In 1900, Louis Bachelier (1870-1946) defended his thesis, Théorie de la Spéculation, before a trio of mathematicians including the eminent Henri Poincaré. It dealt with financial markets
Indeed, in Bachelier's time, Paris was the centre of speculation in bonds and he considered prices as a time series ... with randomness. Alas, any mathematical theory of the stock market was beneath the dignity of "real" mathematicians, so Bachelier's work never received much attention.

Bachelier considered the walk of a drunkard ...
>Aha! That explains why is was ignored, eh?
Pay attention! Bachelier's work predated (by five years) Einstein's theory of Brownian Motion
(named after Robert Brown, who, in 1827 noticed that pollen suspended in water executed random oscillations.)
>Like stock prices and ...
Exactly! But Bachelier used random steps to simulate prices whereas, today, random steps are used to simulate changes in price.

Anyway, Bachelier considered the walk of a drunkard who, in a time T, takes N steps of length d.

>I assume that d is positive.
No, that'd be too easy. The steps might be either positive or negative, with equal probability.
The drunkard moves either up or down, but each step had length d.
So where is he after N steps?

>I give up.
Okay, consider this:

We plot the position of the drunkard at each time. That's Y. The time intervals for each step are T/N. That is, N steps in time T. The slope of the graph is d divided by T/N. That's Nd/T. If the drunkard makes more and more steps in the time T (that is, Ninfinity), then the slope becomes infinite! Although the graph becomes more and more vertical at each point (that's the slope becoming infinite), the position of the drunkard becomes more and more regular! The position of the drunkard (which we'll call P) is the SUM of the steps (which are either +d or -d, so they often cancel each other out). (Note, however, that the sum of the squares of the step lengths is clearly N d2, since the squares are all positive.)

Figure 1

>So how is the position "regular"?
I'm not finished yet! However, I guess it was Bachelier who first thought of the position of the drunkard as a curve, a trajectory ... rather than simply oscillations on the vertical Y-axis. Continuing, we notice that (since the SUM of successive Y-values gives the Y-position):

• The Variance of that SUM (which gives the position of the drunkard) is VAR[P] = VAR[y₁+y₂+...+y_N]
  where y₁, y₂ etc. are the step lengths.
• But if the steps are random and independent, then
            VAR[y₁+y₂+...+y_N] = VAR[y₁] + VAR[y₂] + ... + VAR[y_N]     (See this)
• But all the step lengths are the same length, namely d, so their Variance is d².
• Hence the Variance of the drunkard's position (after N steps) is:
            Variance[P] = N d².
• Since Variance = (Standard Deviation)², then
            Standard Deviation = SQRT(N) d.

>Yeah, so?
See? It's that Square Root thing again!
>Yeah, so?
It had been observed before, with stock prices. Namely: the deviation from the Mean increases as the Square Root of the time interval.
In fact, Regnault noticed it some forty years before Bachelier and his mathematical arguments to explain it are intriguing:

1. Imagine a circle of radius r. Every point within this circle represents a possible stock price.
2. The centre of the circle represents the Mean stock price, so the distance from the centre represents the deviation from the Mean.
3. The totality of possible stock prices is represented by aggregate of ALL points within the circle ... the area of the circle: π r².
4. As time progresses, the aggregate of possible stock prices increases; they can wander farther and farther from the Mean.
   (Stock prices don't move much in a minute, but give 'em a month ...)
5. Imagine, then, that the circle increases in size as time progresses.
6. Regnault then associates the aggregate of possible prices (that's the area of the circle) with time: Area = k T
   (That is, the area is proportional to the time, T ... increasing ... increasing ...)
7. Since the deviation from the Mean is represented by r and the area (that's π r²) is proportional to T, then
   r is proportional to the Square Root of T

"The security varies but is always looking for its real price ... which one can represent as the centre of the circle"

"In order to get an idea of the real premium on each transaction, one must estimate the mean deviation of prices on a given time interval. But following the observations ... made by Mr. Jules Regnault in his book ... for time intervals longer or shorter than a month, the mean deviation is proportional to square root of the number of days."

In the arguments provided by Bachelier (above), he gets the drunkard at a position y₁+y₂+...+y_N.
But, since the y's are either +d or -d with equal probability, the average position is zero (after infinitely many steps)!
Yet the Variance is N d² which becomes infinite as Ninfinity.
There's something fishy going on, eh?

Although it's assumed that the drunkard moves at discrete times, Bachelier attempts to make his motion continuous as follows
(It's natural to have the size of the steps get smaller as the time to make a step gets smaller.):

• The time interval t is subdivided into tiny intervals of size Δt. A step is taken every Δt.
• The number of steps taken in time t is then N = t / Δt.
• The Variance (as we saw above) is then N d² = (t / Δt) d²
• In order for this to remain finite as Δt0, then the drunkard's step during the time Δt must decrease along with Δt.
• That means d² is proportional to Δt.
• The Variance is then proportional to t (and no longer becomes infinite as the number of steps in the time t becomes infinite)
• We then have that d is proportional to SQRT(Δt).