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Does the volatility of a Currency Pair depend on the currency in which the premium is paid? For example- will the Volatility of USDJPY change if the premium is paid in USD instead of JPY. Is there any mathematical formulation for this?

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It depends, on what you mean by returns. For simple returns: no, for log returns yes. To recap, simple returns are given by $$R_\textrm{simple} = \frac{P_{t+1}}{P_t}-1$$ and log returns are given by $$R_\textrm{log} = \log \left(\frac{P_{t+1}}{P_t}\right).$$ The rate of change is given by $$R = \frac{P_{t+1}}{P_t}.$$

A percentage increase in one currency of a pair, implies a decrease in the other of the same size, so $$R^\textrm{USDJPY} = \frac{P_{t+1}}{P_t} = x$$ implies $$R^\textrm{JPYUSD}\frac{P'_{t+1}}{P'_t} = \frac{1}{x}$$ where $P'_t$ is the reverse rate.

In words, if EURUSD is trading at 1.20 today and at 1.212 tomorrow the return from a USD perspective is $1.212 / 1.20 - 1 = 1\%$ as today the USD holder was holding 120 cents of USD and tomorrow he would be holding 1.212 cents of USD. On the other hand, from a EUR perspective the loss is $1.20 / 1.212 - 1 = -0.99\%$.

We can now do a simple experiment to get a feeling of the volatility for these types of returns in R:

> # Simple returns
> set.seed(1)
> returns <- rnorm(10, 1, 0.01) # One added back to R_simple
> returns
 [1] 0.9937355 1.0018364 0.9916437 1.0159528 1.0032951 0.9917953 1.0048743
 [8] 1.0073832 1.0057578 0.9969461
> sd(returns - 1)
[1] 0.00780586
> sd(1/returns - 1)
[1] 0.007769419

Clearly, the volatility of simple returns is not the same. Using the same sample suggests that the volatility of the log returns is equal:

> sd(log(returns))
[1] 0.0077874
> sd(log(1/returns))
[1] 0.0077874

This can be shown to always hold with $x$ defined as above. The log returns for $P_t$ and $P'_t$ are then given by $\log(x)$ and \begin{align} \log(1/x) &=\log{1} - \log{x} \\ &= -\log{x} \end{align}

The standard deviation of sample is equal to standard deviation of the mirrored around its mean.

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  • $\begingroup$ I think in your last code block, you meant to type sd(log(1/returns)) for the second command -- though that does not affect the answer. $\endgroup$ – kurtosis 2 days ago
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    $\begingroup$ Thanks, fixed it! $\endgroup$ – Bob Jansen 2 days ago
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If you're modelling the FX rate as a geometric brownian motion and asking whether the volatility depends on whether you model the rate or the inverse rate, then the answer is no - and we can demonstrate it using Ito's lemma

Assuming the rate $X$ obeys \begin{align} {\frac {dX} X} = rdt + \sigma dW \end{align}

for some rate $r$ and volatility $\sigma$, lemma says that for a function $f(X,t)$

\begin{align} df = \Bigl( {\frac {\partial f} {\partial t}} + r X {\frac {\partial f} {\partial X}} + {\frac {\sigma^2 X^2} 2} {\frac {\partial^2 f} {\partial X^2}} \Bigr) dt + \sigma {\frac {\partial f} {\partial X}} dW \end{align}

Substituting in $f(X,t) = {\frac 1 X}$, we get

\begin{align} d{\frac 1 X} &= \Bigl( rX {\frac {-1} {X^2}} + {\frac {\sigma^2 X^2 } 2} {\frac {2} {X^3}} \Bigr) dt - \sigma X {\frac {1} {X^2}} dW\\ &= - {\frac 1 X} \Bigl( \bigr(r - \sigma^2 \bigl) dt + \sigma dW\Bigr) \end{align}

So the inverse process ${\frac 1 X}$ also follows a geometric brownian motion, with a drift of $-r + \sigma^2$ and a volatility of $\sigma$ (ie. the same volatility as $X$)

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