MGSC 372
Advanced Business Statistics
Lecture 2: Random Walks
Javad Nasiry
© 2025 McGill University
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
1 / 31
Why study random walks?
Review and reinforce several important concepts from MGCR 271
Build up to understanding a commonly used model for stock prices
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
2 / 31
Hopping frogs
A frog starts out at 0 on the number line.
-2
-1
0
1
2
Every second, it jumps one step left or one step right, with equal probability.
Where will the frog be after 100 seconds?
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
3 / 31
Hopping frogs
A frog starts out at 0 on the number line.
-2
-1
0
1
2
Every second, it jumps one step left or one step right, with equal probability.
Where will the frog be after 100 seconds?
Could go all the way to 100 or -100 (extremely improbable)
Could end up back at 0 (most likely outcome, but still not very likely)
Could be somewhere in between
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
3 / 31
A better question: What is a typical outcome for the frog after 100 seconds?
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
4 / 31
A better question: What is a typical outcome for the frog after 100 seconds?
To get a handle on this, let’s simulate some frogs on the computer!
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
4 / 31
A better question: What is a typical outcome for the frog after 100 seconds?
To get a handle on this, let’s simulate some frogs on the computer!
Here is one frog’s path:
12
8
Location
4
0
−4
−8
−12
0
25
50
75
100
Time
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
4 / 31
A better question: What is a typical outcome for the frog after 100 seconds?
To get a handle on this, let’s simulate some frogs on the computer!
Here is a second frog’s path:
12
8
Location
4
0
−4
−8
−12
0
25
50
75
100
Time
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
5 / 31
A better question: What is a typical outcome for the frog after 100 seconds?
To get a handle on this, let’s simulate some frogs on the computer!
Here is a third frog’s path:
12
8
Location
4
0
−4
−8
−12
0
25
50
75
100
Time
After 100 seconds, these three frogs ended up at 4, 10, and -4.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
6 / 31
Now let’s simulate 10 frogs and record where each one ends up after 100 seconds.
The ending positions are:
→2,
→4,
Javad Nasiry
→6,
16,
10,
→14,
MGSC 372 Advanced Business Statistics
→2,
0,
4,
© 2025 McGill University
→28
7 / 31
Now let’s simulate 10 frogs and record where each one ends up after 100 seconds.
The ending positions are:
→2,
→4,
→6,
16,
10,
→14,
→2,
0,
4,
→28
This is easier to visualize in a bar chart:
2.0
Count
1.5
1.0
0.5
0.0
−30
−20
−10
0
10
Location of frog after 100 seconds
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
7 / 31
Why not simulate 10,000 frogs!
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
8 / 31
Why not simulate 10,000 frogs!
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
8 / 31
Why not simulate 10,000 frogs!
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
This is an empirical distribution of the ending location of a frog that follows a
random walk for 100 seconds. It will be (slightly) di!erent every time we run the
simulation.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
8 / 31
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
Let’s use this empirical distribution to address the question: What is a typical
outcome for the frog after 100 seconds?
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
9 / 31
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
Let’s use this empirical distribution to address the question: What is a typical
outcome for the frog after 100 seconds?
As anticipated, 0 was the most likely outcome, but only about 8% of the
frogs ended up exactly at 0.
All 10000 frogs ended up between -40 and 50.
Most frogs ended up between -20 and 20 (to be precise, 9657 out of 10000,
or 96.57%)
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
9 / 31
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
We can also compute the summary statistics of the empirical distribution:
sample median
sample mean
sample standard deviation
Javad Nasiry
0
→0.049
9.885
MGSC 372 Advanced Business Statistics
© 2025 McGill University
10 / 31
800
Count
600
400
200
0
−40
−20
0
20
40
Location of frog after 100 seconds
We can also compute the summary statistics of the empirical distribution:
sample median
sample mean
sample standard deviation
0
→0.049
9.885
Because the distribution looks normal and the mean is close to 0, the standard
deviation is a good numerical measure of the “typical” displacement of the frog
after 100 seconds.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
10 / 31
We’ve considered the case of “binary frogs” (two choices for each step). Let’s now
consider two variations: normal frogs and exponential frogs.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
11 / 31
We’ve considered the case of “binary frogs” (two choices for each step). Let’s now
consider two variations: normal frogs and exponential frogs.
Normal frog: each step is randomly drawn from the standard normal distribution
N (0, 1):
-2
-1
0
1
2
Standard normal distribution
Positive and negative steps are equally likely; the size of most steps is between 0
and 2
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
11 / 31
Exponential frog: each step is randomly drawn from the (shifted) exponential
distribution (with rate 1):
1
-1
0
1
Shifted exponential distribution
Small negative steps are most common (the median step size is -0.307), but large
positive steps can occur
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
12 / 31
Now we’ll learn how to simulate random walks in Python!
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
13 / 31
Results of the simulations
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
14 / 31
Binary steps
Each step comes from the discrete distribution
X
Prob
→1
1
2
1
1
2
Example of a binary random walk with 100 steps:
Location
0
−5
−10
0
25
50
75
100
Time
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
15 / 31
Normal steps
Each step comes from the standard normal distribution N (0, 1)
-2-1 0 1 2
Example of a normal random walk with 100 steps:
Location
10
5
0
−5
0
25
50
75
100
Time
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
16 / 31
Shifted exponential steps
Each step comes from a shifted version of the exponential distribution with rate 1
1
-1 0 1
Example of a shifted exponential random walk with 100 steps:
Location
4
0
−4
0
25
50
75
100
Time
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
17 / 31
Empirical distributions of the endpoints of 10000 simulated random walks,
each of 100 steps.
Binary
800
Count
600
400
200
0
−40
−20
0
20
40
Location after 100 steps
Normal
Exponential
1000
900
Count
Count
750
500
600
300
250
0
−20
0
20
40
0
−40
Location after 100 steps
−20
0
20
40
Location after 100 steps
All three distributions look remarkably similar!
They all have mean close to 0 and standard deviation close to 10
They all look roughly normal
Why are they so similar?
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
18 / 31
Random variables
and the theory of the random walk
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
19 / 31
A random variable assigns a number to the outcome of a random process.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
20 / 31
A random variable assigns a number to the outcome of a random process.
There are two types of random variables: discrete and continuous.
Discrete random variables:
The possible values of the random variable are distinct points on the number
line
We can show the probability of each value in a table (or with a bar graph)
Example: Each step taken by the binary frog is a discrete random variable with
the following probability distribution:
X
Prob
Javad Nasiry
→1
1
2
1
1
2
MGSC 372 Advanced Business Statistics
© 2025 McGill University
20 / 31
Continuous random variables
The possible values of the random variable make up an interval on the
number line
The probability distribution is described by a curve called a density function
The probability of any particular value occurring is 0
The probability of observing a value in the interval [a, b] is the area under the
density function f (x) between a and b:
P (a ↑ X ↑ b) =
! b
f (x) dx
a
Example: Each step taken by a normal or exponential frog is a continuous
random variable. We saw the density functions earlier in the lecture.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
21 / 31
Three parameters associated to a random variable
The expected value or mean of a random variable X is denoted E(X) or µX .
For discrete r.v.’s, it is defined by
"
E(X) =
xi · Prob(xi )
where x1 , x2 , . . . are the values that X can take, and Prob(xi ) is the probability of
the value xi occurring.
(For continuous r.v.’s, the expected value is defined using integrals—you won’t be
asked to use this definition.)
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
22 / 31
Three parameters associated to a random variable
The expected value or mean of a random variable X is denoted E(X) or µX .
For discrete r.v.’s, it is defined by
"
E(X) =
xi · Prob(xi )
where x1 , x2 , . . . are the values that X can take, and Prob(xi ) is the probability of
the value xi occurring.
(For continuous r.v.’s, the expected value is defined using integrals—you won’t be
asked to use this definition.)
The variance of X is denoted var(X) or ω 2 (X). For discrete r.v.’s, it is defined by
"
var(X) =
(xi → E(X))2 · Prob(xi )
The standard deviation of X is denoted sd(X) or ω(X), and is defined by
#
sd(X) = var(X)
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
22 / 31
Back to random walks
-2
-1
0
1
2
Definition
A random walk is a random process that starts at 0 and consists of repeated
steps along the number line, where
each step is independent of the previous steps
each step is drawn from the distribution of a fixed random variable X
In other words, the steps are independent and identically distributed (i.i.d)
copies of a fixed random variable X.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
23 / 31
Theory of the random walk
Consider a random walk of n steps, where each step is drawn from the distribution
of a fixed random variable X. The endpoint of this walk is the random variable
X1 + X2 + . . . + Xn
where the Xi are i.i.d. copies of X.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
24 / 31
Theory of the random walk
Consider a random walk of n steps, where each step is drawn from the distribution
of a fixed random variable X. The endpoint of this walk is the random variable
X1 + X2 + . . . + Xn
where the Xi are i.i.d. copies of X.
How do we express the mean, standard deviation, and variance of this sum in
terms of the parameters of X?
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
24 / 31
Theory of the random walk
Consider a random walk of n steps, where each step is drawn from the distribution
of a fixed random variable X. The endpoint of this walk is the random variable
X1 + X2 + . . . + Xn
where the Xi are i.i.d. copies of X.
How do we express the mean, standard deviation, and variance of this sum in
terms of the parameters of X?
We need to recall the “rules for expected values and variances”...
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
24 / 31
As explained on the board, we end up with the following formulas:
If X1 , . . . , Xn are i.i.d. copies of a fixed random variable X, then
E(X1 + X2 + · · · + Xn ) = n · E(X)
and
sd(X1 + X2 + · · · + Xn ) =
Javad Nasiry
↓
n · sd(X)
MGSC 372 Advanced Business Statistics
© 2025 McGill University
25 / 31
As explained on the board, we end up with the following formulas:
If X1 , . . . , Xn are i.i.d. copies of a fixed random variable X, then
E(X1 + X2 + · · · + Xn ) = n · E(X)
and
sd(X1 + X2 + · · · + Xn ) =
↓
n · sd(X)
The above formulas come from a simple application of the rules for expected
values and variances. The following fact is much deeper:
As n gets larger, the distribution of X1 + · · · + Xn gets closer and
↓ closer to a
normal distribution (with mean n · E(x) and standard deviation n · sd(X))
This is the version of the Central Limit Theorem for sums (do you see how it is
related to the version for averages that you learned in MGCR 271)?
Unless otherwise specified, we will consider the approximation safe to use when
n > 25.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
25 / 31
For the three types of random walks we considered:
E(X) = 0
sd(X) = 1
where X is the distribution of a single step
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
26 / 31
For the three types of random walks we considered:
E(X) = 0
sd(X) = 1
where X is the distribution of a single step
Thus, in all three cases, the location of a random walk after 100 steps has
expected value: 100 · E(X) = 100 · 0 = 0
↓
standard deviation:
100 · sd(X) = 10 · 1 = 10
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
26 / 31
For the three types of random walks we considered:
E(X) = 0
sd(X) = 1
where X is the distribution of a single step
Thus, in all three cases, the location of a random walk after 100 steps has
expected value: 100 · E(X) = 100 · 0 = 0
↓
standard deviation:
100 · sd(X) = 10 · 1 = 10
Moreover, n = 100 is large enough that the Central Limit Theorem gives a very
good approximation. This explains the similarity of the three empirical
distributions!
(Important note: when each step X is drawn from a normal distribution,
X1 + · · · + Xn is normally distributed for EVERY n. There is no need to appeal
to the Central Limit Theorem in this case.)
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
26 / 31
Random walks with drift
If E(X) ↔= 0, the random walk tends to drift in the positive or negative direction,
depending on the sign of E(X).
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
27 / 31
Random walks with drift
If E(X) ↔= 0, the random walk tends to drift in the positive or negative direction,
depending on the sign of E(X).
For example, here are three random walks with steps X drawn from N (0.5, 2):
Random walks with drift
Location
60
40
20
0
0
25
50
75
100
Time
n
The expected value of the location after n steps is n · E(X) = . This is shown
2
by the dashed line.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
27 / 31
Using the Central Limit Theorem
Exercise (solution in appendix): The time it takes for a certain grocery store clerk
to check out a customer follows an exponential distribution with a rate of 0.5
customers per minute. That is, the checkout time has the distribution
0.5
1
f (x) = 12 e→ 2 x
0
(Var = 4 = 36*4
12=sqrt(36*4))
N(2, 12)
where x is measured in minutes. The expected value of this distribution is 2
minutes, and the standard deviation is also 2 minutes.
1 What is the expected time for this clerk to check out 36 customers?72 mins
2 What is the probability that it will take more than 90 minutes for the clerk to
check out 36 customers? =1.5 stds, z-table: 6%
3 How much time must the clerk allow in order to have a 99% chance of
checking out all 36 customers? 2.33 stds away, = 99.96 mins
4 Give a 90% prediction interval for the amount of time it will take the clerk
to check out 36 customers. That is, give an interval that has a 90% chance
of containing the actual amount of time it takes to check out 36 customers.
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
28 / 31
5%
90%
5%
1.64 stds away so +-19.68mins from 72
52.32-91.68 mins
A random walk in two dimensions
40
30
20
10
0
−10
0
Javad Nasiry
10
20
30
MGSC 372 Advanced Business Statistics
40
50
© 2025 McGill University
29 / 31
Appendix: solution to the exercise
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
30 / 31
1
E(X1 + . . . + X36 ) = 36 · E(one customer) = 72
2
By the CLT, X1 + . . . + X
→36 is approximately normal. The mean is 72, and
the standard deviation is 36 · sd(one customer) = 12. So
!
"
90 ↑ 72
P (X1 + . . . + X36 > 90) = P Z >
12
= P (Z > 1.5) = 1 ↑ 0.9332 = 0.0668
3
The z-score with 99% of the area to its left is approximately 2.33. Thus, the
clerk must allow
2.33 ↓ 12 + 72 ↔ 100 minutes
4
The z-score with 5% of the area to its right is 1.645. So a 90% prediction
interval is
[↑1.645 ↓ 12 + 72, 1.645 ↓ 12 + 72] ↔ [52.3, 91.7] minutes
Javad Nasiry
MGSC 372 Advanced Business Statistics
© 2025 McGill University
31 / 31