0
$\begingroup$

I'm a mathematician and I'm newish to trading.

Is it possible to separate or decouple the two currencies in a trading pair?

That is, given the GBP/USD pair, is it possible to create a graph of the GBP value and a graph of the USD value to then attempt to predict movements in the GBP/USD pair based of movements in those decoupled.

Is there a name for this process? I'm interested in researching this idea so any and all pointers are appreciated.

Thanks in advance.

$\endgroup$
7
  • $\begingroup$ Seems like you're looking for concepts such as interest rate parity (tells you how spot and forward quotes of a currency pair are connected - or are supposed to be). $\endgroup$
    – SuavestArt
    Aug 16, 2023 at 21:35
  • $\begingroup$ @SuavestArt I've had a quick Google and that doesn't seem like what I'm after. Though there's always a chance I'm misunderstanding it. $\endgroup$ Aug 16, 2023 at 21:53
  • 5
    $\begingroup$ Currencies only move (and exist, for that matter) relative to other currencies or objects with some perceived value. A Dollar or Pound is only useful with something to compare it to or purchase. $\endgroup$
    – amdopt
    Aug 16, 2023 at 22:01
  • 3
    $\begingroup$ Since you can't measure the value of a currency by that same currency, people look at quotes for currency pairs. $\endgroup$
    – SuavestArt
    Aug 16, 2023 at 22:29
  • 2
    $\begingroup$ Technically, you can do it and the value is always 1 for each currency. 1 USD "exchanges" to 1 USD. Practically, if you are interested in an exchange rate, you need another currency. You can find a decent collection of theories around exchange rates here. $\endgroup$
    – AKdemy
    Aug 17, 2023 at 5:50

1 Answer 1

1
$\begingroup$

Yes, I believe it is possible and I have more or less done something similar to what you ask. Here is a link to a post on my blog which gives Octave code and in this post there are links to other posts on my blog which discuss the rationale behind the code. You can also click on the "currency strength" label at the bottom of the post to find other related blog posts.

I invite you to peruse at your leisure.

EDIT IN RESPONSE TO COMMENTS

Below is a copy and paste of a more recent post on my blog which I believe addresses the points raised in the comments to this answer.

Take, for example, the GBPUSD forex pair, and further give it a current (imaginary) value of 1.2500. What does this mean? Of course it means 1 GBP will currently buy you 1.25 USD, or alternatively 1 USD will buy you 1/1.25 = 0.8 GBP. Now rather than write GBPUSD let's express GBPUSD as a ratio thus:- GBP/USD, which expresses the idea of "how many USD are there in a GBP?" in the same way that 9/3 shows how many 3s there are in 9. Now let's imagine at some time period later there is a new pair value, a lower case "gbp/usd" where we can write the relationship

(1)     ( GBP / USD ) * ( G / U ) = gbp / usd

to show the change over the time period in question. The ( G / U ) term is a multiplicative term to show the change in value from old GBP/USD 1.2500 to say new value gbp/usd of 1.2600,

e.g.    ( G / U ) == ( gbp / usd ) / ( GBP / USD ) == 1.26 / 1.25 == 1.008

from which it is clear that the forex pair has increased by 0.8% in value over this time period. Now, if we imagine that over this time period the underlying, real value of USD has remained unchanged this is equivalent to setting the value U in ( G / U ) to exactly 1, thereby implying that the 0.8% increase in the forex pair value is entirely attributable to a 0.8% increase in the underlying, real value of GBP, i.e. G == 1.008. Alternatively, we can assume that the value of GBP remains unchanged,

e.g.    G == 1, which means that U == 1 / 1.008 == 0.9921

which implies that a ( 1 - 0.9921 ) == 0.79% decrease in USD value is responsible for the 0.8% increase in the pair quote.

Of course, given only equation (1) it is impossible to solve for G and U as either can be arbitrarily set to any number greater than zero and then be compensated for by setting the other number such that the constant ( G / U ) will match the required constant to account for the change in the pair value.

However, now let's introduce two other forex pairs (2) and (3) and thus we have:-

                (1)     ( GBP / USD ) * ( G / U ) = gbp / usd

                (2)     ( EUR / USD ) * ( E / U ) = eur / usd

                (3)     ( EUR / GBP ) * ( E / G ) = eur / gbp

We now have three equations and three unknowns, namely G, E and U, and so this system of equations could be laboriously, mathematically solved by substitution.

However, in my currency strength indicator I have taken a different approach. Instead of solving mathematically I have written an error function which takes as arguments a list of G, E, U, ... etc. for all currency multipliers relevant to all the forex quotes I have access to, approximately 47 various crosses which themselves are inputs to the error function, and this function is supplied to Octave's fminunc function to simultaneously solve for all G, E, U, ... etc. given all forex market quotes. The initial starting values for all G, E, U, ... etc. are 1, implying no change in values across the market. These starting values consistently converge to the same final values for G, E, U, ... etc for each separate period's optimisation iterations.

Having got all G, E, U, ... etc. what can be done. Well, taking G for example, we can write

                (4)     GBP * G = gbp

for the underlying, real change in the value of GBP. Dividing each side of (4) by GBP and taking logs we get

                (5)     log( G ) = log( gbp / GBP )

i.e. the log of the fminunc returned value for the multiplicative constant G is the equivalent of the log return of GBP independent of all other currencies, or as the original question asked, the (change in) value of GBP separated or decoupled the from the pair in which it is quoted.

The full blog post can be read at https://dekalogblog.blogspot.com/2023/08/currency-strength-revisited.html

$\endgroup$
4
  • 2
    $\begingroup$ Pls consider expanding your answer so that one can still understand it in case the link you have provided gets broken or changed. $\endgroup$
    – Alper
    Aug 17, 2023 at 11:52
  • 1
    $\begingroup$ @babelproofreader I've had a skim through your pages and I have only found a single image that resembles what I'm interested in, but no explanation as yet. dekalogblog.blogspot.com/2016/10/… The second image on this page. $\endgroup$ Aug 17, 2023 at 13:07
  • $\begingroup$ So it's in fact a result of a change based on all currencies instead of none other, unless I misunderstand what fminunc does? $\endgroup$
    – AKdemy
    Aug 20, 2023 at 20:11
  • 1
    $\begingroup$ @AKdemy Yes, you are right. The value for G, for example, is based upon all forex pairs for which GBP is a quote currency. This applies to all such values, i.e. U for USD, E for Euro ... etc. The optimisation routine solves for all G, U, E ... simultaneously taking into account all forex pairs where the relevant G, U, E ... are one of the quote currencies. It is, if you like, a total market evaluation of the values for each G, U, E ... and not an evaluation calculated from just a few, select crosses. $\endgroup$ Aug 21, 2023 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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