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Here's the exact question:

Show that for any $s>0$, $\frac{P(t,s)}{P(t,T)}$ is a $Q^T$-martingale.

Here's my attempt:

Let $t^\prime < t$. First consider the case $s>T$. \begin{aligned} \mathbb{E}_{Q^T}\Big[\frac{P(t,s)}{P(t,T)} \lvert \mathcal{F}_{t^\prime}\Big] &= \mathbb{E}_{Q^T}\Big[P(T,s) \lvert \mathcal{F}_{t^\prime}\Big] \\ &= \mathbb{E}_{Q^T}\Big[\frac{P(t^\prime,s)}{P(t^\prime,T)} \lvert \mathcal{F}_{t^\prime}\Big] \\ &= \frac{P(t^\prime,s)}{P(t^\prime,T)} \end{aligned} And then you can use a similar argument for when $T > s$. But this argument has to be wrong surely, as this is not specific to $Q_T$. Could anyone help and point out where I've gone wrong?

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  • $\begingroup$ Please don’t delete useful content. $\endgroup$ – Bob Jansen May 31 '20 at 8:34
  • $\begingroup$ @BobJansen The question has been up for 2 months without a good answer. The answer below completely misses the point of the question and isn't of any use. I fail to see what's useful about this content - especially since the proof can be found elsewhere. $\endgroup$ – R. Rayl May 31 '20 at 15:04
  • $\begingroup$ If you found the solution, it would be nice if you can share it. $\endgroup$ – Bob Jansen May 31 '20 at 15:11
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    $\begingroup$ By definition, the ratio of zero-coupon bonds must a martingale under $Q^T$ because $P(\cdot,T)$ is the numéraire of that measure, that is any traded asset divided by $P(\cdot,T)$ must be a martingale. Any interest rate model needs to be specified such that this relationship holds. $\endgroup$ – Daneel Olivaw Dec 15 '20 at 21:53
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Suppose that $T<S$. Arcording to Girsanov, we have $$\frac{dQ^T}{dQ^S}|_{F_t'}= \frac{P(T,T)/P(t',S)}{P(T,S)/P(t',S)} =\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)}$$ So $$dQ^T|_{F_t'} =\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)} dQ^S|_{F_t'}$$ $$E_{Q^T} (\frac{P(t,S)}{P(t,T)}|F_{t'}) =E_{Q^S} (\frac{P(t,S)}{P(t,T)} *\frac{1}{P(T,S)}\frac{P(t',S)}{P(t',T)}|F_{t'}) =\frac{P(t',S)}{P(t',T)} E_{Q^S} (\frac{P(t,S)}{P(t,T)} *\frac{1}{P(T,S)}|F_{t'}) = \frac{P(t',S)}{P(t',T)}$$ We use similar demonstration for the case $T>S$. We can conclude that $\frac{P(t,S)}{P(t,T)}$ is a $Q^T$ martingale.

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    $\begingroup$ I thought this was immediate, because the ratio of two traded assets is a martingale under the probability measure that is risk neutral with respect to the numeraire (asset in the denominator). Right? $\endgroup$ – dm63 Mar 21 '20 at 0:59

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