# Change of numeraire between t1-forward mesure and t2-forward mesure

Let denote $$\mathbb{Q}_{t_1}$$ the $$t_1$$-forward mesure associated to zero coupon bond $$B(.,t_1)$$.

Let denote $$\mathbb{Q}_{t_2}$$ the $$t_2$$-forward mesure associated to zero coupon bond $$B(.,t_2)$$.

I am trying to deduce $$\frac{d\mathbb{Q}_{t_1}}{d\mathbb{Q}_{t_2}}|_t$$. Here is my reasoning :

Let denote $$V(t)$$ the price of an asset at time t.

By definition $$\frac{V(t)}{B(t,t_1)}$$ is a martingal process under $$\mathbb{Q}_{t_1}$$ for $$t_0\leq t \leq t_1$$.

Similarly, $$\frac{V(t)}{B(t,t_2)}$$ is a martingal process $$\mathbb{Q}_{t_2}$$ for $$t_0\leq t \leq t_2$$.

Hence we could write :

$$\frac{V(t)}{B(t,t_2)}=\mathbb{E}^{\mathbb{Q}_{t_2}}\left ( \frac{V(t_1)}{B(t_1,t_2)} |\mathbb{F}_{t} \right )$$

$$\frac{V(t)}{B(t,t_1)}=\mathbb{E}^{\mathbb{Q}_{t_1}}\left ( \frac{V(t_1)}{B(t_1,t_1)} |\mathbb{F}_{t} \right )$$

Since $$B(t,t_1)$$ and $$B(t,t_2)$$ are $$\mathbb{F}_{t}$$ mesurable we have:

$$\mathbb{E}^{\mathbb{Q}_{t_2}}\left ( \frac{B(t,t_2)}{B(t_1,t_2)}V(t_1) |\mathbb{F}_{t} \right )=\mathbb{E}^{\mathbb{Q}_{t_1}}\left ( \frac{B(t,t_1)}{B(t_1,t_1)}V(t_1) |\mathbb{F}_{t} \right )=\mathbb{E}^{\mathbb{Q}_{t_2}}\left (\frac{d\mathbb{Q}_{t_1}}{d\mathbb{Q}_{t_2}}|_t \frac{B(t,t_1)}{B(t_1,t_1)}V(t_1) |\mathbb{F}_{t} \right )$$

Hence we deduce that :

$$\frac{d\mathbb{Q}_{t_1}}{d\mathbb{Q}_{t_2}}|_t=\frac{B(t,t_2)}{B(t_1,t_2)}\frac{B(t_1,t_1)}{B(t,t_1)}$$

Question : Is my result correct ?

Thanks !