1. The document describes exercises from a probability and statistics lab report, including generating random vectors, estimating distributions, and assessing hypotheses. 2. For the first exercise, random vectors were generated from uniform, normal, and exponential distributions and their histograms, CDFs, and boxplots were represented. Bin sizes were also calculated. 3. Subsequent exercises involved comparing mean and variance, assessing dependence between random variables, modeling loss event data, and applying the central limit theorem.

1. Probability and Statistics
Lab no
1 Report
Nazli Temur - April ,2015
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2. Introduction
This lab includes 5 main exercises that should be completed by the help of R Tool.
I achived to complete all the exercises except 5th one and this report includes a small
brief as per exercises along with R codes&outcomes.
Exercise 1
1.1 Generate 3 random vectors of size 10000 from different distributions .
• A uniform distribution between 0 and 1.
unif <-runif(10000,0.0,1.0)
• AnormaldistributionN(0,10)
norm<-rnorm(10000,0,sqrt(10))
• A exponential distribution of parameter λ = 2
rexp(10000,2)
a) What is the number of bins to be used to represent the corresponding
histograms according to Sturge’s rule?
Technically, Sturges’ rule is a number-of-bins rule rather than a bin-width rule.
> number_of_bin=log(10000,base=2)+1
> number_of_bin
[1] 14.28771
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n=1+log
2
N

3. b) What is the bin size according to the Normal Reference rule?
For Uniform : ((24*(sd(unif)^2)*sqrt(pi))/10000)^(1/3)
0.0706738
For Normal : ((24*(sd(norm)^2)*sqrt(pi))/10000)^(1/3)
0.3470349
For Exponantial : ((24*(sd(exp)^2)*sqrt(pi))/10000)^(1/3)
0.1013582
c) What is the number of bins for each sample vector you have generated
according to the Normal Reference Rule ?
For Uniform :
> unif_n=NULL
> unif_max=length(unif)
> unif_min=0
> unif_n=(unif_max-unif_min)/unif_h
> unit_n [1] 141495.2
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4. For Normal :
> norm_n=NULL //number
> norm_max=length(norm) // number of elements
> norm_max
[1] 10000
> norm_min=0
> norm_n=(norm_max-norm_min)/norm_h
> norm_n //number of elements divided by width of bin equally gives number of bin
[1] 28815.54
For Exponantial :
> exp_n=NULL
> exp_max=length(exp)
> exp_min=0
> exp_n=(exp_max-exp_min)/exp_h
> exp_n
[1] 98660.04
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5. d) Represent the histograms (R is using Sturge’s rule with improvements, hence
you can just use hist(X)) , cdfs and boxplots of each random vector.
hist(unif)
boxplot(unif)
plot.ecdf(unif)
hist(norm)
boxplot(norm)
plot.ecdf(norm)
hist(exp)
boxplot(exp)
plot.ecdf(exp)
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6. 1.2 For each random vector, compute the empirical variance and the empirical IQR
and plot those pairs in a graph.
Varvector=NULL
IQRvector=NULL
for(V in seq(1,1000,by=50))
{
+ x<-rnorm(1000,0,sqrt(V))
+ IQRvector=c(IQRvector,IQR(x))
+ Varvector=c(Varvector,var(x))
}
plot(IQRvector,Varvector)
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7. Exercise 2
2. E[1/X] vs. 1/E[X]
Let us consider the family of uniform distributions in the interval [100 − v, 100 + v] for v > 0
2.1. What are the mean/variance of the family?
x=[a,b] //a =100-v b=100+v
E=[a+b]/2 //mean
V= [b-a]^2/12 //variance
E=(100+v-(100-v))/2 =100 it means the mean is not depend the variance of this uniform
distribution of interval.
V=((100+v) -(100-v))^2 /12 =(2v)^2/12 = v^2/3 which means, the variance is impacted
exponentially depend on the v value.
2.2. For each v ∈ {1, 2, . . . 30}, draw a random vector of size 1000, compute its empirical
variance v[X] as well as E[1/X] (simply mean(1/x) in R). Plot the pairs (E[1/X] − 1/E[X],
> for(v in seq(1,30,by=1))
+ { E=(100-v)+(100+v)/2
+ V=((100+v)-(100-v))^2/12
+ Vector_x<-rnorm(1000,E,V)
+ }
> for(v in seq(1,30,by=1))
+ { E=(100-v)+(100+v)/2
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8. + V=((100+v)-(100-v))^2/12
+ Vector_y<-rnorm(1000,1/E,V)
+ }
> plot(Vector_x,Vector_y)
Exercise 3
3. Dependence vs. similar distribution
3.1. Draw a random variable X and a random variable Y (both of size 10000) from the same
exponen- tial distribution of parameter λ = 2. Plot the qqplot and the scatterplot of X and Y .
The scatterplot is simply obtained by plot(X,Y). In the scatterplot, it might be useful to zoom
in where the mass is. You can adjust the x-axis (resp. y-axis) between the 10-th and 90-th
quantiles of X (resp. Y) with the command :
> X<-rexp(10000,2)
> Y<-rexp(10000,2)
> plot(X,Y,main="Scatter Plot")
> qqplot(X,Y,main="QQ Plot")
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9. For Adjusment :
> min_x=quantile(X,0.1)
> max_x=quantile(X,0.9)
> min_y=quantile(Y,0.1)
> max_y=quantile(Y,0.9)
> X2<-X[X>min_x&X<max_x]
> Y2<-Y[Y>min_y&Y<max_y]
> plot(X2,Y2,main="Adjusted Scatter Plot")
> qqplot(X2,Y2,main="Adjusted QQ Plot")
>
3.2. Let Z = log(X) + 5. Plot the qqplot and the scatterplot of X and Z. Comment the results
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10. The distribution of new vector Z follows the same distribution.We can see this via QQ Plot.
and If we try to draw a scatter plot it will look like line because there is a relation between Z
and X such that Z=a(x)+c , because a is a log of X vector the line will be convergent like
logarithm function.
>Z<-log(X)+5
> qqplot(Z,X,main=" QQ Plot X-Z”)
> Z2<-log(X2)+5
> qqplot(Z2,X2,main="Adjusted QQ Plot X2-Z2")
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11. Exercise 4
4. Loss Events
4.1 Data Cleaning
myfile=scan("~/Desktop/LAB/147.32.125.132.loss.txt")
Read 3439 items
min=quantile(myfile,0.1)
max=quantile(myfile,0.9)
X<-myfile
X2<-X[X>min&X<max]
X2
boxplot(X,X2)
myfile2=scan("~/Desktop/LAB/195.204.26.25.loss.txt")
Read 16091 items
min2=quantile(myfile2,0.1)
max2=quantile(myfile2,0.9)
Y<-myfile2
Y2<-Y[Y>min&Y<max]
Y2
boxplot(Y,Y2)
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12. 4.2 Assessing the exponential hypothesis
4.2.1. For each of the 2 connections (the cleaned versions obtained from the previous
question), estimate the parameter of the exponential distribution that should model it.
First File
> myfile=scan("~/Desktop/LAB/147.32.125.132.loss.txt")
> Read 3439 items
> min=quantile(myfile,0.1)
> max=quantile(myfile,0.9)
> X<-myfile
> X2<-X[X>min&X<max]
> Mean_vector_x=NULL
> for(V in seq(1,1000,by=1)) {
+ x<-rnorm(1000,mean(X2),sqrt(var(X2)))
+ y<-sample(x,10)
+ Mean_vector_x<-c(Mean_vector_x,mean(y))
+ }
+ > hist(Mean_vector_x,main=“Sample Means")
+ > plot(Mean_vector_x,main=“Sample Means”)
Second File
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13. Second File
> myfile2=scan("~/Desktop/LAB/195.204.26.25.loss.txt")
Read 16091 items
> min2=quantile(myfile2,0.1)
> max2=quantile(myfile2,0.9)
> Y<-myfile2
> Y2<-Y[Y>min&Y<max]
> Mean_vector_y=NULL
> for(V in seq(1,1000,by=1)) {
+ x<-rnorm(1000,mean(Y2),sqrt(var(Y2)))
+ y<-sample(x,10)
+ Mean_vector_y<-c(Mean_vector_y,mean(y))
+ }
+ > hist(Mean_vector_y,main=“Sample Means of Second File ")
+ > plot(Mean_vector_y,main=“Sample Means of Second File ")
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14. 4.2.2 For each of the 2 connections, generate a random vector following the exponential
distribution of size 1000, represent the qqplot of each vector and the corresponding trace.
Comment.
qqplot(Mean_vector_x,Mean_vector_y)
Exercise 5
5. Central limit theorem
• A uniform distribution between 0 and 1.
• AnormaldistributionN(0,10)
• A exponential distribution of parameter λ = 2
5.1 Report in a table the empirical (resp. theoretical) mean and standard deviation for each
random vector (resp. random variable).
5.2 Prove that we are in the conditions of the theorem for each vector.
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15. 5.3 Towards which distribution should
︎
(n)(Sn − #) should converge in each case.
5.4 Represent in a table with three columns (one for each original distribution) and two
rows corresponding to:
• the histogram of the original distributions
• S10
5.5 Report also the empirical mean and standard deviation for S10 for all cases.
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