# 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.

Ciao ciao, AM

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