Discrete Random Variables

CIS 2033 based on
Dekking et al. A Modern Introduction to Probability
and Statistics. 2007
Longin Jan Latecki
Discrete Random Variables
Discrete random variables (RVs) are obtained by counting and have sample spaces
which are countable. The values that represent each outcome are usually integers.
Random variables are denoted by capital letters.
For example, a RV X: is the number of times that we flip a coin until H comes up
The possible outcomes are denoted by lower case letters: a=1, a=2, a=3...
Example from Section 4.1
RV S is the sum of two independent throws with a die. The sample space is
Ω = {(ω1, ω2) : ω1, ω2 ∈ {1, 2, . . . , 6} }
= {(1, 1), (1, 2), . . . , (1, 6), (2, 1), . . . , (6, 5), (6, 6)}.
Hence the RV S is the function S : Ω → R, given
S( ω1, ω2 ) = ω1 + ω2 for (ω1, ω2) ∈ Ω
Let {S = k} = {(ω1, ω2) ∈ Ω : S( ω1, ω2) = k }.
We denote the probability of the event {S = k} by P(S = k) , although formally we should
write P({S = k}) instead of P(S = k).
In our example, S attains only the values k = 2, 3, . . . , 12 with positive probability.
For example,
P(S = 2) = P((1, 1) ) = 1/36,
P(S = 3) = P({(1, 2), (2, 1)}) = 2/36,
P(S = 13) = P( ∅) = 0, because 13 is an impossible outcome.
Another example of a RV is the function M : Ω → R, given by
M( ω1, ω2) =max{ω1, ω2} for (ω1, ω2) ∈ Ω.
Probability Mass Function
The probability mass function (pmf) of a discrete random variable
maps each possible outcome in the sample space to it's corresponding probability.
As an example we give the probability mass function p of M
The sum of the probabilities of all possible outcomes will always be equal to 1.
Example 3.1. (Baron)
Consider an experiment of tossing 3 fair coins and counting the number of heads.
Let X be the number of heads . Prior to an experiment, its value is not known. All we
can say is that X has to be an integer between 0 and 3. We can compute
Hence X is a discrete RV with the following pmf:
We know that X as a RV is a function M : Ω → R.
What is Ω here?
Probability Distribution Function
The distribution function of a random variable X, also referred to as the
cumulative distribution function (CDF) yields the probability
that X will take a value less than or equal to a.
Hence, the value of F(a) is equal to the sum of all probabilities
of outcomes less than or equal to a:
F (a)   P( X  ai )  P( X  a)  P( X  a)
ai a
If we are given a CDF, we can get the pmf with the following formula:
P( X  a)  P( X  a)  P( X  a)  P( X  a)  P( X  a   )  F (a)  F (a   )
for some sufficiently small ε>0.
Graphs of pmf and CDF
Probability Mass Function
Cumulative Distribution Function
F (a)   P( X  ai )  P( X  a)  P( X  a)
ai a
Example 3.3 (Baron) (Errors in independent modules)
A program consists of two modules. The number of errors
X1 in the first module has the pmf P1(x), and the number
of errors X2 in the second module has the pmf P2(x),
independently of X1 given by the table. Find the pmf and
cdf of Y = X1 + X2, the total number of errors.
Solution. We break the problem into steps. First, determine all possible values of Y, then
compute the probability of each value. Clearly, the number of errors Y is an integer
that can be as low as 0 + 0 = 0 and as high as 3 + 2 = 5. Since P2(3) = 0, the second
module has at most 2 errors.
Bernoulli Distribution
The Bernoulli distribution is used to model an
experiment with only two outcomes, success
and failure. The parameter p is the chance for
An example is flipping a coin, where “heads”
may be success and “tails” may be failure.
Binomial Distribution
The Binomial Distribution represents multiple
Bernoulli trials. The parameter n is the number of
trials, and the parameter p is the probability of
success as in the Bernoulli distribution.
P(X=k) is the probability of k successful outcomes in
n trials.
Probability mass function and distribution function of the Bin(10, 1/4 ) distribution.
See Section 4.3
Geometric Distribution
A geometric distribution gives information about the
probability of success after k attempts. The parameter
p is the probability of success on the kth try.
This means that all previous k-1 tries failed
An example of this would be finding the probability that
you will hit a bullseye with a dart on your kth toss.
We had a geometric distribution in Ch. 2:
If we flip a coin until it lands on heads, the random
variable X of the experiment is the number of times the
coin needs to be flipped until heads came up.
If the chance of landing on heads is p, the chance of
landing on tails is 1-p. Therefore:
P(X=1) = p. P(X=2) = (1 – p)1p. P(X=3) = (1-p)2p
and in general P(X=n) = (1-p)n-1p
The pmf and CDF
for p=1/4,
i.e., of Geo(1/4)