2
$\begingroup$

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:

enter image description here

enter image description here

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.

$\endgroup$
  • 2
    $\begingroup$ It appears to be a "sign error" and I believe the author meant to write $(S_0-P)e^{rt}$ $\endgroup$ – noob2 Jul 15 '20 at 5:08
  • $\begingroup$ So good to hear. Thank you for confirming. $\endgroup$ – Idrees Jul 15 '20 at 6:10
  • $\begingroup$ This is not a sign error. $\endgroup$ – Tosh Jul 15 '20 at 14:03
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ thanks so much this makes sense! $\endgroup$ – Idrees Jul 16 '20 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.