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I see that Libor $L(t,S,T)$ is a martingale under $T-$forward measure. Where we used argument that zero-coupon bonds are martingales under $T$-forward measure, as zero-coupon bond is a traded security.

We also have arguments that "money account discounted no divident, tradable security is martingale under risk neutral measure". I wonder whether "libor" can be a tradable security.

I find, using $B(T)$ the numeraire of risk neutral measure $\mathbb{Q}$, $P(t,T)$ the zero-coupon bond as numeraire of $T$-forward measure, then money account discounted libor:

$V(t) = B(t)\mathbb{E}^\mathbb{Q}[\frac{L(t,S,T)}{B(T)}]=B(t)\mathbb{E}^\mathbb{T}[\frac{L(t,S,T)}{B(T)}\frac{d\mathbb{Q}}{d\mathbb{T}}]= B(t)\mathbb{E}^\mathbb{T}[\frac{L(t,S,T)}{B(T)}\frac{B(T)P(t,T)}{P(T,T)B(t)} ] = \mathbb{E}^\mathbb{T}[L(t,S,T)P(t,T)]$

Then by the definition of libor $L(t,S,T) = \frac{1}{\delta}(\frac{P(t,S)}{P(t,T)}-1)$, plug in to the last term above, we get

$V(t) = \frac{1}{\delta}\mathbb{E}^{T}[P(t,S)-P(t,T)] =\frac{1}{\delta}\mathbb{E}^{T}[\frac{P(t,S)-P(t,T)}{P(T,T)}] = \frac{1}{\delta} \frac{P(t,S) - P(t,T)}{P(t,T)} $

Where I put $P(T,T)$ in the denominator as we know it's value is 1.

then we got $V(t) = \frac{L(t,S,T)}{B(t)} = \mathbb{E}^\mathbb{Q}[\frac{L(t,S,T)}{B(T)}]$ ? I feel there are some issue somewhere and would like to know how people explain it...

So what's the reason that libor cannot be treated as a tradable asset ? When can we add $P(T,T)=1$ in the denominator ?

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