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.

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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