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?


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
  • 1
    $\begingroup$ Thanks, fixed it! $\endgroup$ – Bob Jansen 2 days ago

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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