libor rate - local martingale

I am a newbie for Libor rates and all these questions...

Let be : $L(t,\delta)$ the Libor rate and $L_{t}(T,\delta)$ the forward Libor rate. Let's define : $Lb(T,\delta):=1+\delta L(T,\delta)=1/B(T,T+\delta)$ and $Lb_{t}(T,\delta):=1+\delta L_{t}(T,\delta)=B(t,T)/B(t,T+\delta)$. The question is to prove that under the forward measure of maturity $T+\delta$ that both $Lb_{t}(T,\delta)$ and $L_{t}(T,\delta)$ are local martingales. I began to define the forward measure of maturity $T+\delta$ (under which the numeraire is $B(T,T+\delta)$ ) :$Q^{T+\delta}$ but it's a lot of calculus. So how can we solve this ? Do we have to start from the model $dB(t,T)/B(t,T)=r_{t}dt+\Gamma (t,T)dW_{t}$ in order to define $dQ(t,T+\delta)=...dQ$ ?

• actually if I write : $Lbis_{t}(T,\delta)=B(t,T)/B(t,T+\delta)=E_{Q^{T+\delta}}[B(T+ \delta,T)|F_{t}]=E_{Q^{T+ \delta}}[B(t,T)/B(T+\delta,T+\delta)|F_{t}]$ I think that answers my question isn't it ? – glork Dec 13 '15 at 18:12

By definition, under the $T+\delta$-forward measure, the price of any tradable asset relative to the bond price $B(t, T+\delta)$ is a martingale. Since \begin{align*} Lb_t(T, \delta) = \frac{B(t, T)}{B(t, T+\delta)}, \end{align*} it is a martingale by definition. Regarding $L_t(T, \delta)$, since \begin{align*} L_t(T, \delta) = \frac{1}{\delta}\left(\frac{B(t, T)}{B(t, T+\delta)} -1 \right) \end{align*} is a linear combination of martingales, it is also a martingale.
• many thanks ! Does my solution work if i write that : $B(t,T)/B(t,T+ \delta):=E_{Q^{T+ \delta}}[B(T+ \delta,T)|F_{t}]=E_{Q^{T+ \delta}}[B(T+ \delta,T)/B(T+ \delta,T+ \delta)|F_{t}]$ (the first equality is for me the definition for $Q^{T + \delta}$ – glork Dec 13 '15 at 19:17
• There is no such thing as $B(T+\delta, T)$, unless you defined it differently. – Gordon Dec 13 '15 at 19:28
• True I wanted to make a paralell with $F_{t}(S,T):=S_{t}/B(t,T)=E_{Q^{T}}[S_{T}/B(T,T)|F_{t}]$ so that we have something like $B(t,T)/B(t,T+h)=E_{Q^{T+h}}[???/B(T+h,T+h)|F_{t}]$ – glork Dec 13 '15 at 19:54
• You can have $B(t, T)/B(t, T+h) = E_{Q^{T+h}}(B(T, T)/B(T, T+h) \mid \mathscr{F}_t)$, for $t \le T$. – Gordon Dec 13 '15 at 19:58