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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 !

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