1
$\begingroup$

I encounter a problem: do we have the following equality : $B(0,T_{i})e^{\int_{0}^{t}r_{s}ds}=B(t,T_{i})$ and if yes why because I am stuck with this ... I try to use that : $B(t,T_{i}) = B(0,T_{i})e^{\int_{0}^{t}r_{s}ds-0.5\int_{0}^{t}vol^{2}ds+\int_{0}^{t}voldW_{s}}$ Thanks !

Question revised:

Actually my question is the following : the forward price at t is defined by :$F_{t,T}=Price(t)/B(t,T)$. Let be $Price(t)=E_{Q}[e^{-\int_{t}^{T}r_{s}ds}Payoff|F_{t}]$. By using the forward measure defined by : $dQ^{T}=(e^{-\int_{0}^{T}r_{s}ds}/B(0,T) )dQ$ we have that : $Price(t)=E_{Q^{T}}[e^{-\int_{t}^{T}r_{s}ds}Payoff*B(0,T)/e^{-\int_{0}^{T}r_{s}ds}|F_{t}]$. I need to prove that this last equality is $E_{Q^{T}}[Payoff|F_{t}]*B(t,T)$ so that $Price(t)/B(t,T)=E_{QT}[Payoff|F_{t}]$ which is a "formula" I know.

$\endgroup$
  • $\begingroup$ Please define $r_s$. Is this the riskfree rate? $\endgroup$ – emcor Dec 12 '15 at 17:57
  • $\begingroup$ Sorry. Yes it is. $\endgroup$ – glork Dec 13 '15 at 17:31
1
$\begingroup$

The first equation looks correct if B(t,T(i)) is a forward price of an instrument paying no dividends. The second equation looks correct if B(t,T(i)) is a random variable being projected by a lognormal process

$\endgroup$
  • $\begingroup$ Actually, $B(t,T)$ is the price of a zero coupon with maturity T $\endgroup$ – glork Dec 12 '15 at 18:14
  • $\begingroup$ But how can we show that with the calculus I began $\endgroup$ – glork Dec 12 '15 at 18:15
1
$\begingroup$

Let $r_t$ be the interest rate. Then \begin{align*} B(t, T_i) &= E\Big[e^{\big(-\int_t^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big]\\ &= e^{\int_0^t r_s ds} E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big]. \end{align*} Note that, for $t>0$, unless $r_t$ is deterministic, \begin{align*} E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big] &\ne E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)}\Big]\\ &= B(0, T_i), \end{align*} as $$E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)} \mid \mathscr{F}_t\Big]$$ is an $\mathscr{F}_t$ measurable random variable, while $$E\Big[e^{\big(-\int_0^{T_i} r_s ds\big)}\Big]$$ is a constant.

In conclusion, the identity $B(0,T_{i})e^{\int_{0}^{t}r_{s}ds}=B(t,T_{i})$ is incorrect, unless the interest rate is deterministic.

Addition based on the revision.

Consider the payoff $Payoff_T$ at time $T$. Then the value at time $t$, where $0 \le t \le T$, under the risk-neutral probability measure $Q$, is given by \begin{align*} Price(t) = E_Q\left(e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right). \end{align*} Let $Q_T$ be the $T$-forward probability measure. Then, \begin{align*} \eta_t &\equiv \frac{dQ}{dQ_T}\big|_{\mathscr{F}_t}\\ &=\frac{e^{\int_0^t r_s ds} B(0, T)}{B(t, T)}. \end{align*} Using the abstract Bayes formula, \begin{align*} E_Q\left(e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right) &= E_{Q_T}\left(\frac{\eta_T}{\eta_t}e^{-\int_t^T r_s ds} Payoff_T \mid \mathscr{F}_t\right)\\ &=B(t, T) E_{Q_T}\left( Payoff_T \mid \mathscr{F}_t\right). \end{align*} That is, \begin{align*} E_{Q_T}\left( Payoff_T \mid \mathscr{F}_t\right) = \frac{Price(t)}{B(t, T)}. \end{align*}

$\endgroup$

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.