1
$\begingroup$

I may be confused. I am looking at the risk neutral vs. physical measures. We know that knowing the short interest rate stochastic process $r$, a bond maturing at time $T$ can be considered as a derivative with payoff \$$1$. Its price at time $t$ is $$P(t) = \mathbf E^Q\big[e^{-\int_t^Tr}\big|\mathcal F_t\big],$$ in the risk neutral measure $Q$. Now if we want to compute the time $t$ value $V(t)$ of a stock at time $T$, should we compute the expectation in the risk neutral measure $Q$ $$V(t)=\mathbf E^Q\big[e^{-\int_t^Tr}S(T)\big|\mathcal F_t\big]$$ or the physical measure $P$ $$V(t)=\mathbf E^P\big[e^{-\int_t^Tr}S(T)\big|\mathcal F_t\big]$$ ?

$\endgroup$
  • 1
    $\begingroup$ If investing in the stock is a self-financing strategy (no dividends, no additional lending costs/opportunities) then yes the equation you write under $\Bbb{Q}$ should be fine. If not, you need to consider the proces where dividends are reinvested (similar for lending costs). But this is a bit pointless though IMO because most models start of the stock dynamics under $\Bbb{Q}$, so you'll just find something trivial by computing the RHS like $S(t)=S(t)$. $\endgroup$ – Quantuple Aug 27 '18 at 7:38
  • 1
    $\begingroup$ The physical probability measure is supposed to give the probability of events that correspond to reality. The risk neutral measure is a hacked, tilted probability measure constructed so that taking expectations (and discounting at the risk free rate) give you prices. If you want prices, use the risk neutral measure. If you want physical expectations, use the physical measure. $\endgroup$ – Matthew Gunn Aug 27 '18 at 14:09
  • $\begingroup$ @Quantuple: You are right. I was indeed confused. A trivial replicating or hedge argument indeed gives the price of the stock at $t$ as $S(t)$. That can also be seen as the $0$ strike call on $S$. $\endgroup$ – Hans Aug 27 '18 at 19:34
  • $\begingroup$ @MatthewGunn: I understand what purpose the risk neutral measure is serving. You are of course right, together with Quantuple. $\endgroup$ – Hans Aug 27 '18 at 19:36
1
$\begingroup$

The equivalent martingale measure (EMM) $\mathbb{Q}$ is a measure under which all the asset prices discounted using a risk-free bond are martingales, i.e. given the bond price $B(t)$ and the asset price $S(t)$ we have $$\frac{S(t)}{B(t)} = E_t^{\mathbb{Q}}\left[\frac{S(T)}{B(T)}\right]$$ If the bond follows the following ODE $dB(t) = r(t)B(t)dt$ then $\frac{B(t)}{B(T)} = e^{-\int_t^T r(s)ds}$ and therefore we get the equation you have suggested under $\mathbb{Q}$

$$ S(t) = E_t^{\mathbb{Q}}\left[e^{-\int_t^T r(s)ds}S(T)\right]$$

If you want to price under the physical measure $\mathbb{P}$, then you know there exists a state price density $\xi(t)$ that (assuming only one asset) evolves according to the following process $d\xi(t) = -\xi(t)\left(r(t)dt + \frac{\mu(t)-r(t)}{\sigma(t)}dW(t)\right)$ and whose product with any asset is a $\mathbb{P}$ martingale, i.e.

$$ \xi(t)S(t) = E_t^{\mathbb{P}}\left[\xi(T)S(T)\right]$$

implying the following pricing equation under the physical measure:

$$ S(t) = E_t^{\mathbb{P}}\left[\frac{\xi(T)}{\xi(t)}S(T)\right]$$

$\endgroup$
  • $\begingroup$ You presume "the bond follows the following ODE $dB(t) = r(t)B(t)dt$". My question assume the short rate $r$ is stochastic, which implies the bond is stochastic as well and $dB(t) = r(t)B(t)dt +\sigma\frac{\partial B}{\partial r}dW^Q$ where $W^Q$ is the Brownian motion under the risk neutral measure $Q$ and $\sigma$ is the instantaneous volatility of the short rate $r$. Does your answer still stand? $\endgroup$ – Hans Aug 27 '18 at 15:48
  • $\begingroup$ I am saying you need not restrict your argument to the bond process following an ODE rather than a general SDE. $\endgroup$ – Hans Aug 27 '18 at 19:38
  • $\begingroup$ That still works and you can use your bond as a numeraire as long as it is default-free. In that case it is not true anymore that $\frac{B(T)}{B(t)} = e^{-\int_t^T r(s)ds}$ but you can still use the pricing equation $$S(t) = E_t^{\mathbb{Q}}\left[\frac{B(T)}{B(t)} S(T)\right]$$ The issue here is whether $\frac{B(T)}{B(t)} S(T)$ is easy to integrate or not. Anyway, $\frac{B(T)}{B(t)}$ is the price of a ZCB with $T-t$ maturity and have zero covariance with $S(T)$ unless $S(T)$ is another fixed income security. $\endgroup$ – fni Aug 27 '18 at 23:45
  • $\begingroup$ And the reason why I mentioned the zero covariance is because $$S(t) = E^{\mathbb{Q}}_t\left[\frac{B(t)}{B(T)}S(T)\right] = E^{\mathbb{Q}}_t\left[\frac{B(t)}{B(T)}\right]E^{\mathbb{Q}}_t\left[S(T)\right] + Cov^{\mathbb{Q}}_t\left(\frac{B(t)}{B(T)},S(T)\right)$$ If the covariance is zero, we have that $$S_t = \text{Price ZCB}(T-t)\times E^{\mathbb{Q}}_t\left[S(T)\right]$$ $\endgroup$ – fni Aug 27 '18 at 23:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.