c4 discrete random variables n.
Skip this Video
Download Presentation

Loading in 2 Seconds...

play fullscreen
1 / 14


  • Uploaded on

C4: 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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

PowerPoint Slideshow about 'C4: DISCRETE RANDOM VARIABLES' - russ

Download Now An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
c4 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

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
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 probabilities

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
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:

If we are given a CDF, we can get the pmf with the following formula:

for some sufficiently small ε>0.

graphs of pmf and cdf
Graphs of pmf and CDF

Cumulative Distribution Function

Probability Mass Function


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
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 success.
  • An example is flipping a coin, where “heads” may be success and “tails” may be failure.
binomial distribution
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
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

Example: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)2pand in general P(X=n) = (1-p)n-1p

We had a geometric distribution in Ch. 2:

The pmf and CDFfor p=1/4, i.e., of Geo(1/4)