This is not a homework or assignment exercise.

I'm trying to evaluate $\displaystyle \ \ I := E_\beta \big[\frac{1}{\beta(T_0)} K \mathbf{1}_{\{B(T_0,T_1) > K\}}\big]$, where $\beta$ is the savings account, $B$ is the present value (at some time $t$) of the zero coupon bond paying $\$1$ at maturity, $\mathbf{1}$ is the indicator function, $r(s)$ is the short rate, and $f(t,T)$ is the instantaneous forward rate (i.e. $f(t,T,T)$). Expanding on this a bit more:

$$ \displaystyle \ \ \beta(T_0) = e^{\int_0^{T_0} r(s)d(s)}$$

$$ \displaystyle \ \ B(t,T) = e^{-\int_t^T f(t,s)d(s)}$$

The first step is to do a measure change as follows:

$$\displaystyle \ \ I = E_{T_0} \big[\frac{1}{\beta(T_0)} K \mathbf{1}_{\{B(T_0,T_1) > K\}} \frac{\beta(T_0)B(0,T_0)}{\beta(0)B(T_0,T_0)}\big]$$

My question is why is this correct? For a density process to be a valid change of measure we need to to be a Doleans-Dade exponential local martingale, but I can't see why this is the case in this example, because:

$$\displaystyle \ \ \frac{\beta(0)B(T_0,T_0)}{\beta(T_0)B(0,T_0)} = 1$$

which is not in the usual form of a Doleans exponential?

  • 2
    $\begingroup$ @ Jase : I have to appologize I misread your last equation last time. It is equal to 1, only conditionally to $\mathcal{F}_{T_0}$, or alternatively if you take expectation of it, as $B(t,T)=e^{-\int_t^T f(s,t)ds}=E_\beta[e^{-\int_t^T r(s)ds}]=E_\beta[\frac{\beta(t)}{\beta(T)}]$. So your last equation (at t=0) is equal to $\frac{e^{-\int_0^{T_0} r(s)ds}}{E_\beta[e^{-\int_0^{T_0} r(s)ds}]}$. The fact that this is a correct measure change is a theorem, a version of which can be found in Brigo and Mercurio's book "Interest Rate Models Theory and Practice". Regards $\endgroup$
    – TheBridge
    Commented Nov 23, 2012 at 7:31


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.