The Cumulative Distribution Function for a Random Variable

The Cumulative Distribution Function for a Random Variable \

Each continuous random variable \ has an associated probability density function (pdf) 0 ?B?. It "records" the probabilities associated with \ as areas under its graph. More precisely,

"the probability that a value of \ is between + and ," oe T ?+ Y \ Y ,? oe '+,0 ?B? .B.

For example,

T T

?" ?$

Y Y

\ Y $? \? oe T

?$

Y

oe '"$0 ?B? \ _? oe

.B '$_ 0

?B?

.B

T ?\

Y

"? oe T ? _

\

Y

"? oe

' "

_

0

?B?

.B

i) Since probabilities are always between ! and ", it must be that 0 ?B? !

(so

that

',

+

0

?B?

.B

can

never

give

a

"negative

probability"),

and

ii) Since a "certain" event has probability ",

T ? _ \ _? oe " oe '__0 ?B? .B oe total area under the graph of 0 ?B?

The properties i) and ii) are necessary for a function 0 ?B? to be the pdf for some random variable \?

We can also use property ii) in computations: since

'_

_

0

?B?

.B

oe

'$

_

0

?B?

'$_0 ?B? .B

oe

"

T ?\ Y $? oe '$_0 ?B? .B oe " '$_0 ?B? .B oe " T ?\ $?

The pdf is discussed in the textbook.

There is another function, the cumulative distribution function (cdf) which records the same probabilities associated with \, but in a different way. The cdf J ?B? is defined by

J ?B? oe T ?\ Y B?.

J ?B? gives the "accumulated" probability "up to B." We can see immediately how the pdf and cdf are related:

J ?B? oe

T ?\

Y

B?

oe

'B

_

0

?>?

.>

(since "B" is used as a variable in the

upper limit of integration, we use some

other variable, say ">", in the integrand)

Notice that J ?B? ! (since it's a probability), and that

a) b)

BBl??ilmi_m_JJ?B??Boe? oeBBl?i?lmi_m'_B'_B0_?0>??>.?>.oe>

'__0 ?>? .> oe '__0 ?>?

oe .>

" oe

and !, and

that

c) J w?B? oe 0 ?B? (by the Fundamental Theorem of Calculus)

Item c) states the connection between the cdf and pdf in another way:

the cdf J ?B? is an antiderivative of the pdf 0 ?B? (the particular antiderivative where the constant of integration is chosen to make the limit in a) true)

and therefore

T ?+

Y

\

Y

,?

oe

',

+

0

?B?

.B

oe

J ?B?l+,

oe

J ?,?

J ?+?

oe

T

?\

Y

,?

T ?\

Y

+?

________________________________________________________________________

Example: Suppose \ has an exponential density function. As discussed in class,

0

?B?

oe

oe

! -/-B

B! B !

(where -

oe

" .

?

If B !, 'B_0 ?>? .> oe '!B0 ?>? .> oe '!B-/-> .> oe /->lB! oe " /-B, so

J

?B?

oe

oe

! "

/-B

B! B !

If

\

has

mean

.

oe

$,

say,

then

-

oe

" .

oe

" $

.

If we want to know T ?\ Y %?, we can either compute

'%_0

?B?

.B

oe

'%

_

" $

/?"?$

?B.B

?

!?($'%!$,

or

(now

that

we

have

the

formula

for

J

?B?

we can simply compute J ?$? oe " /?"?$??% oe " /%?$ ? !?($'%!$?

(The graphs of 0 ?B? and J ?B? are shown on the last page before exercises. In the figure,

notice the values of lim J ?B? and lim J ?B? ??

B?_

B?_

________________________________________________________________________

Example: If \ is a normal random variable with mean . oe ! and standard deviation

5

oe

"?

then

its

pdf

is

0 ?B?

oe

" ?#1

/B#?#,

and

its

cdf

J ?B?

oe

" ?#1

'B

_

/>#

?#

.>.

Because there is no "elementary" antiderivative for />#?#, its not possible to find an

"elementary" formula for J ?B?.

However, for any B, the value of

" ?#1

'B

_

/>#

?#

.>

can

be estimated, so that a graph of J ?B? can be drawn. (See figure on the last page before

exercises.)

Example: More generally, probability calculations involving a normal random variable \ are computationally difficult because again there's no elementary formula for the cumulative distribution function J ?B? that is, an antiderivative for the probability den=ity function ?

0 ?B?

oe

" 5?#1

/?B.?#?#5#

Therefore it's not possible to find an exact value for

T ?+

Y

\

Y

,?

oe

',

+

" 5?#1

/?B.?#?#5# .B

oe

J ?,?

J ?+?

Suppose \ is a normal random variable with mean . oe "?* and standard deviation 5 oe "?(. If we want to find T ? $ Y \ Y #?, we need to estimate

" ?"?(??#1

'2

3

/?B"?*?#?#?"?(?#

.B

oe

J

?#?

J

?

$??

This can be done with Simpson's Rule. However, such calculations are so important that the TI83-Plus Calculator has a built in way to make the estimate:

Punch keys 28. HMWX V

Choose item 2 on the menu: normalcdf

On the screen you see

normalcdf ?

Fill in

normalcdf ? $? #? "?*? "?(?

and the TI-83 gives the approximate value of the integral above: !?&#"480

The general syntax for the command is

normalcdf (lowerlimit,upperlimit,.? 5)

If you enter only then the TI-83 assumes . oe !? 5 oe " as the default values

normalcdf ?lowerlimit,upperlimit?

Note that using the values for .? 5 example given above:

T ?. 5 Y \ Y . 5? ? normalcdf ??#? $?'? "?*? "?(? ? !?')#( T ?. #5 Y \ Y . #5? ? normalcdf ? "?&? &?$? "?*? "?(? ? !?*&%& T ?. $5 Y \ Y . $5? ? normalcdf ? $?#? (? "?*? "?(? ? !?**($

In fact (as may have been mentioned in class) these probabilities come out the same for any normal random variable, no matter what the values of . and 5: for example, the probability that any normal random variable takes on a value between ,, one standard deviation of its mean is ? 0.6827?

Exercises:

1.

A certain "uniform" random variable \ has pdf 0 ?B? oe

oe

"?& !

#YBY( otherwise.

a) What is T ?! Y \ Y $??

b) Write the formula for its cdf J ?B?

c) What is J ?$? J ?!? ?

2.

A certain kind of random variable as density function 0 ?B? oe

1

?"

"

B# ?

.

a) What is T ?\ "??

b) Write the formula for its cdf J ?B?

c) Write a formula using J ?B? that gives the answer to part a). Check that it agrees with your numerical answer in a).

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