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?
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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.
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$\begingroup$ Notice however that the drift (or dt term) changes: $\mu$ changes to $-\mu-\sigma^2$ $\endgroup$– nbbo2Commented Jun 5, 2020 at 20:55
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$\begingroup$ 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 $\endgroup$ Commented Sep 23, 2020 at 10:40