that.

$\bf{SALVA}$. Good. Well to do this, we once again appeal to Girsanov. Drawing our inspiration from what we did in the GBM case, let's define the Radon-Nikodym derivative as $$\left. \frac{d\mathbb{P}}{d\mathbb{Q}} \right\vert_{\mathcal{F}_t} = \mathcal{E}[{\color{blue}{-\lambda W_t^{\mathbb{P}} - \lambda_\perp W_t^{\mathbb{P},\perp}}}]$$ where $$\lambda := \frac{\mu_t-r}{\sqrt{v_t}}$$ and $\lambda_\perp$ is anything you want as long as $W_t^{\mathbb{P},\perp}$ is independent of $W_t^\mathbb{P}$.

Now take the equation for $S_t$. Given the definition of $\lambda$ we get something similar to the GBM case: $$\frac{dS_t}{S_t} = r dt + \sqrt{v_t} d(W_t^{\mathbb{P}} + \lambda t)$$ Now we use Girsanov and see that: $$W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} - \langle W^{\mathbb{P}}, {\color{blue}{-\lambda W^{\mathbb{P}} - \lambda_\perp W^{\mathbb{P},\perp}}} \rangle_t$$ using the bilinarity of quadratic variation and the independence of $W^\mathbb{P}_t$ and $W^\mathbb{P,\perp}_t$ we have $$W_t^{\mathbb{Q}} = W_t^{\mathbb{P}} + \lambda t$$ and that regardless of the value of $\lambda_\perp$. But, if we now look at the equation for the instantaneous variance: $$dv_t = \alpha dt + \beta dB_t^\mathbb{P}$$ Girsanov thoerem tells us that $$B_t^\mathbb{Q} = B_t^\mathbb{P} - \langle B^\mathbb{P}, {\color{blue}{-\lambda W^{\mathbb{P}} - \lambda_\perp W^{\mathbb{P},\perp}}} \rangle_t$$ and using the fact that (Cholesky decomposition) $$B_t^\mathbb{P} = \rho W_t^{\mathbb{P}} + \sqrt{1-\rho^2} W_t^{\mathbb{P},\perp}$$ along with the bilinearity property of quadratic variation we have that: $$\begin{align} B_t^\mathbb{Q} &= B_t^\mathbb{P} - \left\langle \rho W^{\mathbb{P}} + \sqrt{1-\rho^2} W^{\mathbb{P},\perp}, -\lambda W^{\mathbb{P}} - \lambda_\perp W^{\mathbb{P},\perp} \right\rangle_t \\ &= B_t^\mathbb{P} + (\lambda \rho + \lambda_\perp \sqrt{1-\rho^2}) t \end{align}$$ and at the end of the day: $$\begin{gather} \frac{dS_t}{S_t} = r dt + \sqrt{v_t} dW_t^\mathbb{Q} \\ dv_t = (\alpha - \beta(\lambda \rho + \lambda_\perp \sqrt{1-\rho^2}) )dt + \beta dB_t^\mathbb{Q} \tag{2} \\ d\langle W^\mathbb{Q}, B^\mathbb{Q} \rangle_t = \rho dt \end{gather}$$

$\bf{SAGE}$. I think I get it. When I look at the dynamics of the instantaneous variance under the real-world measure versus that under your risk-neutral measure (i.e. when I compare $(1)$ to $(2)$), the instantaneous volatility process is not the same anymore, contrary to the GBM case. And what you were starting to say with "parametrizing the Radon-Nikodym..." here transpires through the fact that the dynamics under $\mathbb{Q}$ really depends on the parameter $\lambda_\perp$, which we have defined as an arbitrary constant: the dynamics under $\mathbb{Q}$ is somewhat defined "up to an arbitrary parameter".

$\bf{SALVA}$. [looking at SAGE carefully again]. You are definitely anything but a jerk. Indeed... and you would have reached the exact same conclusion should you have used a dynamic replication argument rather than the martingale approach + Girsanov. The financial rationale is that, because there exists no unique way of hedging the stochasticity of volatility (remember that the stochasticity of the underlying price, can be addressed by taking positions in the underlying itself, since it trades), you face an indetermination: the ideal hedging strategy depends on you... the same kind of indetermination as with what value of $\lambda_\perp$ to pick when parametrising the Radon-Nikodym derivative. We say that the market model is incomplete: there are less marketed securities than there are independent sources of risk (here the represented by the independent Brownian motions $W^{\mathbb{P}}$ and $W^{\mathbb{P},\perp}$. Now the tricky question is how to choose a particular measure $\mathbb{Q}$ out of this set of admissible measures?

$\bf{SAGE}$. Let's go and get a coffee. I'm tired.