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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