Why Replicate a Future-Starting Investment?
A future-starting investment pays a stated principal at one future date and returns an accumulated amount at a later date. It may be possible to create those same future cash flows using borrowing and lending arranged today. This construction gives a direct cash-flow interpretation of the implied forward accumulation: the borrowing is repaid at the investment’s start date, while the lending produces the amount due at its end date.
The key is to compare amounts at the same dates. A discount factor for an earlier maturity is generally different from one for a later maturity, so the amount borrowed today must be chosen carefully. The construction also makes the direction of the discount-factor ratio clear.
This tutorial assumes that the zero-coupon payments are default-free, that borrowing and lending can be undertaken at the same discount factors, and that there are no transaction costs, taxes, or other market frictions. These assumptions allow each future cash flow to be valued using the same \(D(t)\). If borrowing and lending rates differ, a single discount-factor curve need not describe both sides of the transaction, and this replication formula may not apply.
Constructing the Cash Flows
Suppose an investor wants to invest \(P\) at time \(t_1\) and receive the accumulated amount at time \(t_2\). Borrow \(P D(t_1)\) today, with repayment \(P\) due at \(t_1\). At the same time, lend that same amount \(P D(t_1)\) today until \(t_2\). Since a time-0 payment of \(P D(t_1)\) buys a zero-coupon payment of \(P D(t_1)/D(t_2)\) at \(t_2\), the resulting portfolio has these cash flows:
| Time | Borrowing | Lending | Net cash flow |
|---|---|---|---|
| 0 | Receive \(P D(t_1)\) | Pay \(P D(t_1)\) | 0 |
| \(t_1\) | Repay \(P\) | No payment | \(-P\) |
| \(t_2\) | No payment | Receive \(P D(t_1)/D(t_2)\) | \(P D(t_1)/D(t_2)\) |
There is no net payment at time 0. At \(t_1\), the investor pays \(P\); at \(t_2\), the investor receives the accumulated amount. The synthetic position therefore matches a future-starting investment without requiring an initial net outlay.
If an annual effective rate over the interval is required, it must be annualized over the interval’s length, \(t_2-t_1\). That annual rate, \(f\), satisfies \((1+f)^{t_2-t_1}=D(t_1)/D(t_2)\).
Why the Ratio Has This Direction
The ratio can be derived by pricing the future cash flows of a forward investment. Paying \(P\) at \(t_1\) has time-0 value \(P D(t_1)\). Receiving \(C\) at \(t_2\) has time-0 value \(C D(t_2)\). A zero-initial-value arrangement requires these values to be equal. Solving for \(C\) gives the amount that the synthetic investment must return.
The borrowing-and-lending construction produces exactly this amount: borrowing \(P D(t_1)\) creates a repayment of \(P\) at \(t_1\), and lending \(P D(t_1)\) creates a payment of \(P D(t_1)/D(t_2)\) at \(t_2\). This agrees with the value equation. The result is not \(D(t_2)/D(t_1)\): that reversed ratio would give a different maturity-\(t_2\) payment and generally would not make the time-0 values balance.
Worked Example: Replicating a One-Year Forward Investment
Suppose \(D(1)=0.96\) and \(D(2)=0.88\). Construct a synthetic investment of 1,000 at time 1, payable at time 2. Find its time-2 proceeds and its annual effective accumulation rate. Use the borrowing-and-lending method; in particular, match the time-0 borrowing and lending amounts.
Step 1: Find the amount to borrow and lend today. Borrowing 1,000 to be repaid at time 1 requires a time-0 borrowing of \(1{,}000(0.96)=960\). Lend the same 960 today to time 2. The net time-0 cash flow is \(+960-960=0\).
Step 2: List the future cash flows. At time 1, repay 1,000. At time 2, the loan pays \(960/0.88=1{,}090.909090\ldots\).
Step 3: Calculate the implied accumulation. The factor is \(1{,}090.909090\ldots/1{,}000=1.090909090\ldots\), or \(0.96/0.88\). Since the interval is one year, the annual effective rate is \(1.090909090\ldots-1=0.090909090\ldots\), or 9.091% to three decimal places.
Check. The time-0 value of the time-1 repayment is \(1{,}000(0.96)=960\). The time-0 value of the time-2 receipt is \(1{,}090.909090\ldots(0.88)=960\). The two values balance, and the time-0 net cash flow is zero. This method avoids treating the time-0 borrowing of 960 as if it were itself the principal due at time 1.
Scaling the Position and Annualizing the Factor
The construction works for any positive principal \(P\): each cash flow scales in direct proportion to \(P\). The implied factor \(D(t_1)/D(t_2)\) does not change when the principal changes. For an interval of more than one year, however, that factor is the total accumulation over the whole interval, not the annual effective rate.
Worked Example: Scaling a Two-Year Forward Investment
Suppose \(D(1)=0.90\) and \(D(3)=0.75\). A fund wants a synthetic investment of 50,000 at time 1, payable at time 3. Find the borrowing, lending, time-3 proceeds, and equivalent annual effective rate. Use the discount-factor ratio for the total two-year accumulation before annualizing.
Step 1: Determine the time-0 amount. The amount needed to borrow and lend is \(50{,}000(0.90)=45{,}000\). Borrow 45,000 today, to repay 50,000 at time 1, and lend 45,000 today until time 3.
Step 2: Calculate the proceeds at time 3. The time-3 payment is \(45{,}000/0.75=60{,}000\). Equivalently, the two-year accumulation factor is \(0.90/0.75=1.20\), and \(50{,}000(1.20)=60{,}000\).
Step 3: Annualize over two years. The annual effective rate \(f\) satisfies \((1+f)^2=1.20\). Thus \(f=\sqrt{1.20}-1=0.095445\ldots\), or 9.545% to three decimal places.
Check. At time 0, the borrowing and lending offset: \(+45{,}000-45{,}000=0\). The cash flows are \(-50{,}000\) at time 1 and \(+60{,}000\) at time 3. Their time-0 values are \(-50{,}000(0.90)=-45{,}000\) and \(60{,}000(0.75)=45{,}000\). The values balance. This method avoids reporting 20% as an annual rate: 20% is the accumulation over two years.
A Later Start Date and a Longer Interval
The same reasoning applies when the investment starts after time 1. Both discount factors are still measured from time 0, so the ratio uses the factor at the actual start date in the numerator and the factor at the actual end date in the denominator. The length of the interval matters when converting the total factor to an annual effective rate.
Worked Example: Replicating from Year 2 to Year 5
Let \(D(2)=0.80\) and \(D(5)=0.64\). Construct a synthetic investment of 25,000 at time 2 that pays at time 5. Find the time-0 borrowing and lending amount, the time-5 receipt, and the equivalent annual effective rate. Use the full three-year interval when annualizing.
Step 1: Set the time-0 amount. The time-0 amount for a repayment of 25,000 at time 2 is \(25{,}000(0.80)=20{,}000\). Borrow 20,000 today, due as 25,000 at time 2, and lend 20,000 today until time 5.
Step 2: Find the time-5 receipt. The lending pays \(20{,}000/0.64=31{,}250\) at time 5. The total three-year forward accumulation factor is \(0.80/0.64=1.25\), and \(25{,}000(1.25)=31{,}250\).
Step 3: Find the annual effective equivalent. The annual effective rate satisfies \((1+f)^3=1.25\). Therefore \(f=1.25^{1/3}-1=0.077217\ldots\), or 7.722% to three decimal places.
Check. The borrowing and lending cancel at time 0. The time-0 value of the repayment is \(25{,}000(0.80)=20{,}000\), and the time-0 value of the receipt is \(31{,}250(0.64)=20{,}000\). This method avoids using \(D(1)\) or \(D(3)\) merely because the interval lasts three years; the relevant dates are 2 and 5.
Common Mistakes and Practical Checks
A reliable solution records the dates and signs before calculating. The synthetic investment has a negative cash flow at its start date and a positive cash flow at its end date. Its time-0 borrowing and lending offset; the zero initial net cash flow is a useful check on the construction.
- Reversing the ratio. The forward accumulation is \(D(t_1)/D(t_2)\), not \(D(t_2)/D(t_1)\). The numerator prices the payment at the investment’s start; the denominator prices its payoff.
- Confusing the borrowing amount with its repayment. Borrowing \(P D(t_1)\) today creates a repayment of \(P\) at \(t_1\), not a repayment of \(P D(t_1)\).
- Forgetting the length of the interval. The factor \(D(t_1)/D(t_2)\) is the accumulation over \(t_2-t_1\) years. Annualize it by raising it to the power \(1/(t_2-t_1)\), then subtract 1.
- Mixing valuation dates. Both discount factors in the ratio must refer to prices measured at the same time, here time 0.
- Ignoring the funding assumption. The construction uses the same curve for borrowing and lending. If separate curves or transaction costs are specified, the cash flows must be valued using the applicable terms rather than assuming this single-curve result.
Check Your Understanding
For each problem, identify the borrowing and lending cash flows before calculating the implied accumulation.
- Suppose \(D(1)=0.94\) and \(D(4)=0.70\). For an investment of 1 at time 1, find the total three-year accumulation factor and its equivalent annual effective rate. (Answer: factor \(=0.94/0.70=1.342857\); annual effective rate \(=1.342857^{1/3}-1\approx 10.326\%\).)
- Suppose \(D(2)=0.85\) and \(D(5)=0.68\). A portfolio seeks an investment of 20,000 at time 2, payable at time 5. Find the time-0 borrowing and lending amount, the time-5 payment, and the annual effective rate. (Answer: borrow and lend 17,000; repay 20,000 at time 2; receive 25,000 at time 5; annual effective rate \(=1.25^{1/3}-1\approx 7.722\%\).)
- Suppose \(D(1)=0.92\) and \(D(2)=0.87\). Construct a one-year synthetic investment of 12,500 starting at time 1. Give the time-0 amount and time-2 proceeds to the nearest cent, and check the time-0 values. (Answer: borrow and lend \(12{,}500(0.92)=\$11{,}500.00\); repay \(\$12{,}500.00\) at time 1; receive \(11{,}500/0.87=\$13{,}218.39\) at time 2; both future amounts have time-0 value \(\$11{,}500.00\), to the nearest cent.)
- Suppose \(D(1.5)=0.93\) and \(D(4)=0.80\). Find the cash flows for an investment of 8,000 from time 1.5 to time 4. (Answer: borrow and lend \(8{,}000(0.93)=7{,}440\) at time 0; repay 8,000 at time 1.5; receive \(7{,}440/0.80=9{,}300\) at time 4; the total factor is 1.1625 over 2.5 years.)
- Suppose \(D(2)=0.96\) and \(D(6)=0.72\). For a future-starting investment of 75,000 at time 2, find its time-6 proceeds and equivalent annual effective rate. (Answer: borrow and lend \(75{,}000(0.96)=72{,}000\); repay 75,000 at time 2; receive \(72{,}000/0.72=100{,}000\) at time 6; annual effective rate \(=(4/3)^{1/4}-1\approx 7.457\%\).)