Discounting a Known Future Payment to Its Earlier Value

Summary

The document explains an equivalence between two cash flows involving a bond discount factor. At the earlier date, the bond price for payment at the later date is known, so the positive-part payoff and the payoff scaled by the inverse discount factor can both be evaluated then. Multiplying the later payment by the discount factor converts it to its value at the earlier date.

The argument is an algebraic application of discounting: scaling the later cash amount by the price of a unit payment at that date yields the earlier-date amount in the question. This is a time-value-of-money equivalence under the stated discount-factor convention, rather than a claim that the nominal payments on the two dates are equal. The answer does not discuss uncertainty in rates, transaction costs, or alternative pricing assumptions; it relies on the given discount factor and the quantities being known at the earlier time.

Key ideas

• The discount factor prices a unit payment made at the later date in terms of the earlier date.
• Both payoff expressions are known at the earlier date under the setup described.
• Multiplying the scaled later payment by the discount factor recovers the earlier payment amount.
• The equivalence concerns discounted value, not equality of the cash amounts paid on different dates.

Tags

# Answer by Gordon (score 1, accepted)


# $\frac{1}{p(T_{i-1},T_i)}(A-p(T_{i-1},T_i))^+$ 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.

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/15996

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.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.