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


1 Answer 1


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 $$ \frac{d}{ds}P(xy)=\frac{d}{ds}P\left((x_0 + s) \frac{z_0}{x_0 + s}\right)=\frac{d}{ds}P(z_0)=0.$$

  • $\begingroup$ @g g Thank you for your answer. The geometric interpretation is quite elegant and it's quite cool since it seems like some differential geometry can be applied. Moreover my problem is changed meanwhile so I'm going top post a new question about delta in portfolio...stay foolish $\endgroup$
    – clarkmaio
    Jan 8, 2018 at 9:16

Your Answer

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

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