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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?

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

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