Cash-and-Carry Arbitrage and the Growth of the Cash Difference

Summary

The document discusses a futures and bond arbitrage explanation involving two prices, an initial cash difference, and its value at a later time. The answer describes shorting the more expensive side and using the proceeds to buy the cheaper side. It then says the difference is invested at the risk-free rate, with exponential growth representing the time value of money. This is presented as an intuitive restatement of the setup rather than a derivation from first principles.

The response also describes closing the position at the later date by reversing the short. However, its final expression and explanation of the resulting profit are unclear and appear internally inconsistent with the preceding cash-flow description. The document does not provide the full original arbitrage setup, definitions of the symbols, or a careful accounting of each leg at maturity. Readers should therefore treat it as a partial intuition for financing and closing a cash-and-carry trade, not a reliable formula for profit without checking the underlying derivation.

Key ideas

• The answer frames the trade as shorting the more expensive side and buying the cheaper side.
• It describes investing the initial price difference at the risk-free rate.
• The exponential factor represents the growth of cash over time under continuous compounding.
• The response’s closing cash-flow explanation is unclear, so its stated profit expression needs verification against the original setup.

Tags

# (Self-study) Futures, bonds, and arbitrage


# (Self-study) Futures, bonds, and arbitrage












I'm currently self studying futures, so I'm sorry if this questions comes off a bit stupid. I'm currently reading a book by Walsh, J.B. Knowing the Odds: An Introduction to Probability.

I quote this part of the text:

I want to understand how the bond's worth at time $t$, $(P-S_0)e^{rt}$ came about. If I understood it right, in b) $S_0-P$ was invested, so how come at time $t$, the bond's worth is not $(S_0-P)e^{rt}$.

Sorry about the ignorant question.

## Answer by Tosh (score 0, accepted)

https://quant.stackexchange.com/a/55689

Let me just rephrase in less complex literature. So at $t=0$, you short the expensive side, $S_0$. Use proceed to buy the cheaper side, $P$. You will invest the difference, $(S_0 - P)$ at the risk free rate, where you multiply by $e^{rt}$ due to time value of money, which grows at time $t$.

Now at time $t=t$, You close the position, i.e if you have gone short at $t = 0 $, you will go for long (liquid) position at $t = t$. Hence you should get a net of $(S_0 - P) + (P-S_0)$. This where you get this rissoles profit of $(P-S_0)e^{rt}$.

Hope this helps and good luck for your self study.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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