Continuous distributions - University of Connecticut

CHAPTER 7

Continuous distributions

7.1. Basic theory

7.1.1. Denition, PDF, CDF. We start with the denition a continuous random

variable.

Denition (Continuous random variables)

A random variable X is said to have a continuous distribution if there exists a non-

negative function f = fX such that ?b

P(a X b) = f (x)dx

a

for every a and b. The function f is called the density function for X or the PDF for

X.

M ?-oref

precisely, such

(x)dx = P(-

an

<

X X

is

<

said to

) = 1.

hIanvepaarnticaublsaorl,utPe(lyXc=onati)n=uo?uasa

distribution. Note that

f (x)dx = 0 for every a.

E?-xamf (pxl)edx7=.1.1

Suppose and

we

are given

?

that

f (x) = c/x3 ? 1

for

c

x

c f (x)dx = c

dx = ,

-

1 x3

2

we have c = 2.

1 and 0 otherwise. Since

PMF or PDF?

Probability mass function (PMF) and (probability) density function (PDF) are two

PDF names for the same notion in the case of discrete random variables. We say

or

simply a density function for a general random variable, and we use PMF only for

discrete random variables.

Denition (Cumulative distribution function (CDF))

The distribution function of X is dened as

?y F (y) = FX (y) := P(- < X y) = f (x)dx.

-

It is also called the cumulative distribution function (CDF) of X.

97

98

7. CONTINUOUS DISTRIBUTIONS

We can dene CDF for any random variable, not just continuous ones, by setting F (y) := P(X y). Recall that we introduced it in Denition 5.3 for discrete random variables. In

that case it is not particularly useful, although it does serve to unify discrete and continuous random variables. In the continuous case, the fundamental theorem of calculus tells us,

f provided satises some conditions, that

f (y) = F (y) .

By analogy with the discrete case, we dene the expectation of a continuous random variable.

7.1.2. Expectation, discrete approximation to continuous random variables. Denition (Expectation)

expec- X f For a continuous random variable with the density function we dene its

tation by

?

EX = xf (x)dx

-

integrable X if this integral is absolutely convergent. In this case we call

.

Recall that this integral is absolutely convergent if

? |x|f (x)dx < .

-

In the example above,

EX

=

?

1

2 x dx

x3

=

? 2

1

x-2dx

=

2.

Later in Example 10.1 we will see that a continuous random variable with Cauchy distribution

has innite expectation.

Proposition 7.1 (Discrete approximation to continuous random variables)

X Suppose

is a nonnegative continuous random variable with a nite expectation.

Then there is a sequence of discrete random variables {Xn} n=1 such that

EXn --- EX. n

X Proof. First observe that if a continuous random variable

is nonnegative, then its

density f (x) = 0 x < 0. In particular, F (y) = 0 for y 0, thought the latter is not needed

for our proof. Thus for such a random variable

? EX = xf (x)dx.

0

Suppose n N, then we dene Xn() to be k/2n if k/2n X() < (k+1)/2n, for k N{0}.

X 2 This means that we are approximating

from below by the largest multiple of -n that is

still below the value of X . Each Xn is discrete, and Xn increase to X for each S.

7.1. BASIC THEORY

99

Consider the sequence {EXn} n=1. This sequence is an increasing sequence of positive num-

bers, and therefore it has a limit, possibly innite. We want to show that it is nite and it

is equal to EX .

We have

k

k

EXn =

2n P Xn = 2n

k=1

k

k

k+1

=

2n P 2n

X< 2n

k=1

k ? (k+1)/2n

=

f (x)dx

2n

k=1

k/2n

? (k+1)/2n k

=

f (x)dx.

k=1 k/2n

2n

If x [k/2n, (k + 1)/2n), then x diers from k/2n by at most 1/2n, and therefore

Note that

? (k+1)/2n

? (k+1)/2n k

0

xf (x)dx -

f (x)dx

k/2n

k/2n

2n

? (k+1)/2n

k

1 ? (k+1)/2n

=

x - f (x)dx

f (x)dx

k/2n

2n

2n k/2n

? (k+1)/2n

?

xf (x)dx = xf (x)dx

k=1 k/2n

0

and

1 ? (k+1)/2n

1 ? (k+1)/2n

1?

1

f (x)dx =

f (x)dx =

f (x)dx = .

2n

k=1

k/2n

2n k=1

k/2n

2n 0

2n

Therefore

100

7. CONTINUOUS DISTRIBUTIONS

?

? (k+1)/2n k

0 EX - EXn = xf (x)dx -

0

k=1 k/2n

2n f (x)dx

? (k+1)/2n

? (k+1)/2n k

=

k=1 k/2n

xf (x)dx -

k=1 k/2n

2n f (x)dx

? (k+1)/2n

? (k+1)/2n k

=

k=1

k/2n

xf (x)dx -

k/2n

2n f (x)dx

1 ? (k+1)/2n

1

f (x)dx = -- 0.

2n

k=1

k/2n

2n n0

any X We will not prove the following, but it is an interesting exercise: if m is

sequence of

discrete random variables that increase up to X , then limm EXm will have the same value

EX .

X Y This fact is useful to show linearity, if

and

are positive random variables with nite

expectations, then we can take Xm discrete increasing up to X and Ym discrete increasing up to Y . Then Xm + Ym is discrete and increases up to X + Y , so we have

E(X + Y ) = lim E(Xm + Ym) m

= lim EXm + lim EYm = EX + EY.

m

m

Note that we can not easily use the approximations to X , Y and X + Y we used in the previous proof to use in this argument, since Xm + Ym might not be an approximation of the

same kind.

X If is not necessarily positive, we can show a similar result; we will not do the details.

Similarly to the discrete case, we have

Proposition 7.2

Suppose X is a continuous random variable with density fX and g is a real-valued

function, then

? Eg(X) = g(x)f (x)dx

-

as long as the expectation of the random variable g (X ) makes sense.

variance As in the discrete case, this allows us to dene moments, and in particular the

Var X := E[X - EX]2.

As an example of these calculations, let us look at the uniform distribution.

7.1. BASIC THEORY

101

Uniform distribution

We

say

that

a

random

variable

X

has

a

uniform

distribution

on

[a, b]

if

fX (x)

=

1 b-a

if a x b and 0 otherwise.

X To calculate the expectation of

?

?b 1

EX =

xfX(x)dx =

-

1

?b

x dx a b-a

=

x dx

b-a a

1 b2 a2 a + b

=

-= .

b-a 2 2

2

This is what one would expect. To calculate the variance, we rst calculate

EX 2

=

?

-

x2fX (x)dx

=

?b

a

x2 b

1 -

dx a

=

a2

+

ab 3

+

b2 .

We then do some algebra to obtain

Var X

=

EX 2

-

(EX )2

=

(b

- a)2 .

12

102

7. CONTINUOUS DISTRIBUTIONS

7.2. Further examples and applications

Example 7.2. X Suppose has the following p.d.f.

f (x) =

2 x3

x

1

0 x < 1.

Find the CDF of X , that is, nd FX (x). Use the CDF to nd P (3 X 4).

Solution: we have FX (x) = 0 if x

1 and will need to compute

?x 2

1

FX(x) = P (X x) = 1 y3 dy = 1 - x2

x 1 when

. We can use this formula to nd the following probability

P (3 X 4) = P (X 4) - P (X < 3)

1

1

7

= FX(4) - FX(3) =

1 - 42

- 1 - 32

=. 144

Example 7.3. Suppose X has density

2x 0 x 1

fX(x) = 0

.

otherwise

Find EX .

Solution: we have that

?

?1

2

E [X] =

xfX(x)dx =

0

x ? 2x dx = . 3

Example 7.4. The density of X is given by

fX(x) =

1 2

0

if 0

x

2 .

otherwise

Find E eX .

Solution: using Proposition 7.2 with g(x) = ex we have

EeX

=

?2

0

ex

?

1 dx

2

=

1 2

e2 - 1

.

Example 7.5. Suppose X has density

2x 0 x 1

f (x) =

.

0 otherwise

? Copyright 2017 Phanuel Mariano, Patricia Alonso Ruiz, Copyright 2020 Masha Gordina.

7.2. FURTHER EXAMPLES AND APPLICATIONS

103

Find Var(X).

Solution:

in

Example

7.3

we

found E [X] = ?1

2 3

.

Now

?

1

E X2 = x2 ? 2xdx = 2 x3dx

0

0

=

1 .

2

Thus

1 22 1

Var(X) = -

=.

23

18

Example 7.6. Suppose X has density

ax + b 0 x 1

f (x) =

.

0

otherwise

and

that

E [X2]

=

1 6

.

Find

the

values

of

a

and

b.

Solution:

We

need

to

use

the

fact

that

?

-

f (x)dx

=

1

and

E [X2] =

1 6

.

The

rst

one

gives

us

?1

a

1 = (ax + b) dx = + b,

0

2

and the second one give us

1

=

?

1

x2 (ax + b) dx

=

a

+

b .

60

43

Solving these equations gives us

a = -2, and b = 2.

104

7. CONTINUOUS DISTRIBUTIONS

7.3. Exercises

Exercise 7.1. X Let be a random variable with probability density function

cx (5 - x) 0 x 5,

f (x) =

0

otherwise.

(A) What is the value of c? (B) What is the cumulative distribution function of X ? That is, nd FX (x) = P (X x). (C) Use your answer in part (b) to nd P (2 X 3). (D) What is E [X ]? (E) What is Var(X )?

Exercise 7.2. UConn students have designed the new U-phone. They have determined

X that the lifetime of a U-Phone is given by the random variable (measured in hours), with

probability density function

f (x) =

10 x2

x

10,

0 x 10.

20 (A) Find the probability that the u-phone will last more than hours. (B) What is the cumulative distribution function of X ? That is, nd FX (x) = P (X x). (C) Use part (b) to help you nd P (X 35)?

Exercise 7.3. X Suppose the random variable has a density function

f (x) =

2 x2

x > 2,

0 x 2.

Compute E [X ].

Exercise 7.4. An insurance company insures a large number of homes. The insured value,

X , of a randomly selected home is assumed to follow a distribution with density function

f (x) =

3 x4

x > 1,

0 otherwise.

1.5 Given that a randomly selected home is insured for at least , calculate the probability 2 that it is insured for less than .

Exercise 7.5. X The density function of is given by

a + bx2 0 x 1,

f (x) =

0

otherwise.

If

E [X]

=

7 10

,

nd

the

values

of

a

and

b.

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