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Random Variable

Outcome of random experiment can be
 Categorical or Numerical

However, we often want to represent outcomes as
numbers.

Random variable is a function that associates a
unique numerical value with every outcome of an
experiment.
Discrete vs. Continuous
Random Variable
Discrete
Binomial
Continuous
Uniform
Normal
t
F
χ2
Discrete Random Variable

Discrete random variable, X, is a random
variable with a finite or countable number of
possible outcomes.

Notation:
 X = discrete random variable,
 k = a number that the discrete random variable
could assume,
 P(X=k) is the probability that the random
variable X equals k.
Probability Distribution Function

The probability distribution function (pdf) for a
discrete random variable X is a table or rule that
assigns probabilities to the possible values of X.
X
x1
x2
x3
P(X)
P(X= x1) = p1
p2
p3
…
…
Example:
X – Number of Times Go Out (per week)
X
0
1
2
3
4
5
6
7
Total
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
1.000
Probability Distribution
Two main conditions:
 The sum of all the individual probabilities (p1,
p2, …) must equal to 1.
 The individual probabilities must be between 0
and 1.

Probability histogram can be used to display the
distribution for a discrete random variable
 Use x-axis to represent the values or outcomes
and y-axis for the corresponding probabilities.
Cumulative Distribution

Cumulative distribution function (cdf) for r.v. X
is a rule that provides the cumulative probabilities
P(X≤k), i.e., the probabilities that X is less than or
equal to k.
X
x1
x2
x3
…
P(X)
P(X= x1) = p1
P(X≤x2)=
p 2 + p1
p2 + p1 +p3
…
Example

A psychology experiment on the behavior of
young children involves placing a child in a
designed area with 5 toys.

Over a fixed time period various observations
were made, including the number of toys the child
plays with.
Example:

Based on many kids, the following probability
distribution was determined:
X=# toys
0
1
2
3
4
P(X=k)
0.03
0.16
0.3
0.23
0.17

P(X=5) =
5
?
Example:

Based on many kids, the following probability
distribution was determined:
X=# toys
0
1
2
3
4
P(X=k)
0.03
0.16
0.3
0.23
0.17

5
0.11
P(X=5) = 1 – (0.03 + 0.16 +0.3 +0.23 +0.17) =
= 1 – 0.89 = 0.11
Example:

X=#
Cumulative Distribution of X
0
1
2
3
4
5
0.03
0.03+0.16
0.19+0.3
0.72
0.89
1
=0.19
=0.49
toys
P(X)

What is the probability that a randomly chosen kid
will play with at most 2 toys?

P(X<=2) = P(X=0) + P(X=1) +P(X=2)
Example:

X=#
Cumulative Distribution of X
0
1
2
3
4
5
0.03
0.03+0.16
0.19+0.3
0.72
0.89
1
=0.19
=0.49
toys
P(X)

What is the probability that a randomly chosen kid
will play with at most 2 toys?

P(X<=2) =0.03+0.16 + 0.3 = 0.49
Example:

X=#
Cumulative Distribution of X
0
1
2
3
4
5
0.03
0.03+0.16
0.19+0.3
0.72
0.89
1
=0.19
=0.49
toys
P(X)

What is the probability that a randomly chosen kid
will play with at least 3 toys?

P(X>=3) =
Example:

X=#
Cumulative Distribution of X
0
1
2
3
4
5
0.03
0.03+0.16
0.19+0.3
0.72
0.89
1
=0.19
=0.49
toys
P(X)

What is the probability that a randomly chosen kid
will play with at least 3 toys?

P(X>=3) = [use complement rule] = 1-P(X<=2)
= 1-0.49 = 0.51
Example:

Given the child has played with at least 3 toys,
what is the probability this child will play with all
5 toys?
X=# toys
0
1
2
3
4
P(X=k)
0.03
0.16
0.3
0.23
0.17

Event of interest: X=5

P(X=5|X>=3) = (X=5 and X>=3)/P(X>=3)
5
0.11
Condition X>=3
Example:

Given the child has played with at least 3 toys,
what is the probability this child will play with all
5 toys?
X=# toys
0
1
2
3
4
P(X=k)
0.03
0.16
0.3
0.23
0.17
5
0.11

Event of interest: X=5
Condition X>=3

P(X=5|X>=3) = (X=5 and X>=3)/P(X>=3) =
P(X=5)/ P(X>=3) = 0.11/0.51 =0.2156
Example:
X – Number of Times Go Out (per week)
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
Given that a student goes
out at least 2 times per week,
what is the probability that
goes out 5 or more times per
week?
Event of interest: X>=5
Condition X>=2
P(X>=5| X>= 2) = ?
Example:
X – Number of Times Go Out (per week)
Event of interest: X>=5
Condition X>=2
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
1) P(X>=5| X>= 2) =
P(X>=5 and X>=2)/P(X>=2)
2) P(X>=2) = 1-(P(X=0) +
P(X=1)) = 1 –(0.074+0.167)
= 1- 0.241 = 0.759
Example:
X – Number of Times Go Out (per week)
Event of interest: X>=5
Condition X>=2
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
1) P(X>=5| X>= 2) =
P(X>=5 and X>=2)/P(X>=2)
2) P(X>=2) = 0.759
3) P(X>=5 and X>=2) =
P(X>=5) =
0.083+0.019+0.037 =0.139
4) P(X>=5| X>= 2) =
= 0.139/ 0.759 = 0.1831
Expected Value

Expected value, E(X), of a random variable X is
the mean value in the sample space of possible
outcomes (= population mean).

Expected value is interpreted as the mean value
that would be obtained from an infinite (total)
number of observations on the random variable.
Expected Value

Probability distribution function (pdf) of X.
X
x1
x2
x3
P(X)
P(X= x1) = p1
p2
p3
…
…
Example:
X=# toys
0
1
2
3
4
P(X=k)
0.03
0.16
0.3
0.23
0.17

5
0.11
What is the expected number of toys played with?
E(X) = 0*0.03 + 1*.16 + 2*0.3 + 3*0.23 +4*0.17 +
5*0.11 = 2.68 toys
Note: we never round E(X)! E(X) may not be value that
ever expected on a single random outcome. Instead, it
is the average over the long run.
Example:
X – Number of Times Go Out (per week)
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
E(X) = ?
Example:
X – Number of Times Go Out (per week)
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
E(X) = 0*0.074 + 1*0.167
+2*0.287 + 3*0.213 + 4*0.12
+5*0.083 +6*0.019 +7*0.037
= 2.648 times
Standard Deviation (std)

Probability distribution function (pdf) of X.
X
x1
x2
x3
P(X)
P(X= x1) = p1
p2
p3
…
…
Example (μ =E(X) = 2.68 toys)
X=# toys
0
1
2
X-μ
-2.68
-1.68
-.68
(X-μ)2
7.182
P(X=k)
0.03
0.16
0.3
(X-μ)2
0.215
P(X=k)
3
4
5
0.23
0.17
0.11
Example:
X – Number of Times Go Out (per week)
X
0
1
2
3
4
5
6
7
P(X=k)
0.074
0.167
0.287
0.213
0.120
0.083
0.019
0.037
Std(X) = sqrt( 0^2*0.074
+1^2*0.167 + 2^2 *0.287
+ 3^2*0.213 + 4 ^2*0.12
+5^2*0.083 +6^2*0.019
+7^2*0.037 – 2.648^2) =
Example: Different strategies sampling 2
out of numbers {1,2,3}:



Sampling WITHOUT replacement,
Order IS important
Number of Samples = 6
(1,2)
(2,1)
(3,1)
(1,3)
(2,3)
(3,2)
Example: Different strategies sampling 2
out of numbers {1,2,3}:



Sampling WITHOUT replacement,
Order IS NOT important
Number of Samples = 3
(1,2)
(1,3)
(2,3)

Usually, sample without replacement, thereby
ensuring the sample is comprised of different
individuals!
Permutations

When sampling from the population WITHOUT
replacement of the subjects and the order of
sampled individuals IS important, the number of
different arrangements, permutations of n
subjects from a population of size N:

where N! = [N factorial] =1*2*3*…*(N-1)*N

0! = 1
Example: Permutations

Locker combination (order is important)
comprised of 3 distinct digits. Compute the
number of different locker combination .
N = _____________
n = ______________
P(N_n) = N!/(N-n)! =
Example: Permutations
Locker combination (order is important)
comprised of 3 distinct digits. Compute the
number of different locker combination .
N = 10
n=3
P(N _ n) = N!/(N-n)! = 10!/(10-3)! =
= (1*2*…*7*8*9*10)/ (1*2*…*7) = 720

Permutations
Simple Properties
General Formula
Combinations

When sampling from the population WITHOUT
replacement of the subjects and the order of
sampled individuals IS NOT important, the
number of different samples, combinations of n
subjects from a population of size N:
Example

Consider sampling 3 students from a class of 10
people.

Compute the number of different samples.

In this situation there will be no replacement, and
the order of students in a sample is not important.
N = _____________
n = ______________
C(N choose n) = N!/n!(N-n)! =
Example

Consider sampling 3 students from a class of 10
people.

Compute the number of different samples.
N =10
n=3
C(N choose n) = N!/n!(N-n)! =10!/(3!7!) =
=8*9*10/(2*3) = 120
Combinations
Simple Properties
General Formula