Continuous Compounding Arbitrage
Two banks offer different annual interest rates with continuous compounding. Bank A offers rate $r_1$ and Bank B offers rate $r_2$, where $0 < r_1 < r_2$.
You start with $\$0$ and can borrow up to $\
- Is there a strategy that guarantees a positive profit regardless of market movements? If so, describe it and compute the guaranteed profit as a function of $r_1$ and $r_2$.
- Evaluate your answer for $r_1 = 0.03$ and $r_2 = 0.06$. Round to the nearest cent.
- If you cannot use a calculator, how would you quickly estimate the answer?
Hints
- When two assets offer different risk-free returns, what classic finance strategy applies?
- With continuous compounding at rate $r$, a dollar grows to $e^{r}$ after one year. What is the cost of borrowing vs. the return on lending?
- Borrow at the cheap rate, invest at the expensive rate. Profit $= e^{r_2} - e^{r_1}$. For small rates, approximate with $e^x \approx 1 + x$.
Worked Solution
How to Think About It: This is a pure arbitrage question -- you have two risk-free rates and no constraints on going long one and short the other. Whenever two instruments offer different risk-free returns, you borrow cheap and lend expensive. The profit is locked in at inception because both rates are deterministic. No market risk, no credit risk (they're banks offering guaranteed rates), just a pure free lunch from the rate differential.
Quick Estimate: With $r_1 = 0.03$ and $r_2 = 0.06$, the spread is 3%. On $\
Approach: Borrow $\
Formal Solution:
With continuous compounding, $\
- Borrow $\$ from Bank A. After 1 year, you owe $e^{r_1}$.
- Deposit $\
$ in Bank B. After 1 year, you receive $e^{r_2}$.Guaranteed profit:
$\Pi = e^{r_2} - e^{r_1}$
Since $r_2 > r_1 > 0$, the exponential function is strictly increasing, so $e^{r_2} > e^{r_1}$ and $\Pi > 0$. The profit is guaranteed.
For $r_1 = 0.03$, $r_2 = 0.06$:
$\Pi = e^{0.06} - e^{0.03} = 1.06184 - 1.03045 = 0.03138 \approx \$0.03$
For the quick estimation in part (3): use the linear approximation $e^x \approx 1 + x$ for small $|x|$. Then $\Pi \approx (1 + r_2) - (1 + r_1) = r_2 - r_1 = 0.03$. This gives $\$0.03$, which matches the rounded answer. The second-order correction $e^x \approx 1 + x + x^2/2$ gives the more precise $\$0.03$ as well (the quadratic terms contribute less than half a cent).
Answer: Borrow $\
$ from the low-rate bank and deposit it at the high-rate bank. Guaranteed profit is $e^{r_2} - e^{r_1}$. For the given rates, this is approximately $\$0.03$.Intuition
This problem illustrates the most basic form of arbitrage: a risk-free profit from a pricing inconsistency. In real markets, if two banks offered different risk-free rates, arbitrageurs would instantly borrow from the cheap one and deposit at the expensive one, driving the rates together. The no-arbitrage principle -- that such free lunches cannot persist in equilibrium -- is the foundation of all derivative pricing. Every time you use risk-neutral pricing or put-call parity, you're implicitly assuming that arbitrages like this have been eliminated.
The estimation technique $e^x \approx 1 + x$ is a workhorse on trading desks. For rates, yields, and returns under about 10%, the linear approximation is accurate to within a few basis points. Knowing when and how to use it lets you do mental math that would otherwise require a calculator, which is exactly what interviewers are testing.
- Deposit $\