data - Why is the ratio of Hi-Low range to Open-Close range close to 2?

I tried it in several symbols and timeframes with the same result:

$$\frac {mean(HIGH-LOW)}{mean(|CLOSE-OPEN|)}$$

Symbol       Result
------       ------
EURUSD:W1    1.9725  
EURUSD:D1    2.0023 
EURUSD:H1    2.1766  
USDJPY:W1    1.9949 
USDJPY:D1    2.0622  

USDJPY:H1    2.2327 
SAN.MC:D1    2.0075  
BBVA.MC:D1   2.0075    
REP.MC:D1    2.1320

Answer

There is a very good reason why the ratio $$\frac {mean(HIGH-LOW)}{mean(|CLOSE-OPEN|)} \approx 2$$ on various financial series. If the price of a security evolves according to a Wiener process beginning at the opening bell and throughout the day, and the drift is negligible for that period of time, i.e.$\mu=0$, then the denominator of the above ratio closely approximates the average absolute deviation, $$AAD=\frac{2\sigma}{\sqrt{2\pi}}\int_0^\infty xe^{-x^2/2}dx=\sqrt {2/\pi}\cdot\sigma$$ for a normal distribution, where $\sigma$ is the standard deviation. On the other hand $$\mathbb E(HIGH-OPEN) = \sqrt{2/\pi}\cdot\sigma$$ $$\mathbb E(LOW-OPEN) = -\sqrt{2/\pi}\cdot\sigma$$(See the running maximum of a Wiener process on Wikipedia.) So we have for such an idealized Wiener process: $$\frac {\mathbb E(HIGH-LOW)}{\mathbb E(|CLOSE-OPEN|)} = \frac{\sqrt{2/\pi}\cdot\sigma-\left(-\sqrt{2/\pi}\cdot\sigma\right)}{\sqrt {2/\pi}\cdot\sigma} = 2.$$ It should not be too surprising to see this more or less borne out by observation.