The Cumulative Distribution Function for a Random Variable \

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

the probability that a value of is between and .\ + , T+ \ , 0B .B'+,For example, T" \ $ 0B .B'"$ T$ \ T$ \ _ 0B .B'$_ T\ " T _ \ " 0B .B'_" i) Since probabilities are always between and , it must be that ! " 0B ! (so that can never give a negative probability), and'+,0B .B ii) Since a certain event has probability ," total area under the graph of T _ \ _ " 0B .B 0B'__The properties i) and ii) are necessary for a function to be the pdf for some random0Bvariable \

We can also use property ii) in computations: since

' ' '_ _ $_ $ _0B .B 0B 0B .B " T\ $ 0B .B " 0B .B " T\ $' '_ $$ _The pdf is discussed in the textbook.

There is another function, the (cdf) which records thecumulative distribution functionsame probabilities associated with , but in a different way. The cdf is defined by\ JB

.JB T\ B

JB Bgives the accumulated probability up to . We can see immediately how thepdf and cdf are related:

(since is used as a variable in theJB T\ B 0> .> B'_B upper limit of integration, we use some other variable, say , in the integrand)>

Notice that (since it's a probability), and thatJB !

a) andlim limB_ B_ _ _

B _JB 0> .> 0> .> "' '

b) , and thatlim limB_ B_ _ _

B _JB 0> .> 0> .> !' '

c) (by the Fundamental Theorem of Calculus)J B 0Bw

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

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

and therefore

T+ \ , 0B .B JBl J, J+ T\ , T\ +'+, +,________________________________________________________________________

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

(where 0B - ! B !-/ B ! -B ".

If , , soB ! 0> .> 0> .> -/ .> / l " /' ' '_ ! !B B B -> -> B -B! JB ! B !

" / B ! -B

If has mean , say, then .\ $ - . " "$.

If we want to know , we can either computeT\ %' '_ _% % "

$"$ B0B .B / .B !($'%!$ JB, or (now that we have the formula for

we can simply computeJ$ " / " / !($'%!$"$% %$

(The graphs of and are shown on the last page before exercises. In the figure,0B JBnotice the values of and lim lim

B_ B_JB JB

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Example: If is a normal random variable with mean and standard deviation\ !.5 " 0B / JB / .>then its pdf is , and its cdf ." "

# #B # > #

_B 1 1

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

elementary formula for . However, for any , the value of canJB B / .>"# _

B > # 1 ' #be estimated, so that a graph of can be drawn. (JB See figure on the last page beforeexercises.)

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

0B /"#

B #5 1

. 5 # #

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

T+ \ , / .B J, J+'+, "# B #5 1 . 5 # #Suppose is a normal random variable with mean and standard deviation\ "*.5 "( T $ \ #. If we want to find , we need to estimate

""( #

B"* #"( 1 ' 32 / .B J# J $# #This can be done with Simpson's Rule. However, such calculations are so important thatthe TI83-Plus Calculator has a built in way to make the estimate:

Punch keys 28. HMWXV normalcdfChoose item 2 on the menu: 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 thecommand is normalcdf (lowerlimit,upperlimit, ). 5

If you enter normalcdfonly lowerlimit,upperlimitthen the TI-83 assumes . 5 ! "as the default values

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

normalcdfT \ # $' "* "( !')#(. 5 . 5 normalcdfT # \ # "& &$ "* "( !*&%&. 5 . 5 normalcdfT $ \ $ $# ( "* "( !**($. 5 . 5

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

Exercises:

1. A certain uniform random variable has pdf otherwise.\ 0B "& # B (!

a) What is ?T! \ $

b) Write the formula for its cdf JB

c) What is ?J$ J!

2. A certain kind of random variable as density function .0B ""B 1 #

a) What is ?T\ "

b) Write the formula for its cdf JB

c) Write a formula using that gives the answer to part a). Check that itJB agrees with your numerical answer in a).