# Girsanov theorem and default rates in bond credit rating

Default rates are kind of probabilities, right?

Is it possible to use the Girsanov theorem in that context? For example if we have a table of real world probabilities, could we use the Girsanov theorem to convert those real world probabilities into risk-neutral probabilities? Maybe even forward probabilities?

• Just bootstrap hazard rates from the bonds: orders of magnitude more granular, dynamic, and by definition, risk neutral. – Mehness Dec 2 '16 at 15:12

## 1 Answer

Suppose I give you objective probabilities $\mathbb{P}(S_T \geq K)$ of an equity finishing above a certain level $K$ at a future time $T$ (or in your case a survival probability in the form of default rates). Can you convert these to risk-neutral probabilities $\mathbb{Q}(S_T \geq K)$ ? Not immediately.

First, I need to give you a model for the behaviour of $S_T$, or equivalently to specify the dynamics of the process $(S_t)_{t\in[0,T]}$ under the filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\in[0,T]}, \mathbb{P})$.

Then, simply noticing that $$\mathbb{Q}(S_T \geq K) = \mathbb{E}^\mathbb{Q} [\mathbb{1}(S_T \geq K)] = \mathbb{E}^\mathbb{P} \left[\mathbb{1}(S_T \geq K) Z_T \right]$$ with $$Z_T = \left. \frac{d\mathbb{Q}}{d\mathbb{P}} \right\vert_{\mathcal{F}_T}$$ the Radon-Nikodym derivative of the measure change should do the trick (with a tractability depending on your modelling assumptions of course). Obviously, you will appeal to Girsanov to translate the dynamics I provided you under $\mathbb{P}$ under the equivalent measure $\mathbb{Q}$.

So to answer your question: 'No, it is not possible', at least not without setting up a proper modelling framework. In addition, the conversion will be one to one if and only if you assume a complete market model + no arbitrage (see fundamental theorems of asset pricing).

Also notice that usually in the markets we do the opposite. We imply probabilities from quoted instruments and these are therefore obtained under $\mathbb{Q}$. The question is then, can we translate them under $\mathbb{P}$. The reasoning is the exact symmetric of the above.