0
$\begingroup$

Question:

Vasicek interest rate model: $$dr_t = α(θ−r_t)dt + σdW_t$$

Price at time t of a 0-coupon bond maturing at T is given by: $$dp(t,T) = α_{t,T} . p(t,T)d_t + β_{t,T} . p(t,T)dW_t$$ $$βt,T = −σB(t,T).$$

T-forward price

$$ F_t = \frac{P_{t,S}}{P_{t,T}} $$

Show that the dynamics of the T-forward $F_t$ price with respect to the T-forward measure in the Vasicek model is given by

$$dF_t = σ (B(t,T)−B(t,S))F_tdW_t$$

Solution provided: $$ d(\frac{P_{t,S}}{P_{t,T}}) = \frac{dP_{t,S}}{P_{t,T}} + P_{t,S} d(\frac{1}{P_{t,T}}) + (dP_{t,S}) d( \frac{1}{P_{t,T}} ) $$ $$= \frac{dP_{t,S}}{P_{t,T}} −P_{t,S}\frac{dP_{t,T}}{P^2_{t,T}} + (···)dt$$ $$= (β_{t,S} −β_{t,T}) \frac{P_{t,S}}{P_t}dW_t + (···)d_t. $$

Can anyone explain how to obtain the first line of the solution please. Initially I thought you could just use the product rule here i.e $u (dv/dx) + v (dv/dy)$ but if I use that I don't get the first term $\frac{dP_{t,S}}{P_{t,T}}$

$\endgroup$
0
$\begingroup$

For the first line of the solution it's just Ito's lemma. \begin{align} d f(X_t,Y_t,t)&= \frac{\partial }{\partial t} f(X_t,Y_t,t)dt+\frac{\partial }{\partial X_t} f(X_t,Y_t,t) dX_t +\frac{\partial }{\partial Y_t} f(X_t,Y_t,t) dY_t \\ &+\frac12 \frac{\partial^2 }{\partial X_t^2} f(X_t,Y_t,t) d\langle X_t,X_t\rangle+\frac12 \frac{\partial^2 }{\partial Y_t^2} f(X_t,Y_t,t) d\langle Y_t,Y_t\rangle\\ &+ \frac{\partial^2 }{\partial X_t\partial Y_t} f(X_t,Y_t,t) d\langle X_t,Y_t\rangle \end{align} In your case you have $X_t=P_{t,S}$ and $Y_t=\frac{1}{P_{t,T}}$ so that the first derivative of $f(X_t,Y_t,t)=X_tY_t$ in $t$ and second derivatives in $X_t$ or $Y_t$ are all 0. So you are left with the first line of the solution. \begin{align}dX_tY_t&=Y_t dX_t + X_t dY_t +d\langle X_t,Y_t \rangle\\ &= \frac{1}{P_{t,T}}dP_{t,S}+P_{t,S}d\left(\frac{1}{P_{t,T}}\right)+d\langle P_{t,S},\frac{1}{P_{t,T}}\rangle \end{align} By the way I suggest you go back to your stochastic calculus class notes because Ito's lemma is a must in quantitative finance.

$\endgroup$
  • $\begingroup$ Very useful, thanks a lot, managed to prove the final answer using Ito lemma. $\endgroup$ – Calypso Apr 1 '17 at 11:24

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.