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Why Inverse FX Pairs Have the Same Volatility

Article Quant Q&A · Author: Ussu
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Summary

The document examines the relationship between GBP/USD and USD/GBP, which are reciprocal exchange rates. Their log returns have equal magnitudes and opposite signs when measured over matching intervals, so their historical volatility, calculated from those log returns, is the same under consistent sampling and conventions.

It also applies Itô’s lemma to a geometric Brownian motion model for one currency pair. Inverting the exchange rate changes the drift, including a volatility adjustment, while preserving the diffusion coefficient’s magnitude. Since implied volatility describes that diffusion scale in the model, the two reciprocal pairs have the same implied volatility in theory. The conclusion assumes standard FX modeling and aligned definitions; the document does not investigate market quotes, smile conventions, or empirical data where measurement and quote differences may affect comparisons.

Key ideas

  • Reciprocal exchange rates have opposite log returns over the same interval.
  • Changing the sign of returns preserves their standard deviation, so historical log-return volatility is equal for inverse pairs.
  • Under the geometric Brownian motion setup, inversion changes drift but preserves diffusion volatility.
  • The theoretical implied-volatility equality assumes consistent FX modeling and quote conventions.

Tags

ForexVolatilityStatisticsDerivatives pricing
Full text
# Relation between Implied and Historical Volatility of GBPUSD and USDGBP


# Relation between Implied and Historical Volatility of GBPUSD and USDGBP












Q1. How is the implied volatility of GBPUSD and USDGBP related to each other mathematically? Please explain this intuitively as well.

Q2. How is the historical volatility of GBPUSD and USDGBP related to each other mathematically?

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My take ( Please correct if i am wrong):
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"Historical Volatility or Realized Volatility is calculated as the standard deviation of log returns."

Let the price of GBPUSD at time t1, t2, t3 ....tn be S1, S2, S3...Sn. So, its return would be Log(S2/S1), Log(S3/S2)....Log(Sn/Sn-1). Suppose the standard deviation of these be σ1.

Now, the price of USDGBP at time t1, t2, t3....tn would be 1/S1, 1/S2, 1/S3.....1/Sn. So, its log return would be log(S1/S2), Log(S2/S3)....Log(Sn-1/Sn). These return have same magnitude as that of GBPUSD but the sign is opposite. So, its standard deviation would remain same as σ1.

Based on these calculations is it fair to assume that historical/released volatility would be same for GBPUSD and USDGBP?

## Answer by Daneel Olivaw (score 5, accepted)

https://quant.stackexchange.com/a/49348

Using usual FX modelling techniques, let us assume $\text{USDGBP}_t$ follows Geometric Brownian Motion under the domestic risk-neutral measure, when the domestic currency is USD: $$d\text{USDGBP}_t=(r_{USD}-r_{GBP})\text{USDGBP}_tdt+\color{blue}{\sigma}\text{USDGBP}_tdW_t$$ $r_{USD}$ and $r_{GBP}$ are the USD and GBP risk-free rates respectively. By Itô's Lemma: $$\begin{align} d\text{GBPUSD}_t=d\left(\frac{1}{\text{USDGBP}_t}\right)&=-\frac{d\text{USDGBP}_t}{\text{USDGBP}_t^2}+\frac{(d\text{USDGBP}_t)^2}{\text{USDGBP}_t^3} \\ &=\frac{r_{GBP}-r_{USD}+\sigma^2}{\text{USDGBP}_t}dt-\frac{\sigma}{\text{USDGBP}_t}dW_t \end{align}$$ $W_t$ has the same distribution than $-W_t$ thus we define a new Brownian Motion $\tilde{W}_t=-W_t$: $$\begin{align} d\text{GBPUSD}_t&=\frac{r_{GBP}-r_{USD}+\sigma^2}{\text{USDGBP}_t}dt+\frac{\sigma}{\text{USDGBP}_t}d\tilde{W}_t \\ &=(r_{GBP}-r_{USD}+\sigma^2)\text{GBPUSD}_tdt+\color{blue}{\sigma}\text{GBPUSD}_td\tilde{W}_t \end{align}$$ Hence $\text{USDGBP}_t$ and $\text{GBPUSD}_t$ have the same implied volatility in theory.

## Answer by demully (score 1)

https://quant.stackexchange.com/a/49392

The correlation of USDGBP and GBPUSD is -1! If your sample and measurement thereof, realised or implied, but suggests <>-1, jour jargonistic problem is “Siegel’s Paradox” :-)

In log terms, ie transforming returns such that they are additive, there is no difference. They sum to zero.

The assumption with implieds is also lognormal returns, so these too (not normal) must average to zero. Lest there be arbitrage in them there hills...

If you could give us an example of non-compliance with the above, it would be my great pleasure to point your way to a free lunch... or find the awkward skew in question... with the algorithm h9w to optimise profits if I can’t :-) I

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.