FX Sensitivities Depend on Which Exchange Rates Are Held Fixed
Summary
The document explains an apparent inconsistency in calculating foreign-exchange sensitivity. A cash flow expressed using one currency pair may appear to depend on another pair after rewriting exchange rates through a triangular currency relationship. Differentiating that rewritten expression while treating all rates as independently variable can therefore produce a nonzero result, even though the original payoff contains no exposure to the rate in question.
The resolution is to define the sensitivity precisely: specify which exchange rates can move and which relationships or market quantities must remain fixed. When the currency cross-rate constraint is maintained, changing one rate requires another to adjust, and the payoff in the example remains constant, giving zero sensitivity. The answer says this calculus issue applies regardless of whether the product is linear or nonlinear. It does not provide a broader sensitivity framework for portfolios with multiple currencies, so the relevant held-fixed conventions must be established for each risk calculation.
Key ideas
- A derivative depends on the choice of independent variables and on what is held fixed.
- Currency exchange rates are linked by cross-rate relationships and cannot always be varied independently.
- Rewriting a payoff in constrained rates and then differentiating as if they were independent can give a misleading sensitivity.
- Maintaining the exchange-rate constraint in the example leaves the payoff unchanged.
- The issue is not determined by whether a product is linear or nonlinear.
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# Existential question about currency exchange Risk Factor
# Existential question about currency exchange Risk Factor
Ciao All,
I'm working to a problem about sensitivities for products with several ccy and this questions came out.
For simplicity consider a linear product (a simple cash flow) w.r.t. the ccy exchange $ccy_1/ccy_2$ $$ P\left( \frac{ccy_1}{ccy_2} \right) = N_{ccy_1} \cdot \frac{ccy_1}{ccy_2} $$
For simplicity I will define: $$ \begin{align} ccy_1 & = AUD \\ ccy_2 & = EUR \\ ccy_3 & = RON \end{align} $$
Suppose now I want to compute the $\delta$ sensi w.r.t. the variable $RON/EUR$. Of course there is not this variable in the expression of the product so that one can say that the sensi is $0$. Infact I expect that this product doesn't depend on that fx exchange.
But of course one can write: $$ \frac{AUD}{EUR} = \frac{AUD}{EUR} \left(\frac{RON}{EUR} \right)^{-1} $$ so that if we take the derivative we have: $$ \partial_{\frac{RON}{EUR}}P = -N_{AUD} \frac{AUD}{EUR} \left(\frac{RON}{EUR} \right)^{-2} \not = 0 $$
> Now the problem is that from a "financial" point of view I would say that the sensi is $0$ but from a mathematical point of view (I trust more this philosophy) I can't since there is term in the analytical form which cointains the variable I'm using in the derivative.
Some coworkers of mine are guessing that there is a different behaviour depending on the nature of the product: there is a $0$ sensi for linear products but in the case of NON linear product, for example standard derivative on currency exchange) a contribution must be take in account.
Do you have any ideas or comment about this "ambiguity"?. Thank you in advice!
Ciao ciao, AM
## Answer by g g (score 3, accepted)
https://quant.stackexchange.com/a/37630
There is no contradiction and basically no ambiguity. Furthermore, the kind of product (linear or non-linear) has no bearing on the question. It is really only a question of basic calculus.
Let us call the three FX rates $x, y, z$ which satisfy the relation (or constraint) $z=xy$ and your product $P$, which is a function of $z$ only. You can interpret $P$ as a function of $x,y$. The fact that $P$ only depends on $z$ means that $P$ is constant on the curves $z=xy$ in $(x,y)$-Space.
Your point of confusion is what "delta" with respect to $x$ means in this context. Here it means you observe the change in $P$ varying $x$ such that the constraint $z=xy$ is observed.
Once you do this all paradox is gone: Fix points $z_0, x_0, y_0$ with $z_0=x_0 y_0$ and observe what happens if you vary $x_0$ a bit by setting $x=x_0 + s$. Since you must observe the constraint, $y$ is no more permitted to vary freely, you have $y=\frac{z_0}{x_0 + s}$. Plug this into $P$ and calculate the derivative $$ \frac{d}{ds}P(xy)=\frac{d}{ds}P\left((x_0 + s) \frac{z_0}{x_0 + s}\right)=\frac{d}{ds}P(z_0)=0.$$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.