Download Definition of Random Variable

— Presentation 10
— Stat 1050 for
computer science'
students
—
— 1-12-2012 G
— 17-1-1434 H
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RANDOM VARIABLES AND
PROBABILITY DISTRIBUTIONS
Definition of Random Variable
A random variable is a function that associates a
real number with each element in the sample
space.
Random variables are often designated by letters
and can be classified as discrete, which are
variables that have specific values, or
continuous, which are variables that can have
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any values within a continuous range.
A random variable, usually written X, is a
variable whose possible values are numerical
outcomes of a random phenomenon. There are
two types of random variables, discrete and
continuous.
Example
Consider an experiment where a coin is tossed
three times. If X represents the number of times
that the coin comes up heads, then X is a
discrete random variable that can only have the
values 0,1,2,3 (from no heads in three successive
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coin tosses, to all heads). No other value is
possible for X.
Example
An example of a continuous random variable
would be an experiment that involves measuring
the amount of rainfall in a city over a year, or
the average height of a random group of 25
people.
Example
Two balls are drawn in succession without
replacement from an urn containing 4 red balls
and 3 black balls. The possible outcomes and the
values y of the random variable Y where is the
number of red balls, are
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Sample space
y
RR
2
RB
1
BR
1
BB
0
Y = number of red balls
Possible values of the random variable is y =
0,1,2
The probability mass function is
Y
0
1
2
f(y)
1/4
1/2
1/4
Discrete Sample Space
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If a sample space contains a finite number n of
different values
x 1 , x 2 ,..., x n
or countably infinite
number of different values
x 1 , x 2 ,..., x n
, it is called a
discrete sample space. Examples of discrete
random variables are; the number of bacteria per
unit area in the study of drug control on bacterial
growth, the number of defective television sets
in a shipment of 100.
Discrete Probability Distributions:
Definition:
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The set of ordered pairs (x, f(x)) is a probability
function,
probability
mass
function
or
probability distribution of the discrete random
variable X if for each possible outcome x,
1.
f ( x)  0
2. 
f ( x)  1
3. P( X
 x)  f ( x)
x
Example
. A random variable X assumes a value equal to
the sum of the rolls of two dice. Since there are
6 possible values for each die, the fundamental
principle of counting asserts that there are 36
possible outcomes of the dice roll. With
unbiased dice, each outcome will have a 1/36
chance of occurring.
Dice Roll (x , y)
x - roll of first die
Number of Outcomes
7
y - roll of second die
2
3
4
5
6
7
8
9
10
11
12
(1,1)
(1,2) , (2,1)
(1,3) , (2,2) , (3,1)
(1,4) , (2,3) , (3,2) , (4,1)
(1,5) , (2,4) , (3,3) , (4,2) , (5,1)
(1,6) , (2,5) , (3,4) , (4,3) , (5,2) , (6,1)
(2,6) , (3,5) , (4,4) , (5,3) , (6,2)
(3,6) , (4,5) , (5,4) , (6,3)
(4,6) , (5,5) , (6,4)
(5,6) , (6,5)
(6,6)
Total
36
So the probability mass function is
x
px(x)
2
3
4
5
6
7
8
1 / 36
2/ 36
3/ 36
4/ 36
5/ 36
6/ 36
5/ 36
1
2
3
4
5
6
5
4
3
2
1
8
9
10
11
12
4/ 36
3/ 36
2/ 36
1/ 36
Example
Tossing 4 Coins one time experiment.
What is the probability distribution of the
discrete random variable X that counts the
number of heads in four tosses of a coin? We
can derive this distribution if we make two
reasonable assumptions:
1. The coin is balanced, so each toss is equally
likely to give a H or T.
2. The coin has no memory, so the tosses are
independent.
The outcome of four tosses is a sequence of
heads and tails such as HTTH. There are 16
possible outcomes in all.
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1
P(X = 0) = 16 = 0.0625
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P(X = 1) = 16 = 0.25
And so on
Now if we want the probability of at least two
heads, then we want
P(X  2) = P(X = 2) + P(X = 3) + P(X = 4) =
6 4 1
  = 0.6875
16 16 16
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And if we want the probability of at least one
head we have two choices
1. We can compute the probability as above:
P(X  1) = P(X = 1) + P(X = 2) + P(X = 3) +
P(X = 4) = 164 + 166  164  161 = 0.9375
2. Or, the easier way is to use the compliment
rule
P(X  1) = 1 – P(X< 1) = 1 – P( X = 0) = 1 = 1 – 0.0625 = 0.9375
Exercise
Show that the following function
f ( x) 
x 1
, x  2,1,2,3 is aprobability
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mass function?
Find p(X>1),p(0<X<2.5) , p(X<2)
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1
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Example
. (1) Find the value of c so that the following
function is a probability distribution of a r.v. x:
P  x   c  x  2 , where x  0,1, 2,3
Solution
P(0)+P(1)+P(2)+P(3)=1 (Note that x can only take
0, 1, 2, 3)
2c+3c+4c+5c=1, so c=1/14.
(2) For this probability distribution, find P(x  2)
P(x  2) =P(x=0) + P(x=1) + P(x=2) = 2c+3c + 4c =
9c = 9/14
Alternatively, P(x  2) = 1- P(x>2) = 1- P(x=3) = 15c = 1- 5/14=9/14.
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Mean and Variance of a discrete r.v

x 
 The mean
 xP  x 
x
(The mean of a r.v. is also called as the
expected value.)
 The variances
 x2    x   x   P  x    x 2  P  x    x2
2
x
x
Var( X )  E ( X )  ( E ( X ))
2
2


The standard deviation = x
13
2
Example
Toss a coin twice, x = # of heads. Find
The probability distribution is
0
x
P(x) 0.25
1
0.5
x
and  .
2
x
2
0.25
 x   xP  x   0  0.25  1 0.5  2  0.25  1
x
(the number of heads we can expect to get if we
toss a coin twice).
 x2   x 2  P  x    x2  02  0.25  12  0.5  22  0.25  12  1.5  1  0.5
x
.
Example
A contractor is required by a county planning
department to submit from 1 to 5 different
forms, depending on the nature of the project.
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Let x = # of forms required of the next
contractor, and px  kx for x=1,2,3,4,5.
(a) What is the value of k?
From the form of P(x), we have
P(1)  P(2)  P(3)  P(4)  P(5)  1 (Note x can only take 1, 2,
3, 4, 5)
That is, 1k + 2k + 3k + 4k + 5k = 1. So k =
1/15 (since P(x) = x/15 are between 0 and 1 for
x = 1, 2, 3, 4, 5 and they sum to 1, k=1/15 is a
valid solution.
(b) What is the probability that at most 3
forms are required?
P(x  3) = p(x=1) + P(x=2)+P(x=3) = 1k + 2k +
3k = 6k = 6/15 = 2/5
(c) What is the expected number (i.e.,
mean) of forms required ?
  1P(1) + 2P(2)+3P(3)+4P(4)+5P(5) = 1k +
2*2k + 3*3k + 4*4k+ 5*5k = 55k = 55/15=3.67
x
15
(d) What is the SD of the number of forms
required?
 x2   x2 P( x)  x2  13 k  23 k  33 k  43 k  53 k  3.672  225k  3.672  225/15  3.672  1.53
 x   x2  1.53  1.24
&&& End of Presentation &&&
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