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

$$dr(t)=\mu(t,r(t))dt+\sigma(t,r(t))d\bar{W}(t)$$

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

$$Z(t)=\frac{\Pi(t)}{B(t)}$$

is a $Q$-martingale.?

I appreciate any help.

Thanks.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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