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Suppose $T$ the maturity of a risky bond which defaults with probability $p$ over its lifetime. If it defaults it pays zero. Thus to price this bond in risk neutral terms would give

$$P=\mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)}(1-p)\right].$$

If such bond and its price would be observable in the market we could estimate $p$, that is, the implied probability of default. It is however known from literature that this $p$ is usually an overestimation of the real probability of default, see for example Hull, White and Predescu (http://www-2.rotman.utoronto.ca/~hull/DownloadablePublications/CreditSpreads.pdf). This difference can be understood as the additional risk premium for for example default risk and systematic risk. But in this same paper it is furthermore stated that without risk premiums the implied default probability would coincide with the real default probability. This sounds plausible, but how would I go ahead and show that statement to be true? I was thinking about something like, using for example the above bond price in the real world measure:

$$P=\mathbb{E}^{\mathbb{P}}\left[e^{-\mu(T-t)}(1-p^*)-s_{premiums}\right],$$

where $s_{premiums}$ the additional price of risk premiums and $\mathbb{P}$ the real world measure, $p^*$ the real world probability of default and $\mu$ the correct discount rate. However if $s_{premiums}=0$ then $p^*=p$ only if $r=\mu$. Am i missing something or is this the wrong approach, what would you suggest?

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  • $\begingroup$ How could you show it? $\endgroup$ – SmallChess Jan 26 '15 at 12:04
  • $\begingroup$ @StudentT See my answer :) Does that make sense? $\endgroup$ – Nathan Meibergen Jan 26 '15 at 13:05
  • $\begingroup$ I am afraid I do not fully understand your question. Yes, of course implied and future realized expectations are identical in the absence of risk premiums. They are also identical if the market priced future expectations correctly which is rarely if ever the case. What is your real question? $\endgroup$ – Matthias Wolf Jan 26 '15 at 13:07
  • $\begingroup$ He was obviously studying the paper and simply stated his understanding about the risk premiums. His question was on how to actually show it, not just talk about it. $\endgroup$ – SmallChess Jan 26 '15 at 13:12
  • $\begingroup$ @MattWolf, please check my answer below. $\endgroup$ – Nathan Meibergen Jan 26 '15 at 13:18
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Suppose the above bond and let $\tau$ be the default time, then by definition of the Q-measure

\begin{align} P&=\mathbb{E}^{\mathbb{Q}}\left[e^{-r(T-t)}1_{\tau>T}\right]\\ &=e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}\left[1_{\tau>T}\right]\\ &=e^{-r(T-t)}\mathbb{Q}(\tau>T). \end{align}

To evaluate a fair price in the real-world measure we first have to note that receiving this bond is not risk free. Thus, to compensate the investor with several risk factors he requires an additional premium of $s_{premiums}$. Note these not te be related to risk-aversion if we assume the market to be liquid, as then prices converge to a fair price. However, hedging away risks will cost money. Furthermore, there are transaction costs and even systemic risk for which we cannot hedge, all of these components require an additional premium. Suppose this premium is received at maturity $T$, then the terminal payoff is risk free in the real world, such that

\begin{align} P&=\mathbb{E}^{\mathbb{P}}\left[e^{-r(T-t)}\left(1_{\tau>T}+s_{premiums}\right)\right]\\ &=e^{-r(T-t)}\mathbb{E}^{\mathbb{P}}\left[1_{\tau>T}+e^{-r(T-t)}s_{premiums}\right]\\ &=e^{-r(T-t)}\mathbb{P}(\tau>T)+e^{-r(T-t)}\mathbb{E}^{\mathbb{P}}\left[s_{premiums}\right]. \end{align}

The result now follows from

$$ \mathbb{Q}(\tau>T)=\mathbb{P}(\tau>T)+\mathbb{E}^{\mathbb{P}}\left[s_{premiums}\right], $$

such that

$$ \mathbb{Q}(\tau>T)>\mathbb{P}(\tau>T).$$

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