Zero-Coupon Bond Pricing and the T-Forward Measure

# zero coupon problem calculus


# zero coupon problem calculus












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.

## Answer by dm63 (score 1)

https://quant.stackexchange.com/a/22291

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

## Answer by Gordon (score 1)

https://quant.stackexchange.com/a/22293

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*}