Bond forward arbitrage relationships

Question

I am trying to see if the following statement is true or not and I would really appreciate your help.

The statement is as follows:

$\forall $ Tradable Asset $V(t)$, $$ E[\frac{P(t,T_{i})P(T_{i},T_{i+1})}{P(t,T_{i+1})}V(T_i)|F_t] = E[V(T_i)|F_t]$$ Where the expectency is taken under any probability measure (not necessarily Risk neutral) although a solution with the Risk neutral measure is also more than welcome.

My intuition is that $P(t,T_{i})P(T_{i},T_{i+1}) \approx P(t,T_{i+1})$ especially under expectencies.

PS: $T(t,T_i)$ is the $T_i$ zero coupon bond price at time t.

Many thanks

Answer 1

I don’t see how this can be true in general. For example, if $V(T_i)=P(T_i,T_(i+1))$ then the LHS would be a squared payoff with convexity, whereas the RHS is linear.

Answer 2

I finally managed to find the answer! The statement is False! Because if

$\forall $ Tradable Asset $V(t)$, $$ E[\frac{P(t,T_{i})P(T_{i},T_{i+1})}{P(t,T_{i+1})}V(T_i)|F_t] = E[V(T_i)|F_t]$$

Then

$\forall $ Tradable Asset $V(t)$, $$ E[(\frac{P(t,T_{i})P(T_{i},T_{i+1})}{P(t,T_{i+1})}-1)V(T_i)|F_t] = 0$$

Therfore almost surely $$ \frac{P(t,T_{i})P(T_{i},T_{i+1})}{P(t,T_{i+1})}-1= 0$$

This means that $$P(T_{i},T_{i+1}) = \frac{P(t,T_{i+1})}{P(t,T_{i})}$$ Which is false because the left hand side is stochastic whereas the right hand side is deterministic...

Thanks everyone