ProbabilityDistributionsContinuousRandomVariables
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Transcript ProbabilityDistributionsContinuousRandomVariables
Probability
Distributions
Continuous Random Variables
Continuous Random Variables
A random variable was a numerical value
associated with the outcome of an experiment.
Finite
discrete random variables were ones in which
the values were countable whole numbered values
A continuous random variable is a random
variable that can assume any value in some
interval of numbers, and are thus NOT
countable.
Examples:
The time that a train arrives at a specified stop
The lifetime of a transistor
A randomly selected number between 0 and 1
Let R be a future value of a weekly ratio of closing prices for
IBM stock
Let W be the exact weight of a randomly selected student
Continuous Random Variables
A random variable is said to be continuous if there is a function fX(x)
with the following properties:
Domain: all real numbers
Range: fX(x)≥0
The area under the entire curve is 1
Such a function fX(x) is called the probability density function
(abbreviated p.d.f.)
The fact that the total area under the curve fX(x) is 1 for all X values
of the random variable tells us that all probabilities are expressed in
terms of the area under the curve of this function.
Example: If X are values on the interval from [a,b], then the P(a≤X≤b) =
area under the graph of fX(x) over the interval [a,b]
fX
A
a
b
Continuous Random Variables
Because all probabilities for a continuous
random variable are described in terms of the
area under the p.d.f. function, the P(X=x) = 0.
Why:
the area of the p.d.f. for a single value is zero
because the width of the interval is zero!
That is, for any continuous random variable, X,
P(X = a) = 0 for every number a. This DOES NOT
imply that X cannot take on the value a, it simply
means that the probability of that event is 0.
Continuous Random Variables
Rather than considering the probability of X taking on a
given single value, we look for the probability that X
assumes a value in an interval.
Suppose that a and b are real numbers with a < b.
Recall that X a is the event that X assumes a value in
the interval(, a]. Likewise, a < X b and b < X are the
events that X assumes values in (a, b] and (b, ),
respectively. These three events are mutually exclusive
and at least one of them must happen. Thus,
P(X a) + P(a < X b) + P(b < X) = 1.
Since we are interested in the probability that X takes a
value in an interval, we will solve for P(a < X b).
Continuous Random Variables
P X a P a X b P b X 1
P a X b 1 P X a P b X
P a X b 1 P X a 1 P X b
P a X b 1 P X a 1 P X b
P a X b P X b P X a
P a X b FX (b) FX ( a )
Because X is a continuous random variable, P(X = a) =
0 and P(X = b) = 0. Thus, it makes no difference
whether or not we include the end points in an interval.
A FX (b) FX (a) P(a X b)
P ( a X b)
P ( a X b)
P(a X b).
The cumulative distribution function
The same probability information is often
given in a different form, called the
cumulative distribution function, (c.d.f),
FX(x)
FX(x)=P(Xx)
0 FX(x) 1, for all x
Domain is all real numbers
Example
The p.d.f. of T, the weekly CPU time (in
hours) used by an accounting firm, is
given below.
if t 0
0
3 2
fT (t ) t (4 t ) if 0 t 4
64
1
if t 4
Example (cont)
The graph of the p.d.f. is given below:
0.5
0.4
0.3
f T (t )
0.2
0.1
0
-4
-2
-0.1
0
2
t
4
6
Example (cont)
P(1 T 2) is
equal to the area between the
graph of and the t-axis over the interval.
0.5
0.4
0.3
f T (t )
0.2
0.1
0
-4
-2
-0.1
0
2
t
4
6
Another Example
The c.d.f. of T (for the previous example)
is given below.
if t 0
0
1
FT (t ) t 3 (16 3t ) if 0 t 4
256
1
if t 4
Find
P(1 T 2)
The graph of the c.d.f.
1.2
1.0
0.8
0.6
F T (t )
0.4
0.2
0.0
-4
-2
-0.2 0
2
t
4
6
Solution
P (1 T 2) P (T 2) P (T 1)
P (T 2) P (T 1)
FT (2) FT (1)
1
1 3
3
(2 )(16 3 2)
(1 )(16 3 1)
256
256
0.2617
Expected Value—Continuous
Random Variable
For discrete finite random variables, the
expected value was determined by taking
each value of the random variable and
multiplying it by the corresponding
probability as stated by:
E( X )
x fX ( x)
all x
Expected Value (cont)
For a continuous random
variable, the process is more
tricky.
For example, suppose we
wanted to know the expected
value of X which is defined on
the interval [0,2].
Unfortunately, the P(X=a) for
any number is 0
So when we try to compute the
expected value we’re adding
up a whole bunch of zeros
X=x
fX(x)
x fX(x)
0
fX(0)
0 fX(0)
1
fX(1)
1 fX(1)
2
fX(2)
2 fX(2)
Expected Value (cont)
We need some other way of looking for the
expected value for a continuous random
variable.
Unfortunately, this requires calculus which you
will learn more about in 115b
Basically we need to find the area under the
entire curve for the function x fX(x)
We can’t do this without knowing some calculus
so we’ll give a more geometric interpretation of
the E(X) for a continuous random variable.
Expected Value—Geometric
Interpretation
We can use the probability density function to give a geometric
interpretation for the mean of a continuous random variable, X.
Suppose that we draw the p.d.f. on a thin sheet of metal and cut out
the region between the graph of f(X) and the x-axis. If we place a
knife edge under a line through on the x-axis and perpendicular to
that axis, then the metal sheet will balance on that edge.
fX
X
A more concrete example
Consider the p.d.f. shown below for a
continuous random variable X
p.d.f. of X
0.4
0.3
f X (x )
0.2
0.1
0
0
1
2
3
4
x
5
6
7
8
Expected Value—Geometric
Interpretation
Note: The mean does not occur at the highest
point of the graph!
x-coordinate
of highest point is the mode.
In the previous example, this was at x = 1
Note: The mean does not divide the area in
half!
x-coordinate
that does this is the median.
In previous example, the median is approximately
0.3466.
Special Distribution
A continuous uniform random variable
is a random variable defined on an interval
such that every subinterval of having the
same length has the same probability.
If X is a continuous uniform random
variable on the interval , then
if x a
0
1
f X ( x)
if a x b
b a
if x b.
0
if x a
0
x a
FX ( x)
if a x b
b a
if x b.
1
Example—Uniform Distribution
A bus arrives at a bus stop every 10
minutes. Let W be the waiting time (in
minutes) until the next bus. The p.d.f. and
c.d.f. of W are given below.
0 if w 0
1
fW ( w) if 0 w 10
10
0 if w 10
0 if w 0
w
FW ( w)
if 0 w 10
10
if w 1
1
Uniform Distribution: p.d.f. & c.d.f.
0.120
0.100
0.080
f W (w )
0.060
0.040
0.020
0.000
-5
-0.020 0
5
10
15
10
15
w
1.2
1.0
0.8
F W (w )
0.6
0.4
0.2
0.0
-5
-0.2 0
5
w
Questions:
P(4 W 6)
Find
Find E(X)
Notice
.
that The expected value of W is given
by:
E (W ) W
0 10
5
2
Special Distribution
An exponential random variable may be
used to model the length of time between
consecutive occurrences of some event in
a fixed unit of space or time.
If X is an exponential random variable with
parameter, then
0
f X ( x) 1 x /
e
if x 0
if x 0.
0
FX ( x)
x /
1
e
if x 0
if x 0.
Example
On average, three customers per hour use
the ATM in a local grocery store. Let T be
the time (in minutes) between consecutive
customers. The p.d.f. and c.d.f. of T are
given below.
0
if t 0
fT (t ) 1 t / 20
e
if t 0
20
if t 0
0
FT (t )
t / 20
1
e
if t 0
Exponential Distribution: p.d.f. &
c.d.f.
0.06
0.05
0.04
0.03
f T (t )
0.02
0.01
0.00
-20 -0.01 0
20
40
60
80
100
120
t
1.2
1.0
0.8
F T (t )
0.6
0.4
0.2
0.0
-20 -0.2 0
20
40
60
t
80
100
120
Questions:
P(T 15)
Find
The expected value cannot be determined
from the p.d.f. function without using
calculus so we’ll simply tell you that:
E( X ) X
The expected value of T is given by:
E (T ) T 20
Summary of Distributions
RANDOM VARIABLE, X
Type
Values
Finite
A finite set of numbers
x1, x2, x3, ¼, xn
Probability Mass Function, fX
p.m.f.
P(X = x) = fX(x)
Probability
fX(x)
Continuous
All numbers in an interval
Probability Density Function, fX
p.d.f.
area
under the
P ( a X b)
graph of f X
over [a, b]
fX(x)
x
a
b
x
Summary of Distributions
RANDOM VARIABLE, X
Type
Values
Cumulative
Probability
Finite
A finite set of numbers
x1, x2, x3, ¼, xn
Cumulative Distribution
Function, FX
c.d.f.
P(X x) = FX (x)
Continuous
All numbers in an interval
Cumulative Distribution
Function, FX
c.d.f.
P(X x) = FX (x)
FX(x)
FX(x)
x
x
Expected Values
Finite Discrete Random Variables:
E( X )
x fX ( x)
all x
Continuous Random Variables:
Distribution: E ( X ) X a b
2
Exponential Distribution: E ( X ) X
Uniform
Additional Information
See Distribution Study Guide under
“Worksheets” link on my webpage
Make sure you are able to distinguish
between the various types of distributions,
their expected values, the c.d.f. and p.d.f.