Is Libor a martingale under T-forward measure

Question

We denote discount factor $D(t)$, and zero coupon bond $B(t,T)$ as: $$B(t,T) =\dfrac{1}{D(t)} E_t[D(T)]$$ here $E_t[X] = E[X|\mathcal{F}(t)].$

And we define

Zero curve $Z(t,T)$ $$B(t, T)\cdot e^{Z(t,T)(T-t)} = 1$$

Libor $L(t,T)$ $$B(t, T)\cdot (1 + (T-t) L(t, T)) = 1.$$ Denote $E^{T}[\ ]$ the $T$-forward measure i.e use $B(t,T)$ as numeraire.

I remember that Libor and zero curve should be martingale under the $T$-forward measure. But we see the representations, equivalently $\dfrac{1}{B(t,T)}$ should be martingale under the $T$-forward. This is the $T$-forward price of $1,$ but discounted value of $1$ is not martingale under original measure i.e $$E_t[D(T)\cdot 1] \neq D(t)\cdot1.$$ We can see that forward rate $$B(t,T-\delta) = (1 + (T-t)F(t,T-\delta,T))B(t,T)$$ is really $T$-forward martingale, since $\frac{B(t,T-\delta)}{B(t,T)}$ is $T$-forward martingale, equivalently $D(t)B(t,T-\delta)$ is martingale under the original measure. So, I really confuse here. Can anyone tell where is the mistake?

Answer 1

Your definition of Libor is invalid as you make it cover the period $t, T$.

A Libor with tenor $\delta$ that fixes on $T$ (or to be accurate usually 2 days before $T$) covers the period $T, T+\delta$. Thus the forward Libor rate $L(t, T, T+\delta)$ is computed as $$ B(t, T+\delta) (1 + \delta L(t, T, T+\delta)) = B(t, T) $$ hence $$ L(t, T, T+\delta) = \frac{1}{\delta}\left(\frac{B(t, T) }{B(t, T+\delta)} -1 \right) $$ is a martingale under the $T+\delta$ forward measure.