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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

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  • $\begingroup$ 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
    Commented Aug 24, 2022 at 16:33
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    $\begingroup$ @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
    Commented Aug 24, 2022 at 18:47

2 Answers 2

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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.

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  • $\begingroup$ 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
    Commented Aug 25, 2022 at 7:55
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    $\begingroup$ 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
    Commented Aug 27, 2022 at 3:09
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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

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