1
$\begingroup$

This is a beginner level question.

I have a $spread = aluminium - 0.7*lead $

$s = a - 0.7*l$

I have two methods to calculate return on this spread:

$ return = (s_t - s_{t-1})/(a_t + 0.7*l_t) $

which is nothing but:

$return = change/value_{investment} $

or

$ return = a_t - a_{t-1}/a_{t-1} - 0.7*(l_t - l_{t-1)/l_{t-1} $

which is nothing but:

$return_{aluminium} - 0.7*return_{lead} $

Which return definition is correct assuming I am taking 2 investment along a spread that is long in $1$ unit of aluminium and correspondingly short in $0.7$ unit of lead and vice-versa.

$\endgroup$
2
$\begingroup$

Which return definition is correct

Neither one, and it appears that you are actually interested in the rate of return. In the first definition of rate of return, the denominator should be $s_{t-1}$.

In the second definition, the ratios do not really compute the return of $a$ and $l$. Each term somewhat resembles the logarithm property, except that you are missing the logarithm function in each term. Even if you applied the logarithm function, your calculations would fail as soon as the positions in $a$ and $l$ differ in magnitude.

$\endgroup$
  • $\begingroup$ Please check my edit, also I cannot add s(t-1) in the denominator since it can take negative values $\endgroup$ – Dhruv Mahajan Mar 17 at 5:03
  • 1
    $\begingroup$ Precisely because the spread could take zero and negative values, it does not make sense to compute a return. Why do you need a return? $\endgroup$ – Alex C Mar 17 at 13:11
  • 1
    $\begingroup$ @DhruvMahajan (1) It is unclear to me what you meant to change in the edit. (2) I am not suggesting you to add $s_{t-1}$ in the denominator, but that you replace the denominator with $s_{t-1}$ instead: Rates of return must have the previous (or initial) period in the denominator. (3) Dividing by $a_t+0.7*l_t$ contradicts your premise of having a short position in lead. (4) In line with Alex C's comment, if $s_{t-1}$ were non-positive, then you would have no funds to invest at time=$t$. $\endgroup$ – Iñaki Viggers Mar 17 at 14:20
  • 1
    $\begingroup$ Well I meant trading via futures (which of course also means you obtain leverage), in which case I struggle to see the value of the standard formulae given above $\endgroup$ – ZRH Mar 17 at 20:06
  • 1
    $\begingroup$ @DhruvMahajan I see. But for pairs trading you would consider a relative spread (aluminum/0.7lead) rather than the absolute spread (aluminum-0.7lead). The relative spread makes things more feasible as it avoids negative or zero spreads. $\endgroup$ – Fokko Mar 18 at 4:53

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.