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.

$$F(x)=\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$


$$F(x)=\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)$$

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:

$$\frac{1}{2} \text{erf}\left(\frac{x}{\sqrt{2}}\right)--\frac{1}{2}=0.9$$



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?

  • $\begingroup$ Just gave it a quick glance: What I first noticed, shouldn't you look at $F(x) - F(-x) =0.9$ for your probability band (lets rather say confidence interval) around 0? Also, isn't $F(-\infty) = 0$? $\endgroup$ – vanguard2k Feb 6 '17 at 8:11
  • $\begingroup$ That scaling seems correct to me. When increasing the time increments, the variance (!) increases proportionally with time. That means, the standard deviation, or volatility, increases proportionally with the square root of time. And, of course, it is proportional to $\sigma$. Hence, you have a total factor of $\sigma \sqrt{t}$. $\endgroup$ – Marie. P. Feb 6 '17 at 10:08
  • $\begingroup$ Yes, but I have seen a couple of papers written about big issues with scaling distribution this way. I'll see if I can find it. $\endgroup$ – ALEXANDER Feb 6 '17 at 10:11

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