1
$\begingroup$

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 $dQ(t,T+\delta)=...dQ$ ?

Improve this question
$\endgroup$
1
  • $\begingroup$ 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 ? $\endgroup$
    – glork
    Dec 13 '15 at 18:12
3
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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 $Q^{T + \delta}$ $\endgroup$
    – glork
    Dec 13 '15 at 19:17
  • $\begingroup$ There is no such thing as $B(T+\delta, T)$, unless you defined it differently. $\endgroup$
    – Gordon
    Dec 13 '15 at 19:28
  • $\begingroup$ 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 $B(t,T)/B(t,T+h)=E_{Q^{T+h}}[???/B(T+h,T+h)|F_{t}]$ $\endgroup$
    – glork
    Dec 13 '15 at 19:54
  • 1
    $\begingroup$ 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 $t \le T$. $\endgroup$
    – Gordon
    Dec 13 '15 at 19:58
  • $\begingroup$ all right that's what I was going to write ! Thanks ! $\endgroup$
    – glork
    Dec 13 '15 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.