# Change of numeraire between T-forward and Bank Account

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.

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: $$$$\frac{ d \mathbb{Q} } { d \mathbb{Q}^{T_1} } \vert \mathcal{F}_t = \frac{P(0,T_1)}{P(t,T_1)} \frac{B(t)}{B(0)}$$$$

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