# Bond price under the risk-neutral measure

Could you point out where I am making mistake in the process below?

It follows from the term structure equation and the Feynman-Kac theorem that the bond price is given by

$$p(t,T) = E_t^Q\left[ \exp\left( -\int_t^T r(u) du \right) \right],$$

where $$E_t^Q$$ denotes the expectation at time $$t$$ under the risk neutral measure $$Q$$.

Let the money market account be

$$B(t) = \exp\left( \int_0^t r(u) du \right),$$

and the bond price expression above is written as

$$p(t,T) = E_t^Q\left[\frac{B(t)}{B(T)} \right].$$

Since the numeraire of $$Q$$ is $$B$$, it follows from the martingale property that

$$E_t^Q\left[ \frac{B(t)}{B(T)} \right] = E_t^Q\left[ \frac{B(t)}{B(t)} \right] = 1.$$

Thus, $$p(t,T)=1$$.

No, $$B$$ is not a $$Q$$ martingale, neither is $$1/B$$, which you have assumed in your calculation (try using constant $$r$$, for an example of why this can be the case). The measure $$Q$$ is a risk neutral measure if the stock price processes that are discounted by $$B$$ are martingales.
• Thank you so much. Now I know from your answer that $\frac{B(t)}{B(T)}$ is not a martingale under $Q$. – Inawashiro Hiromichi Dec 26 '20 at 0:45
I think you have your argument slightly mixed up. Assuming the existence of a risk-neutral measure $$Q$$, you know from the risk-neutral pricing formula that the discounted bond price is a martingale, i.e. ($$D(t)$$ being the discount factor) $$D(t)p(t,T) = E_t^Q[D(T)p(T,T)] = E_t^Q[D(T)].$$ With $$D(t) = \mathrm{exp}(-\int_0^t r(u) du)$$, we immediately get $$p(t,T) = \frac{1}{D(t)}E_t^Q[D(T)] = E_t^Q[-\int_t^T r(u) du].$$ Therefore, the equation for the bond price includes already the numeraire. Using it again wouldn't make sense. Also, without further information about $$r(t)$$, it is not possible to derive an exact solution for the bond price.