= Solution
The terminal density is strictly positive and has <expectation> one under the usual deterministic initial bond-price convention. Its density process is
$$
Z_t=\mathbb E^{\mathbb Q}\left[\frac{D_T}{P(0,T)}\,\middle|\,\mathcal F_t\right]
=\frac{D_tP(t,T)}{P(0,T)}.
$$
For $0\leq s\leq t\leq T$, the <Bayes formula for conditional expectation> under a change of measure gives
$$
\begin{aligned}
\mathbb E^{\mathbb Q_T}\left[\frac{B_t}{P(t,T)}\,\middle|\,\mathcal F_s\right]
&=\frac1{Z_s}\mathbb E^{\mathbb Q}\left[Z_t\frac{B_t}{P(t,T)}\,\middle|\,\mathcal F_s\right]\\
&=\frac1{Z_sP(0,T)}=\frac{B_s}{P(s,T)}.
\end{aligned}
$$
The same calculation at $s=0$ gives the finite <expectation> $1/P(0,T)$, so this is a true <martingale>, not just a formal conditional identity. Thus \b[the <continuous-time bank account> measured in units of the maturity-$T$ bond is a $\mathbb Q_T$-<martingale>]. This is the <forward measure> change of <numéraire>. If the initial bond price were random rather than given, <integrability> of its reciprocal would need to be included for this true-martingale assertion.
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