# $\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+$ at time $T_i$ is equivalent to a payment of $(A-p(T_{i-1},T_i))^+$ at time $T_{i-1}$

How can I show that payment of $\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+$ at time $T_i$ is equivalent to a payment of $(A-p(T_{i-1},T_i))^+$ at time $T_{i-1}$ ? Where A is a deterministic constant.

I appreciate any help.

Thanks.

• Can you define the variable names you are using? It will make the question much more clear. – Bob Jansen Dec 27 '14 at 8:13
• Hi @BobJansen sorry for my delay. The price at time $t$ of a bond with maturity date $T$ is denoted by $p(t,T)$ and $A$ is a deterministic constant. – Roozbe Dec 27 '14 at 18:46
• Don't worry, I see you have your answer :) – Bob Jansen Dec 27 '14 at 19:40

Both the quantity $\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+$ and the quantity $(A-p(T_{i-1},T_i))^+$ are known at time $T_{i-1}$. Then the payment $\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+$ at time $T_i$ discount back to time $T_{i-1}$ is the equivalent payment. That is \begin{align*} \bigg[\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+\bigg] \times p(T_{i-1},T_i) = (A-p(T_{i-1},T_i))^+, \end{align*} where $p(T_{i-1},T_i)$ is the discount factor.