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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:

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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.

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    $\begingroup$ It appears to be a "sign error" and I believe the author meant to write $(S_0-P)e^{rt}$ $\endgroup$
    – nbbo2
    Jul 15, 2020 at 5:08
  • $\begingroup$ So good to hear. Thank you for confirming. $\endgroup$
    – Idrees
    Jul 15, 2020 at 6:10
  • $\begingroup$ This is not a sign error. $\endgroup$
    – Tosh
    Jul 15, 2020 at 14:03

1 Answer 1

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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.

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  • $\begingroup$ thanks so much this makes sense! $\endgroup$
    – Idrees
    Jul 16, 2020 at 0:21

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