1
$\begingroup$

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

Improve this question
$\endgroup$
2
  • $\begingroup$ If you are asking if $\frac{P(t,T_{i}) P(T_{i},T_{i+1})}{P(t,T_{i+1})}=1$, I think the answer is yes. $\endgroup$
    – nbbo2
    Aug 24, 2022 at 16:33
  • 1
    $\begingroup$ @nbbo2 That's basically my idea, but $\frac{P(t,T_i)*P(T_i,T_{i+1})}{P(t,T_{i+1})}=1$ is not completely true, what is true is rather $\frac{P(t,T_i)*P(t,T_i,T_{i+1})}{P(t,T_{i+1})}=1$ with $P(t,T_i,T_{i+1})$ being the forward bond price in this case. Note that $P(T_i,T_{i+1})$ is rather a random variable (not known at time t)... $\endgroup$
    – Xman
    Aug 24, 2022 at 18:47

2 Answers 2

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ That's a good point, however I don't think it's true, because if $V(T_i) = P(T_i,T_{i+1})$ then the LHS is not a squared payoff because $\frac{P(t,T_i)∗P(T_i,T_{i+1})}{P(t,T_{i+1})}*P(T_i,T_{i+1})$ has the same dimension as $P(T_i,T_{i+1})$. (Take the case where interest rates are deterministic and even flat for illustration) $\endgroup$
    – Xman
    Aug 25, 2022 at 7:55
  • 1
    $\begingroup$ But that’s the whole point. If rates are deterministic , the statement is true. See your answer below. The last line of your proof assumes rates are not deterministic. $\endgroup$
    – dm63
    Aug 27, 2022 at 3:09
0
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.