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Suppose that x is the yield to maturity with continuous compounding on a discount bond that pays off $1 at time T. Assume that the x follows the process

$dx=a(x_0-x)dt + sxdz$

where $a, x_0$ and $s$ are positive constants and $dz$ is the wiener process. What is the process followed by a bond price?

Solution: $dS=\mu Sdt+\sigma S dz$

where S is the bond price and $\mu$ and $\sigma$ are expected instantaneous return and instantaneous volatility respectively. Yield to maturity is the total return anticipated on a bond if the bond is held until the end of its lifetime.

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  • $\begingroup$ Can you please say me the definition of the yield to maturity. I forgot it. $\endgroup$ – user16651 Dec 25 '16 at 15:56
  • $\begingroup$ Oh , you didn't understand what I said. What is relation between $S_t$ and $x_t$ $\endgroup$ – user16651 Dec 25 '16 at 16:10
  • $\begingroup$ $S_t(x_t,T)$ where T is time to bond expiry. $\endgroup$ – Dhamnekar Winod Dec 25 '16 at 16:30
  • $\begingroup$ The bond price depend on $r_t$ and $t$ and $T$. Indeed $S=S(t,r;T)$ $\endgroup$ – user16651 Dec 25 '16 at 16:35
  • $\begingroup$ You seem to be correct. $\endgroup$ – Dhamnekar Winod Dec 25 '16 at 16:39
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If I understand your question correctly, then the zero-coupon bond price for the maturity $T$ is given by

\begin{equation} B_t = e^{-x_t (T - t)}, \end{equation}

where $t \in [0, T]$ and $x_t$ is the per annum yield-to-maturity. Note that you didn't make the definition of $x$ fully clear in your question.

To get the dynamics of $B$, you just apply the Ito formula to the function $f(t, x) = e^{-x (T - t)}$ with

\begin{equation} f_t(t, x) = x f(t, x), \quad f_x(t, x) = -(T - t) f(t, x), \quad f_{xx}(t, x) = (T - t)^2 f(t, x). \end{equation}

Then

\begin{eqnarray} \mathrm{d}B_t & = & \mathrm{d}f \left( t, x_t \right)\\ & = & x_t B_t \mathrm{d}t - (T - t) B_t \mathrm{d}x_t + \frac{1}{2} (T - t)^2 B_t \mathrm{d} \langle x \rangle_t\\ & = & \left( x_t + \frac{1}{2} (T - t)^2 s^2 x_t^2 \right) B_t \mathrm{d}t - (T - t) B_t \mathrm{d}x_t\\ & = & \underbrace{\left( x_t + \frac{1}{2} (T - t)^2 s^2 x_t^2 - (T - t) a \left( x_0 - x_t \right) \right)}_{\mu \left( t, x_t \right)} B_t \mathrm{d}t + \underbrace{(t - T) s x_t}_{\sigma \left( t, x_t \right)} \mathrm{B}_t \mathrm{d}z. \end{eqnarray}

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  • $\begingroup$ How would you interpret the last step? $\endgroup$ – Dhamnekar Winod Dec 27 '16 at 14:42
  • $\begingroup$ I don't understand what you mean by "interpret". I simply replace $\mathrm{d}x_t$ by the corresponding expression. $\endgroup$ – LocalVolatility Dec 27 '16 at 14:49

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