RANDOM NUMBER GENERATORS USING DYNAMICAL SYSTEMS REU Project Zsuzsanna HORV ´ RANDOM NUMBER GENERATORS USING
DYNAMICAL SYSTEMS
REU Project
Zsuzsanna HORVÁTH∗
Department of Mathematics, University of Utah
Faculty Mentor: Davar Khoshnevisan
∗
Supported in part by NSF and NSF–VIGRE grants
1
Contents
1 Introduction
3
2 Basic Root Finding
2.1 A Simple Example . . . . . . . . . . . . .
2.2 Generalization . . . . . . . . . . . . . . . .
2.3 A Numerical Example for a Nice Function
2.4 An Additional Example . . . . . . . . . .
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3 Simulations
3.1 White Noise? . . . . . . . . . . . . . . . . . .
3.2 Cauchy Distribution . . . . . . . . . . . . . .
3.3 Proof: Sequence Has The Cauchy Distribution
3.4 Distribution for Other Functions . . . . . . . .
3.5 Further Analysis . . . . . . . . . . . . . . . .
3.6 Unique Tangent Line . . . . . . . . . . . . . .
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4 Other Generators
15
4.1 Generating Normals from Uniform Distribution . . . . . . . . . . . . . . . 15
4.2 Generating Uniforms Using Dynamical Systems . . . . . . . . . . . . . . . 16
5 Independence
18
5.1 Testing for Independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
6 Appendix
20
7 Bibliography
25
2
1
Introduction
Anyone who considers arithmetical methods of producing random digits
is, of course, in a state of sin.
John Von Neumann
Numbers that are chosen at random have numerous applications in many fields, such
as simulations, sampling, numerical analysis and computer programming. When we refer
to randomness, we do not mean a random number; there is no such thing. There is nothing
random about any one number. Instead, we refer to a sequence of independent random
variables with a specified distribution. Hence, each number was obtained by chance, and
has no relationship to any other number in the sequence. At any point, all integers have an
equal chance of occurring, and they occur in an unpredictable fashion. To retrieve random
numbers from a table is a relatively inefficient and time–consuming practice. Since the
advent of computer technology, people have been trying to search for more efficient ways
to obtain random numbers using computer programs. There has been, and still is, an
interest in the production of random numbers using a computer’s arithmetic capabilities.
The main challenge is how do you create a random sequence when each number is dependent upon its predecessor? The answer is that the sequence itself is not random; it just
appears to be. The main purpose is to obtain sequences that behave as if they are random.
Statistics allows for quantitative measures for randomness. The two most useful
and well–known test procedures for studying random data, the chi-square test and the
Kolmogorov–Smirnov test, are both used in this paper. The chi-square test allows us to
give a probabilistic answer; we can say how probable or improbable certain types of events
are. The details are given later in the paper. The chi–square test applies to the situation
when observations fall into a finite number of categories. The Kolmogorv–Smirnov test
can be used for random quantities that range over infinitely many values.
An example of this is a random real number on the interval [0,1]. Even though the
computer can only represent finitely many real numbers, we want our random values to
behave as if all real numbers in the interval for equally likely. In order to specify the
distribution of the values of a random quantity, X, we define the distribution function
F (x) = P (X ≤ x). If we take n independent observations of the random quantity, X, i.e.
X1 , X2 . . . Xn , we can create the empirical distribution function,
Fn (x) =
n
X
i=1
I (Xi ≤ x) ,
where I(A) denotes the indicator of event A.
3
1.5
1.0
0.5
0.0
−0.5
−1.0
−1.5
0
1000
2000
3000
4000
5000
Figure 1.1 5000 pseudo-random numbers
As n gets large, Fn (x) should be a better and better approximation to F (x). The
Kolmogorov–Smirnov test is based on the differences between F (x) and Fn (x). A good
source of random numbers will provide empirical distribution functions that approximate
F (x) sufficiently well. The Kolmogorov–Smirnov test is used to determine how improbable
deviations between F (x) and Fn (x) are. The difference between the Kolmogorov–Smirnov
test and the chi–squared test is that the Kolmogorov–Smirnov test applies to distribution
functions, F (x), having no jumps, while the chi-square test allows for points of discontinuity (i.e. jumps). This is because all observations fall into a specified number of categories.
In this paper, we use the Newton–Raphson process to generate a sequence of random
numbers. We show that these numbers are independent using both the chi–squared test
as well as the Kolmogorov–Smirnov test. Furthermore, we show that the process can be
used to generate uniform random variables, which can then be transformed into standard
normal data using the Box–Muller transformation method. As a final test for independence, we analyze the sample variance.
4
2
Basic Root Finding
2.1
A Simple Example
Let us compute
equation
√
7. First, we must consider the problem as finding the solution of the
f (x) = x2 − 7 = 0.
It is difficult to solve this problem numerically, hence we use a method by which to
√
approximate a function near a given point using its tangent line. We know that 7
lies between 2 and 3, so we use one of these points, say x0 = 2, and use the linear
approximation at x0 = 2.
Recall,
f (x) = x2 − 7.
Hence,
f ′ (x) = 2x.
The linear approximation then becomes
f (x) ≈ f (x0 ) + f ′ (x0 )(x − x0 )
Making our substitutions yields:
f (x) ≈ −3 + 4(x − 2) = 4x − 11
We now approximate the solution to f (x) = 0 by the solution to 4x − 11 = 0. This
yields x = 11
= 2.75. This process produces a relatively good approximation, since the
4√
true value of 7 is 2.645751. In order to gain more accuracy, we repeat the process using
the new approximate solution. Hence, we will define x1 = 11
and perform the linear
4
approximation at this new point.
f (x) ≈ f (x1 ) + f ′ (x1 )(x − x1 )
11
11
121
− 7) + (x − )
16
2
4
We define x2 as the solution to f (x) = 0. Solving the above equations gives that x2 =
√
233
7 than x1 . If we
=
note
that
x
is
an
even
better
approximation
for
2
88
continued this process, we would be generating closer and closer approximations. This
process is, in essence, the Newton Method.
f (x) ≈ (
5
2.2
Generalization
We can now generate approximate solution to the equation f (x) = 0 using this algorithm.
Let us assume that x0 is a point that we believe is relatively close to the solution of the
equation. We can take a linear approximation at x0 and solve the corresponding linear
equation. More precisely, we define x1 as the solution to the linear equation
f (x0 ) + f ′ (x0 )(x − x0 ) = 0.
Rearranging the equation above, we obtain that
x1 − x0 = −
f (x0 )
f ′ (x0 )
and hence,
x1 = x0 −
f (x0 )
f ′ (x0 )
where, x1 can be thought of as the root of the line that is tangent to the plot of f over
x0 . Similar calculations show that
x2 = x1 −
f (x1 )
f ′ (x1 )
where f ′ (x1 ) is the slope of the tangent to f at the point (x1 , f (x1 )). Continuing this
process we define
xn = xn−1 −
2.3
f (xn−1 )
,
f ′ (xn−1 )
n = 1, 2, 3 . . .
A Numerical Example for a Nice Function
Let us consider an example where we want to find the root of the function,
f (x) = x3 − x + 4.
Assume we have chosen x0 = −1 for out initial guess. To perform the Newton–Raphson
iteration, we need to know the derivative,
f ′ (x) = 3x2 − 1.
Hence,
xn = xn−1 −
x3n−1 − xn−1 + 4
3x2n−1 − 1
Using our initial guess of x0 = −1 we can produce the following values:
6
x0
x1
x2
x3
x4
x5
= −1.000000
= −3.000000
= −2.230769
= −1.881119
= −1.800478
= −1.796333
x6 = −1.796322
x7 = −1.796322
x8 = −1.796322
x9 = −1.796322
x10 = −1.796322
x11 = −1.796322
Note that the last few values of xn became closer and closer to the same value. Hence,
we have found the approximate solution of xn = −1.796322, after only 5 steps.
2.4
Consider f (x) = sin(x). Let us use the initial value, x0 = 1. We define
xn = xn−1 −
sin(xn−1 )
.
cos(xn−1 )
Using the Newton–Raphson method, we can produce the following values for xn :
x0 = 1.0000000
x1 = −.5574077
x2 = .06593645
x3 = −.0000957
x4 = 0.0000000
In this case, we were able to find the xn that satisfies f (xn ) = 0 in 4 steps, demonstrating
the efficiency of the Newton–Raphson method.
3
3.1
Simulations
White Noise?
Figure 3.2 was produced by applying Newton’s method to the function f (x) = x2 + 1,
which has no roots. In this case, as the approximations get closer and closer to 0, the slope
of f (x) becomes more and more horizontal. This means that the corresponding tangent
line will cross the x-axis at a position that is extremely far from 0. Hence, the iterations
become: a number very close to zero, a number very far from 0, a number very close to 0,
and so on; making the process appear to be chaotic. Figure 3.2 shows that each output
depends on the previous input, but there will be no convergence to any single value, and
no discernible patterns. In essence, it appears to be white noise. In actuality, Newton’s
method produces a nonrandom sequence; but due to it’s complexity, the iterations in this
7
1.5
1.0
0.5
0.0
−0.5
−1.0
−1.5
0
1000
2000
3000
4000
5000
Figure 3.1 5,000 iterations of Newton’s Method
case misleadingly appear random. To demonstrate this effect, Figures 3.2 and 3.1 were
obtained by graphing the arctangent of the iterations:
yn = arctan(xn )
In this manner, the values are restricted to be between -1.5 and 1.5. In other words, the
random variables will take values in a bounded set.
3.2
Cauchy Distribution
Figure 3.3 shows the histogram of 200 iterations of Newton’s Method for the function
f (x) = x2 + 1 using an initial value x0 = 0.35. The histogram of 200 steps using an initial
value of x0 = 0.25 is displayed in figure 3.4. The density of the Cauchy distribution is
plotted over the histogram of the iterations, showing that if a function is not nice, using
the Newton–Raphson method will generate a chaotic sequence, which will have Cauchy
distribution.
3.3
Proof: Sequence Has The Cauchy Distribution
We will prove this claim using 200 iterations of the Newton–Raphson process, with initial
value x0 = 0.35 and f (x) = x2 + 1.
8
1.5
1.0
0.5
0.0
−0.5
−1.0
−1.5
0
1000
2000
3000
4000
5000
0.00
0.05
0.10
0.15
0.20
0.25
0.30
Figure 3.2 5,000 iterations of Newton’s Method
−40
−30
−20
−10
0
10
Figure 3.3 Histogram of 200 iterations of Newton’s Method for f (x) = x2 + 1 with initial
value x0 = 0.35
9
H0 : sequence generated has Cauchy Distribution with parameter: location=0, scale=1
Ha : H0 is not true
We will divide the interval into K cells and calculate the number of observations that fall
into each cell. We define this as Y1 , Y2 , . . . , Y8 . We define the range of our 8 cells as
[−∞, −10), [−10, −8), [−8, −1), [−1, 0), [0, 1), [1, 8), [8, 10), [10, ∞)
Let K = 8 and n = 200.
Recall, the density of the Cauchy distribution:
1
1
f (x) =
.
π 1 + x2
We define
Ei = 200
Z
ti
f (x)dx.
ti−1
Due to symmetry,
−10
1
= E8
E1 = 200
1 + x2
−∞
Z −8 1
1
E2 = 200
= E7
1 + x2
−10 π
Z −1 1
1
E3 = 200
= E6
1 + x2
−8 π
Z 0 1
1
= E5 .
E4 = 200
1 + x2
−1 π
Z
Recall
1
π
K
X
(Yi − Ei )2
i=1
Ei
∼ χ2 (K − 1)
We obtain that:
Y1
Y2
Y3
Y4
=4
=1
= 40
= 55
Y5
Y6
Y7
Y8
Using the definition of Ei gives:
10
= 55
= 40
=3
=2
0.15
0.10
0.05
0.00
−50
0
50
100
150
Figure 3.4 Histogram of 200 iterations of Newton’s Method for f (x) = x2 + 1 with initial
value x0 = 0.25
E1
E2
E3
E4
= E8
= E7
= E6
= E5
= 6.345104
= 1.571581
= 42.08332
= 50
Using the definition of χ2 we get:
(4 − 6.345104)2 (1 − 1.571581)2 (1 − 1.571581)2 (40 − 42.08332)2
+
+
+
+
6.345104
1.571581
1.571581
42.08332
(55 − 50)2 (40 − 42.08332)2 (3 − 1.571581)2 2 − 6.345104)2
+
+
+
∼ χ2 (7)
50
42.08332
1.571581
6.345104
More succinctly, we have just computed
K
X
(Yi − Ei )2
i=1
Ei
= 6.554694.
Referencing a standard χ2 table shows that χ2 (7) can be as large as 14.07 with 95%
probability. Hence, the null hypothesis that these values are Cauchy cannot be rejected
(at a significance level of 5%). Furthermore, the corresponding p-value is 0.476667, which
strengthens our conclusion that the null hypothesis cannot be rejected.
11
3.4
Distribution for Other Functions
Consider the function f (x) = x4 + 1. Figure 3.5 displays the histogram of 200 iterations
of the Newton–Raphson process with initial value x0 = 0.15. The density of the Cauchy
distribution (with parameters: location=0, scale=10) is shown plotted on the histogram.
Hence, the density of the Cauchy distribution is:
1
1
.
f (x) =
x 2
10π 1 + ( 10
)
H0 : sequence generated has Cauchy distribution with parameter: location=0, scale=10
Ha : H0 is not true
We define the range of our K = 8 cells as:
[−∞, −100), [−100, −5), [−5, −1), [−1, 0), [0, 1), [1, 5), [5, 100), [100, ∞).
Using the definitions from the previous section, we obtain:
Y1
Y2
Y3
Y4
E1
E2
E3
E4
= E8
= E7
= E6
= E5
Hence
=5
= 35
= 31
= 15
Y5
Y6
Y7
Y8
= 14
= 35
= 52
= 13
= 6.345104
= 64.13818
= 23.17162
= 6.345104
K
X
(Yi − Ei )2
i=1
Ei
= 52.539 ∼ χ2 (7).
Although Figure 3.5 may misleadingly imply otherwise, a χ2 statistic as large as 52.539
implies that these iterations did not come from a Cauchy distribution; the null hypothesis
must be rejected. Changing the parameters of the Cauchy distribution does not alter this
conclusion.
3.5
Further Analysis
produced by the Newton–Raphson method. To further examine these jumps, let us con12
0.015
0.010
0.005
0.000
−400
−200
0
200
400
600
Figure 3.5 Histogram of 200 iterations of Newton’s Method for f (x) = x4 + 1 with initial
value x0 = 0.15
sider the function f (x) = x4 + 1. We will use an initial value of x0 = 0.15. The first 50
iterations are below:

0.15000000 -73.96157407 -55.47117994 -41.60338349 -31.20253414
 -23.40189238 -17.55139978 -13.16350359 -9.87251809 -7.40412876
 -5.55248066 -4.16290007 -3.11870966 -2.33079058 -1.72834916
 -1.24783955 -0.80721367 -0.13010296 113.42402305 85.06801712
 63.80101243 47.85075836 35.88806649 26.91604446 20.18702052
 15.14023500 11.35510422
8.51615741
6.38671329
4.78907533

3.58953043
2.68674243
2.00216656
1.47047625
1.02423123

0.53550044 -1.22639762 -0.78426488 -0.06993388 730.87980871
 548.15985653 411.11989240 308.33991929 231.25493946 173.44120458
 130.08090338 97.56067742 73.17050780 54.87788021 41.15840865
Note the large jump between the values of x0 and x1 . While our initial estimate was
a small positive number very close to 0, the next iteration has become a large negative
number very far from 0. The iterations from x2 to x18 are progressively becoming closer
and closer to 0; that is, until the jump between x18 and x19 This jump has created a large
positive number that is once again very far from 0. Due to the fact that f (x) does not
have any zeros, we can expect that this process will continue indefinitely; producing a
chaotic sequence. To further test this claim, we will perform 100000 iterations of Newton’s
13
1.0
0.8
0.6
0.4
0.2
0.0
−1.0
−0.5
0.0
0.5
1.0
Figure 3.6 Histogram of 100000 randomly generated uniform variables on [-0.7594792 ,
0.7598336].
Method and examine the values that the process takes immediately before a jump. A
histogram of these values can be found in Figure 3.7. Note that the values lie on the
interval [-0.7594792 , 0.7598336]. For comparison, Figure 3.6 contains a histogram of
100000 randomly generated uniform variables on the same interval.
3.6
Unique Tangent Line
The values that the Newton Raphson process will generate immediately before a jump
are in the interval [-0.7594792 , 0.7598336]. This is because there are two distinct points
where the tangent line of the graph at those points will intersect the origin, which implies
that before those points the value will stay in the same region, and after those points
there will be a jump. We again consider
f (x) = x4 + 1
and hence,
f ′ (x) = 4x3 .
We are interested in the tangent line to the point (x, x4 + 1). This tangent line is defined
as
g(t) = 4x3 (t − x) + x4 + 1.
14
1.0
0.8
0.6
0.4
0.2
0.0
−1.0
−0.5
0.0
0.5
1.0
Figure 3.7 Histogram of the values that are generated before the jump occurs
Solving this for 0, we obtain
g(0) = 0 = −4x4 + x4 + 1
0 = −3x4 + 1
r
4 1
x=&plusmn;
= &plusmn;.759835.
3
4
4.1
Other Generators
Generating Normals from Uniform Distribution
We will use the Box-Muller transformation method to generate normal random variables.
If x1 and x2 are uniform random variables on (0,1), we define y1 and y2 as follows:
y1 =
p
y2 =
p
The Jacobian determinant is then:
1
∂(x1 , x2 ) ∂x
∂y1
J=
= ∂x
∂(y1 , y2 ) ∂y12
−2 log(x1 ) cos 2πx2
−2 log(x1 ) sin 2πx2
∂x1 ∂y2 ∂x2 ∂y2 1
2
= − √ e−y1 /2
2π
15
1 −y22 /2
√ e
.
2π
2
0
−4
−2
Sample Quantiles
4
Normal Q−Q Plot
−4
−2
0
2
4
Theoretical Quantiles
Figure 4.1 Normal QQ plot
Since this is a product of a function of y1 alone and a function of y2 alone, we see that each
y is independently distributed according to the normal distribution. The details for this
method of how to generate independent standard normal random variables is as follows:
• Step 1: Generate two uniform random variables, x1 and x2 .
• Step 2: Apply the formula above to obtain y1 and y2 .
Figure 4.2 displays the histogram of 100000 values that were obtained by using the Box
Muller transformation method. The density function of the standard normal distribution
is plotted over the histogram. Figure 4.1 shows the QQ plot of these 100000 values.
Figures 4.1 and 4.2 both show that using the Box-Muller transformation method will,
indeed, generate standard normal data.
4.2
Generating Uniforms Using Dynamical Systems
The Newton Raphson process can be used to generate uniform random variables. We begin
with an initial value, x0 , and perform 500 iterations of the Newton–Raphson method. We
then take the produced xn value and use this as our next initial value. We continue with
these steps for 1000000 repetitions. After using this procedure of changing the initial value
for each run, we examine the values that occurred before a jump. Figure 4.3 contains the
histogram of these values. The results of the Kolmogorov–Smirnov test are below. The
16
0.4
0.3
0.2
0.1
0.0
−4
−2
0
2
4
Figure 4.2 Histogram of 100000 values using Box Muller transformation method
p-value is 0.2205, hence we cannot reject that null hypothesis that the values came from
a uniform distribution.
&gt; ks.test(resultT,&quot;punif&quot;)
One-sample Kolmogorov--Smirnov test
data: resultT
D = 0.0493, p-value = 0.2205
alternative hypothesis: two.sided
We use the uniform values that were generated using the Newton–Raphson process and
perform the Box–Muller transformation on these values. The QQ–plot can be found in
Figure 4.4. To further ensure that these value can, indeed, follow a normal distribution,
we perform the Kolmogorov–Smirnov test; the results are below.
&gt; ks.test(box, &quot;pnorm&quot;)
One-sample Kolmogorov--Smirnov test
data: box
D = 0.0568, p-value = 0.4586
17
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Figure 4.3 Histogram of values generated before a jump occurs when the initial value is
continuously changed
alternative hypothesis: two.sided
Hence, we can not reject the null hypothesis that these values are normally distributed.
5
5.1
Independence
Testing for Independence
We wish to determine if the process outlined in the previous section will generate independent values. We will accomplish this by analyzing the sample variance. We define
n−k
γ(k) =
1X
(Xi − X̄)(Xi+k − X̄)
n i=1
k = 1, . . . , n − 1
where X̄ denotes the sample mean of X.
The ratio,
γ(k)
γ(0)
denotes the sample variance. If the values are independent, then the plot
of this ratio should be around 0.This plot can be found on Figure 5.1. The value of γ(1)
γ(0)
is 1, all other values fluctuate around 0. Therefore we can make the statement that these
values are independent.
18
1
0
−2
−1
Sample Quantiles
2
3
Normal Q−Q Plot
−3
−2
−1
0
1
2
Theoretical Quantiles
−1.0
−0.5
0.0
0.5
1.0
Figure 4.4 Normal QQ–plot using Dynamical Systems
0
Figure 5.1 Plot of
100
200
γ(k)
γ(0)
19
300
400
3
6
Appendix
zero&lt;-function(x){
x^2+1
}
deriv&lt;-function(x){
2*x
}
newton&lt;-function(x0, n){
x&lt;-numeric(0)
x&lt;-x0
for (i in 1:n){
x[i+1]=x[i]-zero(x[i])/deriv(x[i])
}
return(x)
}
n&lt;-1000
x0&lt;-4
kk&lt;-1000
y&lt;-numeric(0)
for(k in 1:kk) y&lt;-rbind(y,newton(k/1000,n))
yy&lt;-atan(newton(.1,5000))
postscript( ’~/graph1.ps’,horizontal=FALSE,height=6,width=6)
plot(yy, type=’l’, xlab=&quot;&quot;, ylab=&quot;&quot;)
dev.off()
xx&lt;-newton(.35,200)
postscript( ’~/graph2.ps’,horizontal=FALSE,height=6,width=6)
hist(xx, nclass=100, xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;&quot;, prob=T)
curve(dcauchy(x, location = 0, scale = 1, log = FALSE), add=T)
dev.off()
h&lt;-runif(5000,-1.5,1.5)
postscript( ’~/graph3.ps’,horizontal=FALSE,height=6,width=6)
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plot(h, xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;5000 pseudo-random numbers&quot;)
dev.off()
xx&lt;-newton(.25,200)
postscript( ’~/graph4.ps’,horizontal=FALSE,height=6,width=6)
hist(xx, nclass=100, xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;&quot;, prob=T)
curve(dcauchy(x, location = 0, scale = 1, log = FALSE), add=T)
dev.off()
integrand&lt;-function(x){(1/(pi))*(1/(1+x^2))}
integrate(integrand, lower=-Inf, upper=-10)
1-pchisq(6.554694, df=7)
#looks for jumps
m=100000
xx&lt;-newton(.15,m)
yH=numeric(0)
yL=numeric(0)
j=1
for(i in 1:m){
if(xx[i]&lt;0 &amp;&amp; xx[i+1]&gt;0){
yL[j]=xx[i]
#values that function takes before shooting far out
yH[j]=xx[i+1] #value it lands on after it shoots off
j=j+1
}
if(xx[i]&gt;0 &amp;&amp; xx[i+1]&lt;0){
yL[j]=xx[i]
yH[j]=xx[i+1]
j=j+1
}
}
xx&lt;-newton(.15,200)
postscript( ’~/graph6.ps’,horizontal=FALSE,height=6,width=6)
hist(xx, nclass=75, xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;&quot;, prob=T)
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curve(dcauchy(x, location = 0, scale = 10, log = FALSE), add=T)
dev.off()
integrand2&lt;-function(x){(1/(10*pi))*(1/(1+(x/10)^2))}
integrate(integrand2, lower=-Inf, upper=-200)
postscript( ’~/graph7.ps’,horizontal=FALSE,height=6,width=6)
hist(yL, nclass=200, xlim=c(-1,1), ylim=c(0,1), xlab=&quot;&quot;, ylab=&quot;&quot;,
main=&quot;&quot;, prob=T)
dev.off()
postscript( ’~/graph8.ps’,horizontal=FALSE,height=6,width=6)
compare&lt;-runif(100000, -0.7594792,0.7598336)
hist(compare, nclass=100, xlim=c(-1,1), ylim=c(0,1),
xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;&quot;, prob=T)
dev.off()
box&lt;-function(n)
{
res&lt;-cos(2*pi*runif(n))*sqrt(-2*log(runif(n)))
return(res)
}
postscript( ’~/graph9.ps’,horizontal=FALSE,height=6,width=6)
rr&lt;-box(100000)
qqnorm(rr); qqline(rr, col=&quot;green&quot;)
dev.off()
postscript( ’~/graph10.ps’,horizontal=FALSE,height=6,width=6)
rr&lt;-box(100000)
hist(rr, nclass=50, xlab=&quot;&quot;, ylab=&quot;&quot;, main=&quot;&quot;, prob=T)
curve(dnorm(x, mean = 0, sd = 1, log = FALSE), add=T)
dev.off()
k=1
result&lt;-numeric(0)
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while((k*500)&lt;length(yL)){
result[k]=yL[500*k]
k=k+1}
#picks a point, generates 500 values, takes last one, and runs it again using
#the last one as the new starting point
resultT&lt;- (result+max(yL))/(max(yL))
box&lt;-cos(2*pi*resultT[1:226])*sqrt(-2*log(resultT[228:453]))
postscript( ’~/uniform1.ps’,horizontal=FALSE,height=6,width=6)
hist(resultT, nclass=9, xlim=c(0,1), xlab=&quot;&quot;, ylab=&quot;&quot;,
main=&quot;&quot;, prob=T)
dev.off()
postscript( ’~/box1.ps’,horizontal=FALSE,height=6,width=6)
qqnorm(box); qqline(box, col=&quot;green&quot;)
dev.off()
#tests for independece:
n=453
gam&lt;-numeric(n)
for (k in 0:(n-1)){
gam[k+1]&lt;-(1/n)*sum((resultT[1:(n-k)]-mean(resultT))*(resultT[(k+1):n]
-mean(resultT)))
}
gam0=(1/n)*sum((resultT-mean(resultT))^2)
#the ratio of gam[1:n]/gam is the sample variance, it should be
#around 0 if the values are independent.
postscript( ’~/gam2.ps’,horizontal=FALSE,height=6,width=6)
plot(gam[1:n]/gam, ylim=c(-1,1), type=&quot;l&quot;, xlab=&quot;&quot;, ylab=&quot;&quot;)
dev.off()
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7
Bibliography
Bain, L.J., Engelhardt, M.(1992). Introduction to Probability and Mathematical Statistics
(Second Edition). Duxbury Thomson Learning: United Kingdom.109,118,150,268.
Devaney, R.L. (1992). A First Course in Chaotic Dynamical Systems. Addison-Wesley
Publishing Company, Inc: New York. 164-173.
Knuth, D.E. (1998). The Art of Computer Programming (Third Edition) Volume 2.