Special Random Variables
M. Mustafa BAHŞI, PhD
Ayşe KURT BAHŞI, PhD
The Bernoulli random variables
• Suppose that a trial, or an experiment, whose outcome can be classified as either a
“success” or as a “failure” is performed. If we let 𝑋 = 1 when the outcome is a
success and 𝑋 = 0 when it is a failure, then the probability mass function of 𝑋 is given
by
𝑃 𝑋 =0 =1−𝑝
𝑃 𝑋=1 =𝑝
• where 𝑝, 0 ≤ 𝑝 ≤ 1, is the probability that the trial is a “success”. A random
variable 𝑋 is said to be a Bernoulli random variable if its probability mass function is
given by above equations for some 𝑝 ∈ (0,1).
𝑋~𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 𝑝
• The probability mass function can be written as follow:
𝑃 𝑋 = 𝑖 = 𝑝𝑖 (1 − 𝑝)1−𝑖 ,
𝑖 = 0, 1
Expected value of the Bernoulli random variable
• The expected value of the Bernoulli random variable is
𝐸 𝑋 =0×𝑃 𝑋 =0 +1×𝑃 𝑋 =1 =0 1−𝑝 +1×𝑝 =𝑝
• That is, the expectation of a Bernoulli random variable is the probability that the
random variable equals 1.
• The variance value of the Bernoulli random variable is
𝑉𝑎𝑟 𝑋 = 𝐸 𝑋 2 − 𝐸 𝑋 2 = 𝑝 − 𝑝2 = 𝑝(1 − 𝑝)
Binomial random variable
• Suppose now that 𝑛 independent trials, each of which results in a “success” with
probability 𝑝 and in a “failure” with probability 1 − 𝑝, are to be performed. If 𝑋
represents the number of successes that occur in the 𝑛 trials, then 𝑋 is said to be a
binomial random variable with parameters (𝑛, 𝑝).
•
𝑋~𝐵𝑖𝑛𝑜𝑚 𝑛, 𝑝
• The probability mass function of a binomial random variable with parameters 𝑛 and 𝑝
is given by
𝑛 𝑖
𝑃 𝑋=𝑖 =
𝑝 (1 − 𝑝)𝑛−𝑖 ,
𝑖 = 0, 1, 2, … , 𝑛
𝑖
𝑛
𝑛!
• where
=
is the number of different groups of 𝑖 objects that can be chosen
𝑛−𝑖 !𝑖!
𝑖
from a set of 𝑛 objects.
Expected value of the Binomial random variable
• Since a binomial random variable 𝑋, with parameters 𝑛 and 𝑝, represents the number
of successes in 𝑛 independent trials, each having success probability 𝑝, we can
represent 𝑋 as follows:
𝑛
𝑋 = ෍ 𝑋𝑖
𝑖=1
• The expected value of the Binomial random variable is
𝑛
𝐸 𝑋 = ෍ 𝐸 𝑋𝑖 = 𝑛𝑝
𝑖=1
• The variance value of the Binomial random variable is
𝑛
𝑉𝑎𝑟 𝑋 = ෍ 𝑉𝑎𝑟 𝑋𝑖 = 𝑛𝑝(1 − 𝑝)
𝑖=1
Example
• It is known that disks produced by a certain company will be defective with
probability 0.01 independently of each other. The company sells the disks in packages
of 10 and offers a money-back guarantee that at least 1 of the 10 disks is defective.
• a) What proportion of packages is returned?
• b) If someone buys three packages, what is the probability that exactly one of them
will be returned?
The Poisson random variable
• A random variable 𝑋, taking on one of the values 0, 1, 2, . . . is said to be a Poisson
random variable with parameter 𝜆, 𝜆 > 0, if its probability mass function is given by
𝑖
𝜆
𝑃 𝑋 = 𝑖 = 𝑒 −𝜆 , 𝑖 = 0, 1, …
𝑖!
• The symbol 𝑒 stands for a constant approximately equal to 2.7183.
• The probability distribution of a Poisson variable is defined as the Poisson probability
distribution.
𝑋~𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆
The Poisson random variable
• Some examples of random variables that usually obey, to a good approximation, the
Poisson probability law are:
1. The number of misprints on a page (or a group of pages) of a book.
2. The number of people in a community living to 100 years of age.
3. The number of wrong telephone numbers that are dialled in a day.
4. The number of transistors that fail on their first day of use.
5. The number of customers entering a post office on a given day.
6. The number of α-particles discharged in a fixed period from some radioactive particle.
Example
• Suppose that the average number of accidents occurring weekly on a particular
stretch of a highway equals 3. Calculate the probability that there is at least one
accident this week.
Moment generating function of the Poisson random
variable
𝜙 𝑡 = 𝐸 𝑒 𝑡𝑋
∞
𝑖
𝜆
= ෍ 𝑒 𝑡𝑖 𝑒 −𝜆
𝑖!
𝑖=0
∞
𝑖
𝜆
= 𝑒 −𝜆 ෍ 𝑒 𝑡𝑖
𝑖!
𝑖=0
𝑡
−𝜆
𝜆𝑒
=𝑒 𝑒
𝑡 −1)
𝜆(𝑒
=𝑒
Expected value of the Poisson random variable
• The expected value of the Poisson random variable is
𝐸 𝑋 = 𝜙′ 0 = 𝜆
• The variance value of the Poisson random variable is
𝑉𝑎𝑟 𝑋 = 𝜙′′ 0 − 𝜙′ 0
2
=𝜆
Example
• Suppose the probability that an item produced by a certain machine will be defective
is 0.1. a) Find the probability that a sample of 10 items will contain at most one
defective item. Assume that the quality of successive items is independent. b)
Compare the desired probability and the Poisson approximation result.
Example
• 5% of the tools produced by a certain process are defective. Find the probability that
in a sample of 40 tools chosen at random, exactly three will be defective. Calculate a)
using the binomial distribution, and b) using the Poisson distribution as an
approximation.
Other Special Discrete Distributions
• Although the binomial distribution and the Poisson distribution are probably the most
common and useful discrete distributions, a number of others are found useful in
some engineering applications. Among them are the negative binomial distribution
and the geometric distribution.
The uniform random variable
• A random variable 𝑋 is said to be uniformly distributed over the interval [𝛼, 𝛽] if its
probability density function is given by
1
, 𝑖𝑓𝛼 ≤ 𝑥 ≤ 𝛽
𝑓 𝑥 = ቐ𝛽 − 𝛼
0
, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
• The probability that 𝑋 lies in any subinterval of [𝛼, 𝛽] is equal to the length of that
subinterval divided by the length of the interval [𝛼, 𝛽]. This follows since when [𝑎, 𝑏]
is a subinterval of [𝛼, 𝛽]
𝑏
1
𝑏−𝑎
𝑃 𝑎<𝑥<𝑏 =
න 𝑑𝑥 =
𝛽−𝛼
𝛽−𝛼
𝑎
Expected value of the uniform random variable
• The expected value of the uniform random variable is
𝛼+𝛽
𝐸𝑋 =
2
The variance value of the uniform random variable is
𝛽−𝛼 2
𝑉𝑎𝑟 𝑋 =
12
Example
• A temperature sensor is known to produce an error between the true temperature
and the measured temperature with equal probability in the range [−0.5, 0.5]. What is
the probability that the sensor's measurement error is in the range [−0.1, 0.2]?
Example
• Buses arrive at a specified stop at 15-minute intervals starting at 7 A.M. That is, they
arrive at 7, 7:15, 7:30, 7:45, and so on. If a passenger arrives at the stop at a time that
is uniformly distributed between 7 and 7:30, find the probability that he waits
• (a) less than 5 minutes for a bus;
• (b) at least 12 minutes for a bus.