LessonThis is a lesson restated in new words. Any figure in it is an example from the source, not this desk's record.
A standard Brownian motion starts at zero, and the change between two times is normal with mean zero and variance equal to the time gap. On a grid, that change is a standard normal draw scaled by the square root of the gap. With a unit gap and a horizon of 252 steps, the lecture’s many paths ended with a sample mean near zero and a sample variance near 252. Building those paths as one array is the version you keep. A Python loop is only there to show the sum.
Why it matters
The continuous definition the lecture uses has three working parts. The value at time zero is zero. Increments over non-overlapping intervals are independent. An increment from an earlier time to a later time is normal, centered at zero, with variance equal to the length of the gap. Stationarity is the “centered at zero for every gap of the same length” piece. The lecture calls this enough for simulation and skips a stricter almost-sure construction.
The idea in plain words
A standard normal draw Z has mean 0 and variance 1. Multiply it by the square root of the first time step and the product is normal with mean 0 and variance equal to that step. The mean stays zero because a constant scale comes out of the expectation and the expectation of Z is zero. The variance picks up the square of the scale, which cancels the square root and leaves the time step. So the scaled draw has the same distribution as the first Brownian increment. The lecture writes that match as equality in distribution, not as equality of two particular random numbers.
The next step repeats the argument with a fresh standard normal and with the square root of the gap between t2 and t1. Add it to the first level. After k steps the level is the sum of every scaled draw so far, and the value at zero contributes nothing. If every gap is 1, the square root is 1 and the path is a cumulative sum of plain standard normals.
The loop version appends that sum one day at a time for a 252-day horizon, using 251 increments after the starting zero so the list has the length they wanted. Redraw the list and the picture changes. To check the definition, they build 10,000 independent paths and keep only the final value of each. The average of those finals sat near zero. The variance sat near 252. One pass they mention was about 249, and further passes moved around 252. That is the sample variance estimating the population variance of the increment from 0 to 252. The finals are statistics. They have a distribution of their own, and the lecture expects that distribution to be centered on mean 0 and variance 252.
How to use it
Choose the grid before you choose the code. Unit steps are the lecture’s simplification so the scale factor is 1. If your step is 1/252 of a year, multiply each standard normal by the square root of that step instead. The terminal variance should then match the sum of the step lengths, not automatically 252. Know which clock you are on.
What usually goes wrong
Forgetting the leading zero makes a correct cumulative sum of increments look like a path that does not start at zero. The lecture separates those pictures on purpose. The correlation-style “it wiggles, so it is Brownian” test is not the test. The terminal mean and variance are the test.
Questions people ask
What is added at each step?
A normal increment with mean zero and variance equal to the time gap. On a unit grid that increment is just a standard normal draw. On a finer grid it is a standard normal times the square root of the gap.
What should 10,000 terminal values look like after 252 unit steps?
A mean near zero and a variance near 252. The lecture saw one variance near 249 and describes the cloud as centered on 252. A histogram of repeated variance estimates was bell-shaped around that same 252.
Why not use the loop?
It matches the recurrence and it is slow once you want many paths. The array of shocks, a cumulative sum, and a column of starting zeros produce the same terminal check.