If an interest rate model with the following $P$-dynamics for the short rate.


Now consider a $T$-claim of the form $\chi = \Phi(r(T))$ with corresponding price process $Π(t)$.

Can anyone help me to find stochastic differential of $Π(t)$ ?

and show that the normalized price process


is a $Q$-martingale.?

I appreciate any help.


  • $\begingroup$ My teacher told me that stochastic differential of $Π(t)$ is of the form $dΠ(t) = r(t)Π (t) dt + σ_Π Π(t)dW(t)$ but I don't know how to show that . $\endgroup$ – Roozbe Jan 1 '15 at 16:17
  • 1
    $\begingroup$ Your equation is simply the risk neutral dynamics. Do you know how to change a measure to the risk-neutral? $\endgroup$ – SmallChess Jan 2 '15 at 0:01
  • $\begingroup$ Hi @StudentT , Yes I know how to change a measure to the risk-neutral. thanks for your help. $\endgroup$ – Roozbe Jan 2 '15 at 2:07

You Know that $dB_t=r_tB(t)dt$ . Ito's formula give us \begin{align} dZ(t)=\frac{1}{B(t)}d\,\Pi(t)-\frac{\Pi(t)}{B\,^2(t)}dB(t)+0 \end{align} As your teacher mentioned, $d\Pi(t)=r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)$,Thus we have \begin{align} & dZ(t)=\frac{1}{B(t)}[r(t)\Pi(t)dt+\sigma(\Pi(t),t)dW(t)]-\frac{\Pi(t)}{B\,^2(t)}r(t)B(t)dt\\ & dZ(t)=\frac{1}{B(t)}r(t)\Pi(t)dt+\frac{1}{B(t)}\sigma(\Pi(t),t)dW(t)-\frac{1}{B(t)}r(t)\Pi(t)dt\\ \end{align} then \begin{align} dZ(t)=\frac{1}{B(t)}\sigma(\Pi(t),t)dW(t) \end{align} Martingale Representation Theorem shows that $Z(t)$ is a Martingale.


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.