# Show that a zero-coupon bond discounted by a bond with mautrity $T$ is a martingale under the $T$-Forward measure

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?

• Please don’t delete useful content. – Bob Jansen May 31 '20 at 8:34
• @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. – R. Rayl May 31 '20 at 15:04
• If you found the solution, it would be nice if you can share it. – Bob Jansen May 31 '20 at 15:11
• 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. – Daneel Olivaw Dec 15 '20 at 21:53

Suppose that $$T. Arcording to Girsanov, we have $$\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.