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STAT 430/510 Lecture 10

STAT 430/510 Probability
Lecture 10: Continuous Random Variable
Pengyuan (Penelope) Wang

June 13, 2011


STAT 430/510 Lecture 10

Introduction

The set of possible values for discrete random variable is
discrete.
However, there also exist random variables who can take
values on a whole interval.


STAT 430/510 Lecture 10


Definition of Continuous Random Variable

X is a continuous random variable if it take continuous
values.


STAT 430/510 Lecture 10

Definition of Continuous Random Variable

X is a continuous random variable if it take continuous
values.
There exists a nonnegative function f , having the property
that, for any set B of real numbers
P(X ∈ B) =

f (x)dx
B

for example, for set B = (a, b), P(X ∈ B) =

b
a

f (x)dx

The function f is called the probability density function of
random variable X .


STAT 430/510 Lecture 10

Definition of Continuous Random Variable

X is a continuous random variable if it take continuous
values.
There exists a nonnegative function f , having the property
that, for any set B of real numbers
P(X ∈ B) =


f (x)dx
B

for example, for set B = (a, b), P(X ∈ B) =

b
a

f (x)dx

The function f is called the probability density function of
random variable X .
f must satisfy: f ≥ 0,

f (x)dx = 1.

Important: how to interpret f ?


STAT 430/510 Lecture 10

How to represent the probability distribution of such random
variables?

Think of PMF.
It is sort of like the continuous version of PMF.
For any set B of real numbers
P(X ∈ B) =

f (x)dx
B

for example, for set B = (a, b), P(X ∈ B) =

b
a

f (x)dx


STAT 430/510 Lecture 10

Example

The amount of time in hours that a computer functions
before breaking down is a continuous random variable with
probability density function given by
f (x) =

1 −x/100
,
100 e

x ≥0
0, x < 0

Confirm that it is a density function.


STAT 430/510 Lecture 10

Probability on an interval

P(a ≤ X ≤ b) =

b
a

f (x)dx

The amount of time in hours that a computer functions
before breaking down is a continuous random variable with
probability density function given by
f (x) =

1 −x/100
,
100 e

x ≥0
0, x < 0

What is the probability that
(a) it will function for 100 - 150 hours?


STAT 430/510 Lecture 10

The probability that a continuous random variable would take
exactly one number is 0.

P(a ≤ X ≤ b) =

b
a

f (x)dx

Then for any value u,
P(X = u) = (u ≤ X ≤ u) =

u
u

f (x)dx = 0

The probability that the computer would function for exactly
100 hours is 0.


STAT 430/510 Lecture 10

Cumulative Probability Function

Cumulative Probability Function:
F (x) = P(X < x) = P(X ≤ x) =

x
∞ f (u)du

The amount of time in hours that a computer functions
before breaking down is a continuous random variable with
probability density function given by
f (x) =

1 −x/100
,
100 e

x ≥0
0, x < 0

What is the probability that
(b) it will function for fewer than 100 hours?


STAT 430/510 Lecture 10

Summary of Properties of Continuous Random Variable

1 = P(X ∈ (−∞, ∞)) =
P(a ≤ X ≤ b) =

b
a


−∞ f (x)dx

f (x)dx

P(X = u) = 0
Cumulative Probability Function:
F (x) = P(X < x) = P(X ≤ x) =

x
−∞ f (u)du


STAT 430/510 Lecture 10

Expected Value

If X is a continuous random variable with probability
density function f (x), then its expected value is,


E[X ] =

xf (x)dx
−∞

And, for any function g,


E[g(X )] =

g(x)f (x)dx
−∞

It is totally analogous to the discrete case.


STAT 430/510 Lecture 10

Expected Value and Variance of Continuous R.V.

For continuous random variable X with pdf f (x),

−∞ xf (x)dx

E[X 2 ] = −∞ x 2 f (x)dx

Var (X ) = −∞ (x − µ)2 f (x)dx

E[X ] =

µ = EX .
SD(X ) =

Var (X )

= E[X 2 ] − (E[X ])2 , where


STAT 430/510 Lecture 10

Properties of Expected Value and Variance: totally the same as the
discrete case

E[aX + b] = aE[X ] + b, where a and b are constants
Var (aX + b) = a2 Var (X ) , where a and b are constants.
E[X + Y ] = E[X ] + E[Y ], where X and Y are random
variables.
Var [aX + bY ] = a2 Var [X ] + b2 Var [Y ], if X and Y are
independent.
They are the same as the discrete case.


STAT 430/510 Lecture 10

Example

Find E[X ] and Var(X) when the density function X is
f (x) =
1
0 x2xdx = 2/3
1
E[X 2 ] = 0 x 2 2xdx = 1/2
Var (X ) = E[X 2 ] − (E[X ])2

2x, 0 ≤ x ≤ 1
0, otherwise

E[X ] =

= 1/18


STAT 430/510 Lecture 10

Example

The density function X is given by
f (x) =
Find E[eX ].
E[eX ] =

1 x
0 e 1dx

=e−1

1, 0 ≤ x ≤ 1
0, otherwise


STAT 430/510 Lecture 10

Uniform Random Variable

A continuous random variable X is said to have a uniform
distribution on the interval [A, B], if X can take any value
in [A, B] and the distribution is flat on every points.
Probability density function:
f (x) =

1
B−A ,

A≤x ≤B
0, otherwise


STAT 430/510 Lecture 10

cdf

For uniform r.v. X on [A, B], the cdf is

0, m < A

x−A
, A≤x ≤B
F (x) =
 B−A
1, x > B


STAT 430/510 Lecture 10

Expected Value and Variance

X is uniform random variable on [A, B].
E[X ] =
Var (X )

A+B
2
2
= (B−A)
12


STAT 430/510 Lecture 10

Example

If X is uniformly distributed over (0,10), calculate the
probability that
(a) X < 3
(b) X > 6
(c) 3 < X < 8
P(X < 3) =

3 1
0 10 dx

=

3
10


STAT 430/510 Lecture 10

Example

If X is uniformly distributed over (0,10), calculate the
probability that
(a) X < 3
(b) X > 6
(c) 3 < X < 8
P(X < 3) =
P(X > 6) =

3 1
3
0 10 dx = 10
10 1
2
6 10 dx = 5


STAT 430/510 Lecture 10

Example

If X is uniformly distributed over (0,10), calculate the
probability that
(a) X < 3
(b) X > 6
(c) 3 < X < 8
3 1
3
0 10 dx = 10
10 1
dx = 25
P(X > 6) = 6 10
8 1
P(3 < X < 8) = 3 10
dx = 12

P(X < 3) =


STAT 430/510 Lecture 10

Example

Buses arrive at a specific stop at 15-minute interval
starting at 7 A.M. That is, they arrive at 7, 7:15, 7:30, 7:45,
and so on. If a passenger arrives at the stop at a time that
is uniformly distributed between 7 and 7:30, find the
probability that he waits
(a) less than 5 minutes for a bus
(b) more than 10 minutes for a bus


STAT 430/510 Lecture 10

Example: Solution

Let X denote the number of minutes past 7 that the
passenger arrives at the stop.
{waiting for < 5min} = {10 < X < 15} ∪ {25 < X < 30}
P(10 < X < 15) + P(25 < X < 30) =
15 1
30 1
1
10 30 dx + 25 30 dx = 3
{waiting for > 10min} = {0 < X < 5} ∪ {15 < X < 20}
5 1
20 1
P(0 < X < 5) + P(15 < X < 20) = 0 30
dx + 15 30
dx =

1
3


STAT 430/510 Lecture 10

last Example

A stick of length 1 is split at a point U that is uniformly
distributed over (0,1). Determine the expected length of the
piece that contains the mid point.


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