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

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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.

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  • $\begingroup$ Thank you so much. Now I know from your answer that $\frac{B(t)}{B(T)}$ is not a martingale under $Q$. $\endgroup$ – Inawashiro Hiromichi Dec 26 '20 at 0:45
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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.

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