Normal-Return Bands and Square-Root-of-Time Scaling
Summary
The document derives a one-sided probability threshold for a standardized normal variable, using the inverse error function to obtain a value of about 1.28155 for 90% cumulative probability. It then asks whether this threshold, multiplied by daily volatility and the square root of elapsed time, can define a price band through an exponential transformation of log returns. This connects normal quantiles, log returns, and the familiar square-root-of-time volatility scaling idea.
The derivation as presented has a material limitation: it standardizes a single-period return, but extending that result to a longer horizon requires assumptions about the return process, including independent increments with stable variance; normality alone does not establish square-root scaling. The displayed probability is one-sided, so it does not describe a central interval containing 90% of outcomes. The document poses these issues but includes no answer or empirical validation, and its density and cumulative-distribution formulas should be checked before implementation.
Key ideas
- The document uses a normal quantile to form a one-sided probability threshold for standardized log returns.
- It proposes scaling daily volatility by the square root of the horizon before converting the log-return band into prices.
- Square-root-of-time scaling requires assumptions about how returns and variances behave across periods.
- A one-sided 90% cumulative threshold is different from a two-sided band containing 90% of outcomes.
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# Scaling Z value/or voll with square root of time
# Scaling Z value/or voll with square root of time
Let's say that I have a historical price trace of future prices. I convert this variable to get it normally distributed. (Let's assume that the random variable after this conversion turns out to become normally distributed.)
The conversion is done as follows:
Let's say that we define these prices as Subscript[P, t], at time t.
$$\frac{\log \left(\frac{P_{t+1}}{P_t}\right)-\mu }{\sigma }$$
I now have a random variable with mean zero and variance 1 that is normally distributed. (assumption)
$$\frac{\int e^{-\frac{(x-\mu )^2}{2 \sigma ^2}} \, dx}{\sqrt{2 \pi \sigma ^2}}=\frac{\sigma (\text{erf} (x-\mu ))}{\left(\sqrt{2} \sigma \right) \left(2 \sqrt{\sigma ^2}\right)}$$
Then setting $\sigma =1$ and $\mu =0$
I want this probability band to contain 90% of the distribution. Using fundamental theorem of calculus:
$$F(x)-F(-\infty )=0.9$$
From this I solve for x.
$$F(x)=\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$
Note:
$$F(x)=\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$
and limit as x approaches infinity is:
$$\lim_{x\to -\infty } \, \frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)=-\frac{1}{2}$$
and is calculated as follow using the continuity of Erf(x) at x = -inf.
$$\frac{1}{2} \text{erf}\left(-\frac{\infty }{\sqrt{2}}\right)=-1*\frac{1}{2}=-\frac{1}{2}$$
Now from these changes we have this:
$$\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)--\frac{1}{2}=0.9$$
Hence:
$$\frac{x}{\sqrt{2}}=\text{erf}^{-1}(0.8)$$
Then from this:
$x=\sqrt{2} \text{erf}^{-1}(0.8)$ $$x=1.28155$$
My variable is in daily increments: and I want to scale this probability band as the timeframe increases.
$$\frac{\log \left(\frac{P_{t+1}}{P_t}\right)-\mu }{\sigma }=1.28155$$
Then with mean = 0
$$P_{t+1}=P_t e^{1.28155 \sigma \sqrt{t}}$$
Would this scaling of z value by volatility and square root of time be correct? And if so which assumptios has to be satisfied for it to hold?Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.