Week 7: Arena notes Week 8 notes:
Arena view: ● Some Lousy Generators:
Discrete Probability: DISC(CumulativeP, V) CONT(P,V), POIS(λ), EXPO(λ), NORM(µ; σ), 1-random devices like a coin toss (0 and 1 values randomly). Roll a die to generate random
TRIA(a,b,c), UNIF(a,b), WEIB(β; α) numbers between 1 and 6. More sophisticated random devices include Geiger counters and
atomic clocks. Naturally, random devices have strong randomness properties; however, we
Advanced Process Panel: Blocks(Hold, seize, search, delay, release, store, unstore,
cannot reproduce their results easily.
remove, match, pickup, gather, dropoff, assign attribute, adjust variable, readWrite, signal,
2-Random Number Tables: Published by the RAND corporation, contains one million random
search and remove) spreadsheets like(Advanced Set, Expression, Failure, file, StateSet, digits and one hundred thousand normal random variates.
Statistic, Storage) 3-The mid-square method was created by the famous mathematician John von Neumann. The
Basic Process panel: create, dispose, process, decide, batch, clone, separate, assign, main idea here is we take an integer, square it, and then use the middle part of that integer as
record, label, go to label, adjustable batch. Spreadsheets: Attribute, entity, queue, resource, our next random integer, repeating the process as many times as we need. To generate the
variable, schedule, set, picture Uniform (0,1) random variable, we would divide each generated integer by the appropriate
Advanced Transfer panel: ENTER, LEAVE, PICKSTATION, ROUTE, STATION, ACCESS, power of ten.
4-Fibonacci Additive Congruential Generators:
CONVEY, EXIT, START, STOP, ACTIVATE, ALLOCATE, FREE, HALT, MOVE, REQUEST, TRANSPORT.
SPREADSHEETS (STATION DATA, SEQUENCE, CONVEYOR, SEGMENT, TRANSPORTER, DISTANCE,
NETWORK, NETWORK LINK, ACTIVITY AREA)
X i=( X i−1+ X i−2 ) mod m, i=1,2 …∧R i=X i /m
Trick 1: Pre-assign the service times as an attribute (Sealer Time) in an Assign module ● Linear Congruential Generator
1-LCGs:
immediately after each customer arrives.
Trick 2: While we’re at it, use the Assign to store each customer’s arrival time as an attribute.
Use the Arena variable TNOW to do so.
X i=( aX i−1+ c ) mod m, where X 0 isthe seed∧Ri =X i /m
- Sometimes you need to do things that are too complicated for the Process module version of
Seize–Delay–Release Now you can handle things like… 1) Seize–Assign–Delay–Release 2) Non-
Example: X i=( 5 X i−1+3 ) mod 8. X 8=X 0=0. This generator is a full-
symmetric multiple Seize’s and Release’s 3) Complicated Seize’s and Release’s that might period generator since it has a cycle length equal to me
depend on sets of servers
Resource spreadsheet in the Basic Process template Click on Failures column Add a 31
Failure Name.
2- “Desert island” generator: X i=16807 X i−1 mod 2 −1. This generator
Go to the Failure spreadsheet in the Advanced Process template, where you’ll see your new has a full-period cycle length > 2 billion
failure name. Choose type of failure: Count (failure after a certain # of arrivals) Time (after a
certain amount of time) Choose downtime (for repair): Can be any expression.
X i and Ri
Let's look at the algorithm for generating each :
-Certain primitive blocks such as the Queue block can’t even connect to a Seize module from
the Advanced Process template or a Process module from the Basic Process template
K ← ⌊ X i−1 /127773 ⌋
-Using the Alter block to change the number of resources
X i ←16807 (X i−1−127773 K )−2836 K
Advanced Transfer Panel ● Various Seize Selection Rules: Cyclical, Random,
Modules related to movement: Preferred Order, Specific Member, Largest Remaining if X i <0 , then set X i ← X i +2147483647
Station, Route, Enter, Leave. Capacity, Smallest Number Busy.
Sets and Advanced Sets (e.g.,
sets of Sequences). ENTER –
Ri ← X i∗4.656612875E-10
LEAVE modules for Station-to- ● Station: Somewhere you can go to Route: Tells you
31
Station movement.
Resource (e.g., an assistant)
where you’re going. Similar to Connect, except it
takes time. Enter: General way to get into a Station.
3-RANDU Generator (bad) X i=65539 X i−1 mod 2 The random numbers
-Seize-Release Leave: General way to depart a Station generated by this generator fall entirely on 15 hyperplanes (not iid)
Transporter (e.g., a car)
-Request-Free ● A sequence goes from station to station. Maybe ● Tausworthe Generator:
-Requires Distance Set have different service time distribution at each station
Conveyor (depending on customer type and place in visitation B i=( Bi−r + Bi−q ) mod 2=Bi−r XOR Bi −q where< r <q . The
- Access-Exit sequence
q
-Requires Segment Set
● Use the Set spreadsheet in the Basic Process
sequence has a period of
2 −1
template to define sets. A “vanilla” resource has
identical, interchangeable servers. But a resource set
How do we get Unif (0,1) random variables from the Bi' s? We can take a sequence
l
can have distinct servers, with different schedules,
service speeds, service specialties, etc
of l bits and divide them by
2 to compute a real number between zero and one.
● Combined generators:
2191
● Have to be a little careful with Seize-Delay-Release 1- L’Ecuyer’s generator of cycle length :
for a resource set. The problem is that you have to
make sure that you Release the same guy that you 2- Mersenne Twister: Matsumoto and Nishimura have developed the "Mersenne Twister"
originally Seize’d. (If you release a random server, 19937
some other customer may lose his server!) generator, which has a period of
2 −1 . This period is beyond sufficient for any
modern application; we will often need several billion PRNs, but never more than even
100
Questions: 2 . The period only describes how long it takes for the generator to cycle, but doesn’t
True: In Arena, a DECIDE module can route customers probabilistically or conditionally to have an effect on generating the PRNs or the quality of the PRNs.
multiple locations. ● Choosing a Good Generator - Theory
False: In Arena, a primitive QUEUE block can connect with a PROCESS module
True: In Arena, you can pre-assign a service time as an attribute before you actually get to the ● Statistical tests for randomness:
server that you’ll be using. 1- Goodness of fit test - Chi-squared: Are the PRNs approximately Unif (0,1)?? When we design
True: In Arena, it’s perfectly OK to have a CREATE module generating one-at-a-time customer the test, we set the level of significance α = P (Reject H_0 | H_0 true). Typically, α = 0.05 or 0.1,
and is the probability of Type I error.
arrivals, while also having a different CREATE module with4-at-a-time customer arrivals
elsewhere in the model.
We can also specify the probability of Type II error, β = P (Accept H_0 | H_0 false). We
True: In Arena, \fake" customers can be used to schedule machine breakdowns, keep track of
which time period the simulation is currently in, and carry out other duties not associated with regard Ho as the status quo - "that which we currently believe to be true". In this
actual customers
True: In Arena, the SEIZE, DELAY, RELEASE functionalities are in multiple templates. case, Ho refers to the belief that the numbers we are currently generating are, in fact,
True: In Arena, you can schedule failures to occur after a random number of customers have i.i.d Unif (0,1).
used a resource or after a random amount of time has passed.
False: In Arena, you are only allowed to schedule one type of failure for a particular resource.
True: In Arena, it is possible for the resource Joey to be a member of three different resource
Oi−E i ¿ 2
sets. ¿
True: For easy display and organizational purposes, our Arena Call Center simulation is divided ¿
into several sub models.
True: The customer arrival pattern for our Arena Call Center simulation changes over the day ¿ where Oi ∼ Bin(n , 1/ k ),i=1,2,... , k . And
and is therefore modeled as a nonhomogeneous Poisson process. k
False: All of the Tech Support servers for our Arena Call Center simulation are equally skilled
and have the same concurrent work schedules.
χ 20=∑ ¿
-In Arena, what kind of module would you use to change an entity’s picture? ASSIGN i=1
-In Arena, where would you find the Expression spreadsheet? Advanced Process panel.
n
-Consider the Arena Inventory model that we did in class as a demo. Did this model allow for
Ei=E [ Oi ] = , i=1,2,... , k .(1−α ) quantiles for χ 2
backlogs? YES
k
● The name of the automatically supplied ARENA attribute that stores a particular entity's
sequence is Entity. Sequence
distributions with varying degrees of freedom, n , where we define a quantile,
2 2 2
● Where can you set the buffer size of a certain queue? In a Queue block in the Blocks
template
χ α ,n as: P( χ n < χ α , n)=1−α
● There are some great PRN generators out there with incredible cycle
2 2
lengths
2191 ∧even 219937
We reject the null hypothesis Ho ifχ 0 > χ α , k−1 , and we fail to reject
● Attributes are properties for individual customers; variables are properties of the entire
system.
H0 if χ 20 ≤ χ 2α , k−1 . Ei=n/ k should be at least five, and n
● In arena, NORM(a,b) has a variance of b^2 2
should be at least 30. When k is so large, we cant find tables with values for χ α ,k −1 ,
This study source was downloaded by 100000899606070 from CourseHero.com on 02-22-2026
then we can22:23:49 GMT -06:00 below:
use the approximation
On average, A-r algo requires Geom(1/c) trials with expected value = c
Polynomial example: Let’s generate a random variable with pdf
https://www.coursehero.com/file/81716187/simulation-final-exam-cheatsheetdocx/
, 1−x ¿2 ,0 ≤ x ≤ 1 . the maximum of 3
f ( x)=60 x 3 ¿
f ( x) occurs at x=0.6 :
f (0.6)=2.0736 . With this knowledge, we can generate a basic majorizer,
χ 2
α ,k −1 ≈(k−1) 1−
2
9(k −1)
2- Runs tests for independence:
[+ za
2
9( k−1) √ ]
● Runs up and down
t ( x)=2.0736 . We know that c equals the integral of t ( x) from
zero to one, so, in this case,
inefficient majorizer.
c=2.7036 . All this to say, t ( x) is a relatively A ≈ Nor ( 2 n−1
3
,
16 n−29
90 )
downs out of n observations. We reject
where A denote the total number of ups and
Like we said, h(x )=t ( x)/c . In this case, h( x )=1 , since H o if A istoo big∨small . the test statistic is
t ( x)=2.7036=c . This result means that Y is a Unif(0,1) random A−E[ A]
Z0= and we reject H o if |Z 0|> z¿ α / 2 here we look for
variable, since it’s pdf is one. Finally, let’s compute g( x) : √ Var ( A)
1-(alpha/2) in the std normal table
1−x ¿ 2 ● Runs above and below the mean:
¿ 2n 1 n2 1 2 n1 n2( 2n 1 n2−n)
60 x 3 ¿
f (x)
B ≈ Nor
( n
+ ,
2 n2 (n−1) ) that n1 refers to the
g( x)= =¿ number of observations greater than or equal to the mean and n2=n−n1 .
t (x) B−E [B]
Let’s look at a simple example. Let’s draw two uniforms, U=0.13 and Z0= where B is the # runs. we reject H0 if the absolute value
Y =0.25 . If we plug and chug, we see that g(Y )≈ 0.25 . Therefore,
√ Var (B)
of Z0 is greater than the α /2 standard normal quantile
U ≤ g( Y ) and we take X ←0.25
2
● Autocorrelation test (bonus): We define the lag-1 correlation of the Ri 's by
−x
2 2 ρ ≡Corr ( Ri , R i+1 ) . In other words, the lag-1 correlation measures the
Half-normal example: with the pdf : f (x)= e , x≥0 . We can use
√2π correlation between one PRN and its immediate successor. Ideally, if the PRNs are uncorrelated,
ρ should be zero. A good estimator for ρ is given by:
∞
2 e −x 2e
t ( x)=
√ π
e ≥ f (x) & c=∫ t ( x) dx=
0
t( x ) −x
√ π ^ρ ≡
( 12
n−1
∑ R R −3
n−1 k=1 k 1+ k )
Now, let’s compute h(x ) : h( x )= =e easy exp(1) pdf and n−1 ¿2
c (¿ ¿)
x−1 ¿2 /2 In particular, if n is large, and H0 is true:
13 n−19 Let's transform
−¿ 0,
. To generate a half normal, we simply generate
¿
f (x ) ¿ ^ρ ≈ Nor ¿
g( x)= =e
t (x) ^ρ into a standard normal random variable, Z0 , which we accomplish with the
U ∼U (0,1) and Y ∼ exp(1) and accept Y if ρ^ −E[ ^ρ ] ^ρ
following manipulation: Z0= = We reject H0 if the
U ≤ g( Y ) . We can use the half-normal result to generate a Nor(0,1) random √Var ( ^ρ) √ Var ( ^ρ )
variable. We simply have to “flip back” half of the X values over the y -axis. absolute value of Z0 is greater than the α /2 standard normal quantile:
Week 9 notes:
Given U ∼U (0,1) and X from the half-normal distribution, we can see
● Inverse transform method: The inverse transform method states that, if X is a
that:
continuous random variable with cdf F( x ) , then F( X )∼U (0,1) . In
Z= {−XX U ≤ 1/2
U >1/2
∼ Nor(0,1) As always, we can generate a Nor( other words, if we plug a random variable, from any distribution, into its own cdf, we get a
Unif(0,1) random variable.
● Proof: Let Y =F ( X ) . Since Y is a random variable, it has a cdf, which we
μ ,
σ2 ) by applying the transformation μ+σZ .
Poisson example: The Poisson distribution has the following pmf: can denote G( y ) . By definition: G( y )=P(Y ≤ y)
λ −λ
n Since Y =F (X ) : G( y )=P( F (X )≤ y )
P( X =n)=e , n=0,1,.. . Since X is a continuous random variable, its cdf is continuous. Therefore, we can apply
n! −1 , to both sides of the inequality:
x=n ⟺ see exactly n Pois ( λ ) arrivals by t=1.
the inverse,
F
n n +1 G( y )=P( F−1 (F( X )) ≤ F−1 ( y ))
X =n ⇔ ∑ Ai ≤1< ∑ A i What is F−1( F ( X)) ? Simply, X : G( y )=P( X ≤ F−1( y))
i=1 i =1
X ≤ x ), where −1
n n+1 Notice that we have an expression of the form x=F ( y) .
X =n ⇔ ∑
i=1
[ −1
λ ]
ln U i ≤ 1<∑
i=1
−1
λ
ln U i [ ] |
We know, by definition,
−1
F( x )=P( X ≤ x) , so:
P¿
n n+1 G( y )=F (F ( y))= y
X =n ⇔
−1
λ
ln (∏ )i=1
U i ≤ 1<
−1
λ
ln ( )
∏ Ui
i=1
|
In summary, the cdf of
get the pdf, we see that
Y is G( y )= y . If we take the derivative of the cdf to
g( y )=1 Let’s remember the pdf for a uniform random
n n+ 1 variable:
X =n ⇔ ∏ U i ≤ e− λ < ∏ U i 1
Ex: to obtain Pois(2) we use
i=1
e− λ=e−2by=0.1353>
This study source was downloaded
n+ 1
100000899606070
i=1
i=1
If
f (x)= b−a
∏ Ufromi CourseHero.com on 02-22-2026 22:23:49
a=0 , b=1 ,
x ∈[a , b ]
0 GMTotherwise
-06:00
then f (x)=1=g ( y )
{ . Therefore,
Consider the following table:
https://www.coursehero.com/file/81716187/simulation-final-exam-cheatsheetdocx/
Y ∼ U (0,1) .
Arena view: ● Some Lousy Generators:
Discrete Probability: DISC(CumulativeP, V) CONT(P,V), POIS(λ), EXPO(λ), NORM(µ; σ), 1-random devices like a coin toss (0 and 1 values randomly). Roll a die to generate random
TRIA(a,b,c), UNIF(a,b), WEIB(β; α) numbers between 1 and 6. More sophisticated random devices include Geiger counters and
atomic clocks. Naturally, random devices have strong randomness properties; however, we
Advanced Process Panel: Blocks(Hold, seize, search, delay, release, store, unstore,
cannot reproduce their results easily.
remove, match, pickup, gather, dropoff, assign attribute, adjust variable, readWrite, signal,
2-Random Number Tables: Published by the RAND corporation, contains one million random
search and remove) spreadsheets like(Advanced Set, Expression, Failure, file, StateSet, digits and one hundred thousand normal random variates.
Statistic, Storage) 3-The mid-square method was created by the famous mathematician John von Neumann. The
Basic Process panel: create, dispose, process, decide, batch, clone, separate, assign, main idea here is we take an integer, square it, and then use the middle part of that integer as
record, label, go to label, adjustable batch. Spreadsheets: Attribute, entity, queue, resource, our next random integer, repeating the process as many times as we need. To generate the
variable, schedule, set, picture Uniform (0,1) random variable, we would divide each generated integer by the appropriate
Advanced Transfer panel: ENTER, LEAVE, PICKSTATION, ROUTE, STATION, ACCESS, power of ten.
4-Fibonacci Additive Congruential Generators:
CONVEY, EXIT, START, STOP, ACTIVATE, ALLOCATE, FREE, HALT, MOVE, REQUEST, TRANSPORT.
SPREADSHEETS (STATION DATA, SEQUENCE, CONVEYOR, SEGMENT, TRANSPORTER, DISTANCE,
NETWORK, NETWORK LINK, ACTIVITY AREA)
X i=( X i−1+ X i−2 ) mod m, i=1,2 …∧R i=X i /m
Trick 1: Pre-assign the service times as an attribute (Sealer Time) in an Assign module ● Linear Congruential Generator
1-LCGs:
immediately after each customer arrives.
Trick 2: While we’re at it, use the Assign to store each customer’s arrival time as an attribute.
Use the Arena variable TNOW to do so.
X i=( aX i−1+ c ) mod m, where X 0 isthe seed∧Ri =X i /m
- Sometimes you need to do things that are too complicated for the Process module version of
Seize–Delay–Release Now you can handle things like… 1) Seize–Assign–Delay–Release 2) Non-
Example: X i=( 5 X i−1+3 ) mod 8. X 8=X 0=0. This generator is a full-
symmetric multiple Seize’s and Release’s 3) Complicated Seize’s and Release’s that might period generator since it has a cycle length equal to me
depend on sets of servers
Resource spreadsheet in the Basic Process template Click on Failures column Add a 31
Failure Name.
2- “Desert island” generator: X i=16807 X i−1 mod 2 −1. This generator
Go to the Failure spreadsheet in the Advanced Process template, where you’ll see your new has a full-period cycle length > 2 billion
failure name. Choose type of failure: Count (failure after a certain # of arrivals) Time (after a
certain amount of time) Choose downtime (for repair): Can be any expression.
X i and Ri
Let's look at the algorithm for generating each :
-Certain primitive blocks such as the Queue block can’t even connect to a Seize module from
the Advanced Process template or a Process module from the Basic Process template
K ← ⌊ X i−1 /127773 ⌋
-Using the Alter block to change the number of resources
X i ←16807 (X i−1−127773 K )−2836 K
Advanced Transfer Panel ● Various Seize Selection Rules: Cyclical, Random,
Modules related to movement: Preferred Order, Specific Member, Largest Remaining if X i <0 , then set X i ← X i +2147483647
Station, Route, Enter, Leave. Capacity, Smallest Number Busy.
Sets and Advanced Sets (e.g.,
sets of Sequences). ENTER –
Ri ← X i∗4.656612875E-10
LEAVE modules for Station-to- ● Station: Somewhere you can go to Route: Tells you
31
Station movement.
Resource (e.g., an assistant)
where you’re going. Similar to Connect, except it
takes time. Enter: General way to get into a Station.
3-RANDU Generator (bad) X i=65539 X i−1 mod 2 The random numbers
-Seize-Release Leave: General way to depart a Station generated by this generator fall entirely on 15 hyperplanes (not iid)
Transporter (e.g., a car)
-Request-Free ● A sequence goes from station to station. Maybe ● Tausworthe Generator:
-Requires Distance Set have different service time distribution at each station
Conveyor (depending on customer type and place in visitation B i=( Bi−r + Bi−q ) mod 2=Bi−r XOR Bi −q where< r <q . The
- Access-Exit sequence
q
-Requires Segment Set
● Use the Set spreadsheet in the Basic Process
sequence has a period of
2 −1
template to define sets. A “vanilla” resource has
identical, interchangeable servers. But a resource set
How do we get Unif (0,1) random variables from the Bi' s? We can take a sequence
l
can have distinct servers, with different schedules,
service speeds, service specialties, etc
of l bits and divide them by
2 to compute a real number between zero and one.
● Combined generators:
2191
● Have to be a little careful with Seize-Delay-Release 1- L’Ecuyer’s generator of cycle length :
for a resource set. The problem is that you have to
make sure that you Release the same guy that you 2- Mersenne Twister: Matsumoto and Nishimura have developed the "Mersenne Twister"
originally Seize’d. (If you release a random server, 19937
some other customer may lose his server!) generator, which has a period of
2 −1 . This period is beyond sufficient for any
modern application; we will often need several billion PRNs, but never more than even
100
Questions: 2 . The period only describes how long it takes for the generator to cycle, but doesn’t
True: In Arena, a DECIDE module can route customers probabilistically or conditionally to have an effect on generating the PRNs or the quality of the PRNs.
multiple locations. ● Choosing a Good Generator - Theory
False: In Arena, a primitive QUEUE block can connect with a PROCESS module
True: In Arena, you can pre-assign a service time as an attribute before you actually get to the ● Statistical tests for randomness:
server that you’ll be using. 1- Goodness of fit test - Chi-squared: Are the PRNs approximately Unif (0,1)?? When we design
True: In Arena, it’s perfectly OK to have a CREATE module generating one-at-a-time customer the test, we set the level of significance α = P (Reject H_0 | H_0 true). Typically, α = 0.05 or 0.1,
and is the probability of Type I error.
arrivals, while also having a different CREATE module with4-at-a-time customer arrivals
elsewhere in the model.
We can also specify the probability of Type II error, β = P (Accept H_0 | H_0 false). We
True: In Arena, \fake" customers can be used to schedule machine breakdowns, keep track of
which time period the simulation is currently in, and carry out other duties not associated with regard Ho as the status quo - "that which we currently believe to be true". In this
actual customers
True: In Arena, the SEIZE, DELAY, RELEASE functionalities are in multiple templates. case, Ho refers to the belief that the numbers we are currently generating are, in fact,
True: In Arena, you can schedule failures to occur after a random number of customers have i.i.d Unif (0,1).
used a resource or after a random amount of time has passed.
False: In Arena, you are only allowed to schedule one type of failure for a particular resource.
True: In Arena, it is possible for the resource Joey to be a member of three different resource
Oi−E i ¿ 2
sets. ¿
True: For easy display and organizational purposes, our Arena Call Center simulation is divided ¿
into several sub models.
True: The customer arrival pattern for our Arena Call Center simulation changes over the day ¿ where Oi ∼ Bin(n , 1/ k ),i=1,2,... , k . And
and is therefore modeled as a nonhomogeneous Poisson process. k
False: All of the Tech Support servers for our Arena Call Center simulation are equally skilled
and have the same concurrent work schedules.
χ 20=∑ ¿
-In Arena, what kind of module would you use to change an entity’s picture? ASSIGN i=1
-In Arena, where would you find the Expression spreadsheet? Advanced Process panel.
n
-Consider the Arena Inventory model that we did in class as a demo. Did this model allow for
Ei=E [ Oi ] = , i=1,2,... , k .(1−α ) quantiles for χ 2
backlogs? YES
k
● The name of the automatically supplied ARENA attribute that stores a particular entity's
sequence is Entity. Sequence
distributions with varying degrees of freedom, n , where we define a quantile,
2 2 2
● Where can you set the buffer size of a certain queue? In a Queue block in the Blocks
template
χ α ,n as: P( χ n < χ α , n)=1−α
● There are some great PRN generators out there with incredible cycle
2 2
lengths
2191 ∧even 219937
We reject the null hypothesis Ho ifχ 0 > χ α , k−1 , and we fail to reject
● Attributes are properties for individual customers; variables are properties of the entire
system.
H0 if χ 20 ≤ χ 2α , k−1 . Ei=n/ k should be at least five, and n
● In arena, NORM(a,b) has a variance of b^2 2
should be at least 30. When k is so large, we cant find tables with values for χ α ,k −1 ,
This study source was downloaded by 100000899606070 from CourseHero.com on 02-22-2026
then we can22:23:49 GMT -06:00 below:
use the approximation
On average, A-r algo requires Geom(1/c) trials with expected value = c
Polynomial example: Let’s generate a random variable with pdf
https://www.coursehero.com/file/81716187/simulation-final-exam-cheatsheetdocx/
, 1−x ¿2 ,0 ≤ x ≤ 1 . the maximum of 3
f ( x)=60 x 3 ¿
f ( x) occurs at x=0.6 :
f (0.6)=2.0736 . With this knowledge, we can generate a basic majorizer,
χ 2
α ,k −1 ≈(k−1) 1−
2
9(k −1)
2- Runs tests for independence:
[+ za
2
9( k−1) √ ]
● Runs up and down
t ( x)=2.0736 . We know that c equals the integral of t ( x) from
zero to one, so, in this case,
inefficient majorizer.
c=2.7036 . All this to say, t ( x) is a relatively A ≈ Nor ( 2 n−1
3
,
16 n−29
90 )
downs out of n observations. We reject
where A denote the total number of ups and
Like we said, h(x )=t ( x)/c . In this case, h( x )=1 , since H o if A istoo big∨small . the test statistic is
t ( x)=2.7036=c . This result means that Y is a Unif(0,1) random A−E[ A]
Z0= and we reject H o if |Z 0|> z¿ α / 2 here we look for
variable, since it’s pdf is one. Finally, let’s compute g( x) : √ Var ( A)
1-(alpha/2) in the std normal table
1−x ¿ 2 ● Runs above and below the mean:
¿ 2n 1 n2 1 2 n1 n2( 2n 1 n2−n)
60 x 3 ¿
f (x)
B ≈ Nor
( n
+ ,
2 n2 (n−1) ) that n1 refers to the
g( x)= =¿ number of observations greater than or equal to the mean and n2=n−n1 .
t (x) B−E [B]
Let’s look at a simple example. Let’s draw two uniforms, U=0.13 and Z0= where B is the # runs. we reject H0 if the absolute value
Y =0.25 . If we plug and chug, we see that g(Y )≈ 0.25 . Therefore,
√ Var (B)
of Z0 is greater than the α /2 standard normal quantile
U ≤ g( Y ) and we take X ←0.25
2
● Autocorrelation test (bonus): We define the lag-1 correlation of the Ri 's by
−x
2 2 ρ ≡Corr ( Ri , R i+1 ) . In other words, the lag-1 correlation measures the
Half-normal example: with the pdf : f (x)= e , x≥0 . We can use
√2π correlation between one PRN and its immediate successor. Ideally, if the PRNs are uncorrelated,
ρ should be zero. A good estimator for ρ is given by:
∞
2 e −x 2e
t ( x)=
√ π
e ≥ f (x) & c=∫ t ( x) dx=
0
t( x ) −x
√ π ^ρ ≡
( 12
n−1
∑ R R −3
n−1 k=1 k 1+ k )
Now, let’s compute h(x ) : h( x )= =e easy exp(1) pdf and n−1 ¿2
c (¿ ¿)
x−1 ¿2 /2 In particular, if n is large, and H0 is true:
13 n−19 Let's transform
−¿ 0,
. To generate a half normal, we simply generate
¿
f (x ) ¿ ^ρ ≈ Nor ¿
g( x)= =e
t (x) ^ρ into a standard normal random variable, Z0 , which we accomplish with the
U ∼U (0,1) and Y ∼ exp(1) and accept Y if ρ^ −E[ ^ρ ] ^ρ
following manipulation: Z0= = We reject H0 if the
U ≤ g( Y ) . We can use the half-normal result to generate a Nor(0,1) random √Var ( ^ρ) √ Var ( ^ρ )
variable. We simply have to “flip back” half of the X values over the y -axis. absolute value of Z0 is greater than the α /2 standard normal quantile:
Week 9 notes:
Given U ∼U (0,1) and X from the half-normal distribution, we can see
● Inverse transform method: The inverse transform method states that, if X is a
that:
continuous random variable with cdf F( x ) , then F( X )∼U (0,1) . In
Z= {−XX U ≤ 1/2
U >1/2
∼ Nor(0,1) As always, we can generate a Nor( other words, if we plug a random variable, from any distribution, into its own cdf, we get a
Unif(0,1) random variable.
● Proof: Let Y =F ( X ) . Since Y is a random variable, it has a cdf, which we
μ ,
σ2 ) by applying the transformation μ+σZ .
Poisson example: The Poisson distribution has the following pmf: can denote G( y ) . By definition: G( y )=P(Y ≤ y)
λ −λ
n Since Y =F (X ) : G( y )=P( F (X )≤ y )
P( X =n)=e , n=0,1,.. . Since X is a continuous random variable, its cdf is continuous. Therefore, we can apply
n! −1 , to both sides of the inequality:
x=n ⟺ see exactly n Pois ( λ ) arrivals by t=1.
the inverse,
F
n n +1 G( y )=P( F−1 (F( X )) ≤ F−1 ( y ))
X =n ⇔ ∑ Ai ≤1< ∑ A i What is F−1( F ( X)) ? Simply, X : G( y )=P( X ≤ F−1( y))
i=1 i =1
X ≤ x ), where −1
n n+1 Notice that we have an expression of the form x=F ( y) .
X =n ⇔ ∑
i=1
[ −1
λ ]
ln U i ≤ 1<∑
i=1
−1
λ
ln U i [ ] |
We know, by definition,
−1
F( x )=P( X ≤ x) , so:
P¿
n n+1 G( y )=F (F ( y))= y
X =n ⇔
−1
λ
ln (∏ )i=1
U i ≤ 1<
−1
λ
ln ( )
∏ Ui
i=1
|
In summary, the cdf of
get the pdf, we see that
Y is G( y )= y . If we take the derivative of the cdf to
g( y )=1 Let’s remember the pdf for a uniform random
n n+ 1 variable:
X =n ⇔ ∏ U i ≤ e− λ < ∏ U i 1
Ex: to obtain Pois(2) we use
i=1
e− λ=e−2by=0.1353>
This study source was downloaded
n+ 1
100000899606070
i=1
i=1
If
f (x)= b−a
∏ Ufromi CourseHero.com on 02-22-2026 22:23:49
a=0 , b=1 ,
x ∈[a , b ]
0 GMTotherwise
-06:00
then f (x)=1=g ( y )
{ . Therefore,
Consider the following table:
https://www.coursehero.com/file/81716187/simulation-final-exam-cheatsheetdocx/
Y ∼ U (0,1) .
