Change of numeraire between T-forward and Bank Account

Question

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.

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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)

Answer 1

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, $\forall t>0$ $$ \left. \frac{d\Bbb{Q}^T_1}{ d\Bbb{Q}} \right\vert_{\mathcal{F}_t} = \left( \left. \frac{d\Bbb{Q}}{d\Bbb{Q}^{T_1}} \right\vert_{\mathcal{F}_t}\right)^{-1} $$