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 !
