# 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?

My take ( Please correct if i am wrong):


"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?

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.

• So, is it fair to assume that (implied vol of GBPUSD) = -( implied vol of USDGBP)? Intuitively, the spot distribution would be different but log return would stay the same? For example: if spots of GBPUSD are S1, S2, ...Sn, then the spot of USDGBP would be 1/S1, 1/S2, .....1/Sn at time t1, t2,....tn respectively. And log return for GBPUSD between time t1 and tn is log(S1/Sn) which would be same as return of USDGBP i.e. Log((1/S1)/(1/Sn)) except for a change of sign from positive to negative. – Ussu Oct 24 '19 at 11:49
• @Ussu no, volatility cannot be negative, thus the change of Brownian Motion. – Daneel Olivaw Oct 24 '19 at 11:55
• Thanks. It's a good point. Is it fair to conclude that returns would be different in sign but distribution pattern would be the same, so implied vol would stay the same? – Ussu Oct 24 '19 at 12:11
• By careful in distinguishing implied volatility from the volatility (i.e. standard deviation) of the FX process itself. $\text{USDGBP}_t$ and $\text{GBPUSD}_t$ have different location (expectation) and scale (standard deviation), see Wikipedia for example. Even the variance of infinitesimal returns is different: $V(d\text{GBPUSD}_t)=\sigma^2V(\text{GBPUSD}_t)dt\not=\sigma^2V(\text{USDGBP}_t)dt=V(d\text{USDGBP}_t)$. Intuitively, I do not know how to interpret the fact that the diffusion coefficient $\sigma$ remains unchanged. – Daneel Olivaw Oct 24 '19 at 12:35
• How about the relationship between their realised volatility? I think the way I calculated vol for GBPUSD and USDGBP, these are realised volatility. So, now is it fair to conclude that their realised volatility will also be the same. – Ussu Oct 24 '19 at 18:35

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