Transcript Lecture_10
Everything existing in the universe is the fruit of chance.
Democritus
Lecture 10: Continuous RV
12
Probability Theory and Applications
Fall 2005
September 29
September 29
To Come
• More on MGF after we learn continuous
distributions.
• Test 10/2
– Covers materials up through this lecture
– Bring Calculator
– Bring one page of notes (both sides fine)
– Sample exam on web (sorry no answers)
WARNING
To properly specify a CDF you give it for all
possible values.
RIGHT
F(x)
x
0
x<1
1/55
1≤x<2
5/55
2 ≤ x<3
14/55
3 ≤ x<4
30/55
4 ≤ x<5
55/55
5≤x
WRONG WRONG WRONG
F(x)
x
0
0
1/55
1
5/55
2
14/55
3
30/55
4 ≤ x<5
55/55
5
• WRONG
• WRONG
• WRONG
Outline
• Motivating Example for CRV
• Continuous Random Variables
• Sample types of problems
Imagine….
Driving down on a 10 mile stretch of
highway near Roswell New Mexico.
Suddenly
a UFO appears
The spaceship
bathes you in bright light.
Then
the spaceship, you and your car disappears.
?
Alien Abduction Problem
Imagine Fox and Mulder are driving down a
10 mile stretch of highway and they will be
abducted by aliens stretch of highway.
What is the probability they will be abducted
in the first 10 miles assuming that their
chance of getting abducted as any point of
the road is equally likely?
Discrete Version
There are mile markers that divide highway
into 10 segments.
Let X={1,2,..,10} be the probability you
vanish after x-1 and up to mile marker x.
X is discrete uniform.
P(X=x)=1/10 x=1,..,10
x0
0
Note and sketch CDF x
F ( x) P( X x)
10
1
0 x 10
o.w
Continuous Version
Let X needs to be a real random variable
since we could disappear at about 3.2
miles and that is different than 3.9 miles.
Uniform assumption
x0
0
x
F ( x) P( X x)
0 x 10
10
o.w
1
1
0
0
10
Looks good
Probability they disappear in the first half
P(X≤5)=5/10=1/2
Seem like right cdf. What would the “pdf”
be?
Probability of small interval
Probably disappear between point a and b =
F(b)-F(a).
Probability disappear on a very small
interval.
Let 0
P( x X x )
F ( x ) F ( x)
f ( x) (deriv. of F )
by fundamental theorem of calculus
Alien Abduction PDF
• Differentiate CDF to get PDF
F ( x ) 1
x
10
.1
0
10
Continuous RV
X is a continuous R.V., if and only if
F(x)=P(X≤x) is a continuous function from
the reals to [0,1]
If F(x) is an integral of
some function f(x)≥0
x
of the form F ( x) f ( x)dx
then f(x) is
called a probability density function p.d.f
Back to example
Cdf
pdf
0
x
F ( x)
200
1
x0
0 x 200
x 200
1/ 200 0 x 200
f ( x)
otherwise
0
PDF
If F(X) is the cdf of a random variable and
F(x) is an integral of some function f ( x) 0
x
of the form F ( x)
f (t )dt
then f ( x) is called the probability density
function (pdf)
Recall
Fundamental theorem of Calculus
a
F (a) F (b) f ( x)dx
b
P( x X x ) F ( x ) F ( x)
P( x X x )
F ( x ) F ( x)
f ( x)
dF ( x)
f ( x)
is the pdf of X
dx
Alien Abudction
• CDF
x0
0
x
F ( x) P( X x)
0 x 10
10
o.w.
1
• PDF
1/10 0 x 10
f ( x)
otherwise
0
Probability of Event
Let X be a continuous R.V. with cdf F(x) and
pdf f(x).
Let A be an event (subset of R).
p( A)
xA
f (t )dt
Alien Abduction
Probability abducted in 1.3 to 2.4 miles
2.4
P(1.3 X 2.4)
1
2.4 1.3 1.1
dt
1.3 10 10 10 10 .21
F (2.4) F (1.3) .24 .13 .21
Alien Abduction
Probability abducted at 1.3 miles
1.3
P (1.3 X 1.3)
1
1.3 1.3
dt
1.3 10 10 10 0
F (1.3) F (1.3) 0
The probability X=x for any x is 0!!
Note
Let X be a R.V. with pdf f(x)
f (t )dt 1
Problem Type I
Given that x has pdf
f ( x) cx(1 x)
0
for 0 x 1
0.w.
Find c
1
1
f ( x)dx c ( x x 2 )dx
0
2
3 1
x
x
c c / 6 1
2 3 0
c6
continued
6 x(1 x)
f ( x)
0
pdf of X is
for 0 x 1
otherwise
f(x)
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
-1
-0.5
0
0.5
x
1
1.5
2
Problem Type 2
Find cdf of X for previous problem
0
F ( x)
x
f (t )dt 6 ( x x 2 )dt
0
x
t2 t3
6 3x 2 2 x3
2 30
0
F ( x) 3x 2 2 x3
1
x0
0 x 1
o.w.
Problem Type 3
Find P(1/4<x<4)
Using cdf
Using pdf
F(4)-F(1/2)=1-(3/4-1/4)=1/2
4
1
1/2
1/2
6 f (t )dt 6 ( x x 2 )dt
1
t2 t3
6 1/ 2
2 3 1/2
Problem Type 3
Find P(X≤1/3|X≤1/2)
P( X 1/ 3)
P( X 1/ 3 | X 1/ 2)
P( X 1/ 2)
F (1/ 3)
P( X 1/ 3 | X 1/ 2)
14 / 27
F (1/ 2)