I follow a course, and get to the point that one bond price discounted by another one is a martingale: $$ \frac{P(t,T_0)}{P(t,T_1)} - \text{ is a } \mathbb{Q}^{T_1} \text{ martingale } $$ I can not comprehend the proof below, mainly what is measurable inside of expectations: $$ \mathbb{E}^{T_1}_t \big[ \frac{P(T,T_0)}{P(T,T_1)} \big] = \frac{1}{P(t,T_1)} \mathbb{E}_t \big[ e^{-\int^T_t r(s)ds } P(T,T_1) \frac{P(T,T_0)}{P(T,T_1)} \big] = \frac{P(t,T_0)}{P(t,T_1)} $$ for $t<T< \min (T_0,T_1) $.

Here is my attempt: I look at Brigo and Mercurio book and say that my RN derivative( from $T_1$ forward measure $ \mathbb{Q}^{T_1}$ to usual RNM $\mathbb{Q}= \mathbb{Q}^B$ or without stating $\mathbb{Q}$ as I do below) is: \begin{equation} \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \vert \mathcal{F}_t = \frac{P(0,T_1)}{P(t,T_1)} \frac{B(t)}{B(0)} \end{equation}

applying Bayes Rule: $$ \mathbb{E}^{T_1}_t \big[ \frac{P(T,T_0)}{P(T,T_1)} \big] = \frac{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \big]} = \frac{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \big]} $$ and substituting RN derivative inside expectation and to the denominator we obtain: $$ \frac{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \big]} = \frac{ \mathbb{E}_t \big[ \frac{P(0,T_1)}{P(t,T_1)} \frac{B(t)}{B(0)} \frac{P(T,T_0)}{P(T,T_1)} \big]}{ \frac{P(0,T_1)}{P(t,T_1)} \frac{B(t)}{B(0)} } $$ Now, to me it seems that all inside the expectation is measurable besides $\frac{P(T,T_0)}{P(T,T_1)}$ but I am not sure if that reasoning is correct nor how to proof that $\frac{P(T,T_0)}{P(T,T_1)}$ under $\mathbb{Q}$ is a measurable.

After seeing the answer below, I finish the application of the change of measure:

$$ \frac{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q}^{T_1} } { d \mathbb{Q} }\vert \mathcal{F}_T \frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{ d \mathbb{Q}^{T_1} } { d \mathbb{Q} } \vert \mathcal{F}_T \big]} = \frac{ \mathbb{E}_t \big[ \left( \frac{P(T,T_1)}{P(0,T_1)} \frac{B(0)}{B(T)} \right)\frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{P(T,T_1)}{P(0,T_1)} \frac{B(0)}{B(T)} \big] } = \frac{B(0)}{P(0,T_1)} \frac{ \mathbb{E}_t \big[ \left( \frac{P(T,T_1)}{1} \frac{1}{B(T)} \right)\frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{P(T,T_1)}{P(0,T_1)} \frac{B(0)}{B(T)} \big] } $$ knowing that discounted with the bank account Zero Coupon Bond price is a martingale: $$ \frac{P(t,T_x)}{B(t)} = E^{\mathbb{Q}}_t\left[ \frac{P(T,T_x)}{B(T)} \right] $$ we obtain: $$ \frac{ \mathbb{E}_t \big[ \left( \frac{P(T,T_1)}{1} \frac{1}{B(T)} \right)\frac{P(T,T_0)}{P(T,T_1)} \big]}{ \mathbb{E}_t \big[ \frac{P(T,T_1)}{1} \frac{1}{B(T)} \big] } = \frac{ \mathbb{E}_t \big[ \frac{P(T,T_0)}{B(T)} \big]}{ \mathbb{E}_t \big[ \frac{P(T,T_1)}{B(T)} \big] } = \frac{P(t,T_0)}{B(t)} \frac{B(t)}{P(t,T_1)} = \frac{P(t,T_0)}{P(t,T_1)} $$

(I also recommend to see a nice answer here and the link to the paper in that answer)


Your expression for the RN derivative is correct indeed $$ \left. \frac{d\Bbb{Q}}{d\Bbb{Q}^{T_1}} \right\vert_{\mathcal{F}_t} = \frac{P(0,T_1)}{P(t,T_1)} \frac{B(t)}{B(0)} $$ Your problem comes the application of the (abstract) Bayes rule. More specifically you should have $$ \Bbb{E}_t^{T_1}[ X_T ] = \frac{ \Bbb{E}_t \left[ X_T \left. \frac{d\Bbb{Q}^T_1}{ d\Bbb{Q}} \right\vert_{\mathcal{F}_T} \right] } { \Bbb{E}_t \left[ \left. \frac{d\Bbb{Q}^T_1}{ d\Bbb{Q}} \right\vert_{\mathcal{F}_T} \right] } $$ for any measurable $X_T$, with here $$ X_T = \frac{P(T,T_0)}{P(T,T_1)} $$ So you had 2 problems:

  • The RN derivatives must be evaluated at $\mathcal{F}_T$ not $\mathcal{F}_t$ because $X_T$ is deemed $\mathcal{F}_T$-measurable.
  • You have used the wrong RN derivative for the measure change: you should use the inverse of that of your post. Note that, $\forall t>0$ $$ \left. \frac{d\Bbb{Q}^T_1}{ d\Bbb{Q}} \right\vert_{\mathcal{F}_t} = \left( \left. \frac{d\Bbb{Q}}{d\Bbb{Q}^{T_1}} \right\vert_{\mathcal{F}_t}\right)^{-1} $$
  • $\begingroup$ Dear @Quantuple, I personally liked the paper you offered in the answer I linked to. Could you please suggest how to choose numeraire in general case? In case of Interest Rate products numeraire is usually ZC bond price, but when we have collateral, or some option is a combination of some underlying priced in foreign currency we choose different numeraires.. Is there any hint how do we find which numeraire makes a process in expectation a martingale? Hopefully you understood what I mean :) $\endgroup$
    – Ant
    Oct 8 '18 at 9:35
  • 1
    $\begingroup$ Dear @Ant, not sure I understand. Usually switching numéraires is a mathematical trick to obtain results in a simpler, less hand wavy, form. That being said, by definition, any numéraire should make the value of any self-financing strategy (when expressed in that numéraire) a martingale. Not sure I was clear either :) $\endgroup$
    – Quantuple
    Oct 8 '18 at 9:53

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.