Generating Random Numbers: Some Examples
Example 1:
The following 10 numbers are realizations from a Standard Uniform distribution:
.54463 .15389 .85961 .61149 .05219 .41417 .28357 .17783 .40950 .82995
Explain how to use these numbers to generate 10 iid Bernoulli random numbers with a
success probability p = .4. Generate a sample of size 10 from that distribution.
Solution:
Let
(
1 if 0 < Ui ≤ .4,
Yi =
0 if .4 < Ui < 1,
for i = 1, . . . , 10. Thus by inverse cdf method Y1 , . . . , Y10 are iid Bernoulli(.4) Here the
sample generated is 0, 1, 0, 0, 1, 0, 1, 1, 0, 0
Example 2:
Generate a random sample of size 10 from the Bin(n = 10, p = .4) using the above values
from U (0, 1)
Solution:
X ∼ Bin(n, p) means that X is the number of successes in n iid Bernoulli trials with a
success probability p. Thus X = Y1 + Y2 , . . . + Yn where Y1 , . . . , Yn are iid Bernoulli(p).
In Problem 1, we generated Y1 , . . . , Y10 ∼ iid Bernoulli(.4). Thus set X = Y1 + Y2 , . . . + Y10
to obtain X ∼ Bin(10, .4).
Here, a value from Bin(10, .4) is generated as X = 0 + 1 + 0 + 0 + 1 + 0 + 1 + 1 + 0 + 0 = 4. To
generate another value from Bin(10, .4), one needs to repeat the whole procedure, beginning
with obtaining a new sample from Standard Uniform. Thus this method is the least efficient
method for generating random samples from the Binomial distribution, partcularly if n is
even moderately large. An efficient method is the table look-up method. We need the cdf
for the Bin(10, .4) which is available on the Tables page. We reproduce it below:
x
P (X ≤ x)
0
0.006046618
5
0.833761382
1
0.046357402
6
0.945238118
2
0.167289754
7
0.987705446
3
0.382280602
8
0.998322278
4
0.633103258
9
0.999895142
Using this table, it is possible to produce 10 values using the 10 standard uniforms from
Problem 1 and applying the inverse cdf method. For example, the first value to exceed
U1 = .54463 in the above table is 0.633103258; thus the value of the Binomial r.v. returned
is 4. Repeating this procedure to transform the 10 values from U (0, 1), the 10 values from
Bin(10, .4) obtained are 4, 2, 6, 4, 2, 4, 3, 3, 4, 5, respectively, obviously a more efficient method
than summing Bernoulli’s.
1
Example 3:
A continuous random variable Y has pdf
(
2 2
y , −1 ≤ y ≤ 1,
f (y) = 3
0,
otherwise
Use the first random number u1 = .54463 to generate a value y from this distribution.
Solution:
We can use the inverse method for continuous random variables, but first we need to find
the cdf of Y . The cdf of Y is
FY (y) = P (Y ≤ y)
For y < −1, FY (y) = 0 since P (Y ≤ 0) = 0. For −1 ≤ y ≤ 1
Z y
1
3 2
1 y
1
FY (y) =
t dt = y 3 = y 3 +
2 −1 2
2
−1 2
For y > 1, FY (y) = 1 Thus the cdf of Y is given by
FY (y) =


0
1 3
y
2
+
1
2

1
To use the inverse cdf method set
y < −1
−1 ≤ y ≤ 1,
y>1
1
1
u = y3 +
2
2
and solve for y
1 3 1
y +
2
2
3
y = 2u − 1
1
y = (2u − 1) 3
u =
So to obtain a realization y1 from a distribution with cdf FY , substitute u1 = .54463 in the
1
above; so we have y1 = (2 × .54463 − 1) 3 = .44691
Example 4:
Using the first two random numbers u1 and u2 explain how to generate values of a random
variable X where X ∼ Erlang(k = 2, λ = 3). Generate a value from this distribution using
u1 = .54463, u2 = .15389.
Solution:
By definition of the Erlang distribution, X ∼ Erlang(k, λ) if X = Y1 + . . . + Yk where
Y1 , . . . , Yk are iid Exp(λ). Thus, first generate two random variables from Exp(3) using the
inverse cdf method:
1
Yi = − log(1 − Ui ),
i = 1, 2
λ
Then set X = Y1 + Y2 to obtain an Erlang(2, 3) random variable.
For this example, y1 = − 13 log(1 − .54463) = .26221 and y2 = − 13 log(1 − .15389) = .05570.
Thus a value from Erlang(2, 3) is x = .26221 + .05570 = .32191
2