# Preliminary Discussion

For a time series xi the deltas are assumed to be normally distributed, i.e.,

$\Delta x_i := x_{i+1} - x_i \sim \mathcal{N} ( 0, \sigma^2), \qquad (1)$

with mean zero and variance σ2. Then for a sample consisting of N + 1 ticks

$\sum_{i=1}^N | \Delta x_i |^2 \approx N \sigma^2. \qquad (2)$

Also note that

$\sum_{i=1}^{N/k} | \Delta x_i^k |^2 := \sum_{i=1}^{N/k} | x_{1+ik} - x_{1+(i-1)k} |^2 \approx N \sigma^2, \qquad (3)$

assuming that N is divisible by k and there are N + 1 ticks. Introducing time stamps for every tick, i.e., x(ti): = xi, the deltas become a function of fixed time intervals Δt:

$\Delta x_i (\Delta t) := x(t_i+ \Delta t) - x(t_i ) \sim \mathcal{N} ( 0, \sigma^2). \qquad (4)$

Note that the number of observations n depends on the sample size and the implicit Δt. E.g., if there are N observations for a specified $\Delta \hat t$, then for Δt there are

$n = \frac{\Delta \hat t}{\Delta t} N, \qquad (5)$

observations and k = N / n. The variable $\Delta \hat t$ can be set to one second without loss of generality because the series Δxi can always be interpreted as Δxit = 1). Hence Eq. (3) becomes

$\sum_{i=1}^n | \Delta x (\Delta t) |^2 \approx n \Delta t \sigma^2.\qquad (6)$

Furthermore

$\langle | \Delta x(\Delta t) | \rangle_2 = \sqrt{\frac{\sum_{i=1}^n | \Delta x_i (\Delta t) |^2}{n}} \approx \sigma \sqrt{\Delta t}. \qquad (7)$

It is apparent that the only source of variability in Eq. (7) is the number of observations $n \propto \Delta t^{-1}$.

If one takes the logarithm of xi, the numerical result is

$\sum_{i=1}^n | \ln(x_{i+1}) - \ln(x_{i})|^2 \approx n |\ln(1+\frac{\sigma}{\bar x})|^2 =: n \hat \sigma^2,\qquad (8)$

where $\bar x$ is the median value of xi. Note that the series of delta log (mid) prices, $x_{i+1}^{\ln} - x^{\ln}_i$ with $x_i^{\ln} := \ln((a_i+b_i)/2)$, is distribute equivalently to a series of $\hat x_i^{\ln} := (\ln a_i+ \ln b_i)/2$ for constant spread, so the right-hand side of Eq. (8) is applicable, where ai and bi are the ask and bid price at time ti, respectively.

It is interesting to note that compared to Eq. (2), Eq. (8) still has an explicit reference to the price series, although switching to log space eliminates all absolute price references.

# Scaling Laws - Part I

For the scaling law mentioned in U. Müller et. al, "Statistical study of foreign exchange rates, empirical evidence of a price change scaling law, and intraday analysis", Journal of Banking & Finance, Elsevier, vol. 14(6), 1990, relating the average Δxt), or the volatility, to the time interval of sampling Δt,

$\langle |\Delta x (\Delta t)| \rangle_p = \left( \frac{ \Delta t }{ C (p)} \right)^{E (p)},\qquad (9)$

the above mentioned calculations can be employed. Setting p = 2, Eqs. (9) and (7) yield

$\sigma \sqrt{\Delta t} = \left( \frac{\Delta t}{C} \right)^E,\qquad (10)$

or

$C= \frac{1}{\sigma^2} , \qquad E=0.5.\qquad (11)$

In terms of logarithmic deltas, Eq. (8) yields

$C_{\ln} = \frac{1}{\hat \sigma^2} , \qquad E_{\ln}=0.5.\qquad (12)$

# Scaling Law Relations - Part II

Recall from Eqs. (2), (6) and (8) that fixed time intervals are assumed, i.e., Δt is constant. In analogy to Eq. (1), a new dimension of randomness can be added to the process by letting the time intervals vary as well:

$\Delta t_i = t_{i+1} - t_i \sim \mathcal{N} ( \mu_t, \sigma_t^2). \qquad (13)$

So this new time series, Δxiti), has two independent sources of variability: one for the price and another for the time increments. Note however that for the calculations, like the left-hand side of Eq. (6), there is still a fixed sampling period assumed, Δt. This requires an interpolation scheme, if there is no price tick for the sampling time x(ti + Δt). For the current analysis, previous-tick interpolation is employed.

It follows from Eq. (13) that the scaling properties will be impacted, as the number of observations in Eq. (6) depends on Δt and the scaling law of Eq. (9) now reads

$\sigma \sqrt{\frac{\Delta t}{\mu_t + \sigma_t}} = \left( \frac{\Delta t}{C_x(2)} \right)^{E_x(2)}, \qquad (14)$

yielding

$C_x(2)= \frac{\mu_t + \sigma_t}{\sigma^2} , \qquad E_x(2)=0.5.\qquad (15)$

For logarithmic returns, the corresponding result is

$C^{\ln}_x(2)= \frac{\mu_t + \sigma_t}{\hat \sigma^2} , \qquad E_x(2)=0.5. \qquad (16)$