Why Inverted FX Pairs Have the Same Implied Volatility

Summary

The document addresses whether an exchange rate and its reciprocal, such as a currency pair and the inverted pair, should have the same at-the-money implied volatility. It models one exchange rate as a diffusion with proportional volatility and applies Itô's lemma to its reciprocal. The reciprocal has a changed drift, while its proportional diffusion term has the same volatility magnitude.

This supports equal volatility for the pair and its inverse under the stated continuous-time model. The argument is conceptual and does not provide market quotes or empirical comparisons. It does not separately analyze option conventions, market quoting practices, or tenor-specific effects, so its conclusion should be read as a model result rather than a full treatment of FX volatility surfaces.

Key ideas

• Applying Itô's lemma to the reciprocal exchange rate changes its drift.
• The proportional volatility magnitude is unchanged when an exchange rate is inverted in the stated diffusion model.
• The model therefore implies equal volatility for a currency pair and its inverse.
• The document offers no empirical evidence or detailed discussion of market quoting conventions.

Tags

# FX ATM-volatility quotes


# 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?

## Answer by SachaTheBrave (score 3)

https://quant.stackexchange.com/a/54687

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.

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.