# FX ATM-volatility quotes

Is the implied volatility ATM the same for a currency pair as for the inverted currency pair. I.e, can I expect the same volatility quote ATM for (for an instance) EURUSD as for USDEUR? And does this result hold for all tenors?

Here are my thoughts. Let's take for example the pair EURUSD and USDEUR. The fx rate for EURUSD will be $$X_t$$ and USDEUR $$1/X_t$$. Now assume that $$d{X_t} = \mu{X_t} dt + \sigma{X_t} dW_t$$ then thanks to Ito's Lemma you have $$d\bigl(\frac{1}{X_t}\bigr) = 0dt -\frac{1}{X_t^2}dX_t +\frac{1}{2}\frac{-2}{X_t^3}(\sigma X_t)^2 dt = -\frac{1}{X_t^2}dX_t -{X_t}\sigma^2 dt$$ finally $$d\bigl(\frac{1}{X_t}\bigr) = -\frac{1}{X_t^2}(\mu{X_t} dt + \sigma{X_t} dW_t) -\frac{\sigma^2}{X_t} dt = (-\frac{\mu}{X_t}-\frac{\sigma^2}{X_t})dt -\frac{1}{X_t} \sigma W_t$$
So you can see that $$1/X_t$$ has the same vol as $$X_t$$.The answer is yes then.
• Notice however that the drift (or dt term) changes: $\mu$ changes to $-\mu-\sigma^2$ – noob2 Jun 5 at 20:55
• I am under the impression that the final form should be $(-\frac{\mu}{X_t^2}+\frac{\sigma^2}{X_t^3})dt-\frac{1}{X_t^2}\sigma dW_t$. Please correct me if I am wrong – Whitebeard13 Sep 23 at 10:40