Download Discrete Random Variables

Discrete Random Variables
Numerical Outcomes
• Consider associating a numerical value with each
sample point in a sample space.
(1,1)
(2,1)
(3,1)
(4,1)
(5,1)
(6,1)
(1,2)
(2,2)
(3,2)
(4,2)
(5,2)
(6,2)
(1,3)
(2,3)
(3,3)
(4,3)
(5,3)
(6,3)
(1,4)
(2,4)
(3,4)
(4,4)
(5,4)
(6,4)
(1,5)
(2,5)
(3,5)
(4,5)
(5,5)
(6,5)
(1,6)
(2,6)
(3,6)
(4,6)
(5,6)
(6,6)
:
9
10
11
12
• For example, associate each roll of 2 dice with
the sum of their faces.
Random Variable
• A real-valued function f : S 
whose domain is the sample space S
is a random variable for the experiment.
S
:
(3, 6)
(4, 6)
(5, 5)
(6, 4)
(5, 6)
:
f
:
9
10
11
12
Y
• We refer to values of the random variable as
events. For example, {Y = 9}, {Y = 10}, etc.
Probability Y = y
• The probability of an event, such as {Y = 9}
is denoted P(Y = 9).
• In general, for a real number y,
the probability of {Y = y} is denoted
P(Y = y), or simply, p( y).
• P(Y = y) is the sum of probabilities for sample
points which are assigned the value y.
• When rolling two dice,
P(Y = 10) = P({(4, 6)}) + P({(5, 5)}) + P({(6, 4)})
= 1/36 + 1/36 + 1/36
= 3/36
Discrete Random Variable
• A discrete random variable is a random variable
that only assumes a finite (or countably infinite)
number of distinct values.
• For an experiment whose sample points are
associated with the integers or a subset of integers,
the random variable is discrete.
• For an experiment whose sample points are
associated with the reals or an interval of real
numbers, the random variable is not discrete.
(Chapter 4 considers “continuous random variables”.)
“Towards Statistics”
• Studying probability, we learn about certain types
of random variables that occur frequently in
practice and their probability distributions.
Knowledge about the probabilities of these
common random variables may help us make
appropriate inferences about a population.
(page 84).
• Remember, our theoretical models represent the
distribution that may be expected but may differ
from the actual frequencies resulting from an
experiment. (page 87)
Probability Distribution
• A probability distribution describes the probability
for each value of the random variable.
y
2
3
4
5
6
7
8
9
10
11
12
p(y)
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
2
3
4
5
6
7
8
9
10
11
12
Presented as a table, formula, or graph.
Probability Distribution
• For a probability distribution:
y
2
3
4
5
6
7
8
9
10
11
12
p(y)
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
= 1.0
 p( y)  1
y
Here we may take the sum
just over those values of y
for which p(y) is non-zero.
And, of course,
0  p( y )  1, for all y.
Expected Value
• The “long run theoretical average”
• For a discrete R.V. with probability function p(y),
define the expected value of Y as:
E (Y )   y p( y )
y
• In a statistical context,
E(Y) is referred to as the mean and so
E(Y) and m are interchangeable.
An average circle?
• Suppose a class has cut some circles out of
construction paper, as described below.
Radius(inches)
3
4
6
frequency
150
200
125
Let radius be the random variable, R.
Compute E(R).
Function of a Random Variable
• Suppose g(Y) is a real-valued function of a
discrete random variable Y.
s
y
S
g(y)
Y
g(Y)
• It follows g(Y) is also a random variable with
expected value
E[ g (Y )]   g ( y) p( y)
y
Expected Circumference
• Consider the function C = 2pR.
• C is a function of a random variable, and so C is
also a random variable.
Radius(inches)
3
4
6
circumference
6p
8p
12p
frequency
150
200
125
• Compute the expected value, E(C).
For a constant multiple…
• Of course, a constant multiple may be factored out
of the sum
E (cY )   (c y ) p ( y )
y


 c   y p ( y )   cE ( y )
 y

• Thus, for our circles, E(C) = E(2pR) = 2pE(R).
For a constant function…
• In particular, if g(y) = c for all y in Y,
then E[g(Y)] = E(c) = c.
E (c )   c p ( y )
y


 c   p ( y )   (c)(1)  c
 y

For sums of variables…
• Also, if g1(Y) and g2(Y) are both functions of the
random variable Y, then
E[ g1 (Y )  g 2 (Y )]   ( g1 (Y )  g 2 (Y )) p( y )
y
 [ g1 (Y ) p ( y )  g 2 (Y ) p( y )]
y
  g1 (Y ) p ( y )   g 2 (Y ) p ( y )
y
y
E[ g1 (Y )]  E[ g 2 (Y )]
All together now…
• So, when working with expected values, we have
E[ g1 (Y )  g 2 (Y )]  E[ g1 (Y )]  E[ g 2 (Y )]
E (cY )  cE ( y), and E (c)  c.
• Thus, for a linear combination Z = c g(Y) + b,
where c and b are constants:
E ( Z )  E[cg (Y )  b]
 E[cg (Y )]  E (b)
 c E[ g (Y )]  b
Variance, V(Y)
• For a discrete R.V. with probability function p(y),
define the variance of Y as:
V (Y )  E[(Y  m )2 ]
• Here, we use V(Y) and s2 interchangeably to
denote the variance.
• As previously, the positive square root of the
variance is the standard deviation of Y.
Computing V(Y)
• Based on our definition of the variance of Y
V (Y )  E[(Y  m ) ]
2
• And applying our rules for expected value, we
find variance may be expressed as
V (Y )  E[(Y  m ) 2 ]  E[Y 2  2Y m  m 2 ]
 E[Y 2 ]  (2 m ) E[Y ]  E[ m 2 ]
(as the mean is
a constant)
 E[Y 2 ]  (2m )( m )  m 2
 E[Y ]  m or E[Y ]   E[Y ]
2
2
2
2
“Moments and Mass”
• Note the probability function p(y) for a
discrete random variable is also called a
“probability mass function”.
• The expected values E(Y) and E(Y2) are
called the first and second moments,
respectively.
Did you compute mass and moments in Calculus?
Variance of R?
• For our collection of circles, determine the
variance of the random variable R
Radius(inches)
3
4
6
frequency
150
200
125
It’s your choice. Which formula is easier?
V ( R)  E[( R  m R ) 2 ] or
V ( R)  E[ R ]   E[ R]
2
2
Variance of C?
• For our collection of circles, determine the
2
2
variance, V (C )  E[C ]   E[C ]
Radius(inches)
3
4
6
circumference
6p
8p
12p
frequency
150
200
125
• It can be shown that
V (cY )  c V (Y )
2
Verify that
V (2p R)  (2p )2V ( R)