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Simulating Continuous Random Behaviour (cont.)
MAS1302:Computational Probability and Statistics:
Week 7
Dr. David Walshaw
March 14, 2007
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
3. Continuous Random Behaviour (cont.)
3.7 Checking for Randomness
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
3. Continuous Random Behaviour (cont.)
3.7 Checking for RandomnessIf we are going to use random number generators to simulatesystems such as queues, birth–death processes, etc. it is importantthat our psseudo–random number generators are not producingsequences of numbers which are non–random in any systematicway.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
3. Continuous Random Behaviour (cont.)
3.7 Checking for RandomnessIf we are going to use random number generators to simulatesystems such as queues, birth–death processes, etc. it is importantthat our psseudo–random number generators are not producingsequences of numbers which are non–random in any systematicway.
Here we consider the issues involved in checking the ‘randomness’of a random number generator. This sounds quite straightforward.However, as we shall see, a sequence of pseudo–random numberscan exhibit non–random behaviour in a surprising variety of ways.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
3. Continuous Random Behaviour (cont.)
3.7 Checking for RandomnessIf we are going to use random number generators to simulatesystems such as queues, birth–death processes, etc. it is importantthat our psseudo–random number generators are not producingsequences of numbers which are non–random in any systematicway.
Here we consider the issues involved in checking the ‘randomness’of a random number generator. This sounds quite straightforward.However, as we shall see, a sequence of pseudo–random numberscan exhibit non–random behaviour in a surprising variety of ways.
Using a variety of examples, we will consider ways in which asequence of supposedly independent U(0, 1) random numbers candeviate.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1: Checking the marginal distribution
The marginal distribution simply refers to the distribution ofindividual simulated values.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1: Checking the marginal distribution
The marginal distribution simply refers to the distribution ofindividual simulated values.
Suppose we simulate 10,000 observations from a U(0, 1)distribution. We might check the marginal distribution by tallyingthe frequencies in intervals of width 0.2.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1 (cont.)
a) Suppose we obtain the observed frequencies below:
Range (0,0.2) [0.2,0.4) [0.4,0.6) [0.6, 0.8) [0.8, 1.0)Probability 0.2 0.2 0.2 0.2 0.2
Expected Frequency 2000 2000 2000 2000 2000Observed Frequency 1783 1992 2274 1948 2003
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1 (cont.)
a) Suppose we obtain the observed frequencies below:
Range (0,0.2) [0.2,0.4) [0.4,0.6) [0.6, 0.8) [0.8, 1.0)Probability 0.2 0.2 0.2 0.2 0.2
Expected Frequency 2000 2000 2000 2000 2000Observed Frequency 1783 1992 2274 1948 2003
Does it look as though these have the correct marginaldistribution? Let’s have a look at a histogram...
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
Range0
0.2 0.4 0.6 0.8 1.0
2000
0
Frequency
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
Range0
0.2 0.4 0.6 0.8 1.0
2000
0
Frequency
In fact these observed frequencies are much too far from theexpected frequencies - this random number generator is very poor!
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1 (cont.)
b) Now suppose we obtain:
Range (0,0.2) [0.2,0.4) [0.4,0.6) [0.6, 0.8) [0.8, 1.0)Observed Frequency 2001 1998 1999 1999 2003
Do these look OK?
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
2000
00.2 0.4 Range0.6 0.8 1.00
Frequency
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
2000
00.2 0.4 Range0.6 0.8 1.00
Frequency
These observed frequencies are much too close to the expectedones! Again, a poor random number generator!
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.1 (cont.)
c) Now suppose we obtain:
Range (0,0.2) [0.2,0.4) [0.4,0.6) [0.6, 0.8) [0.8, 1.0)Observed Frequency 1954 2022 1977 2031 2016
What about these?
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
00 0.2 0.4 0.6 0.8 1.0 Range
Frequency
2000
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 1)
00 0.2 0.4 0.6 0.8 1.0 Range
Frequency
2000
In fact these are typical of what we would see from a genuineU(0, 1) random variable. This generator looks OK . . . for themarginal distribution at least!
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 Checking for dependence: runs test
Even if the marginal distribution is OK, other things can go wrong.
a) E.g. consider the following sequence:
0.134, 0.768, 0.233, 0.984, 0.746, 0.865, 0.321, 0.432, 0.250, 0.777
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 Checking for dependence: runs test
Even if the marginal distribution is OK, other things can go wrong.
a) E.g. consider the following sequence:
0.134, 0.768, 0.233, 0.984, 0.746, 0.865, 0.321, 0.432, 0.250, 0.777
Here the consecutive numbers increase, decrease, increase, ... andso on. It looks as though they are not independent.
One way of checking this is via a runs test.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
b) E.g. consider the following sequence:
(0.134, 0.279, 0.886), (0.197), (0.011, 0.923, 0.990), (0.876)
Here each run of ascending numbers has been identified inbrackets. We get:
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
b) E.g. consider the following sequence:
(0.134, 0.279, 0.886), (0.197), (0.011, 0.923, 0.990), (0.876)
Here each run of ascending numbers has been identified inbrackets. We get:
# of ‘run–ups’ with length 1 is R1 = 2# of ‘run–ups’ with length 2 is R2 = 0# of ‘run–ups’ with length 3 is R3 = 2
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
b) E.g. consider the following sequence:
(0.134, 0.279, 0.886), (0.197), (0.011, 0.923, 0.990), (0.876)
Here each run of ascending numbers has been identified inbrackets. We get:
# of ‘run–ups’ with length 1 is R1 = 2# of ‘run–ups’ with length 2 is R2 = 0# of ‘run–ups’ with length 3 is R3 = 2
Now in a long sequence of n ‘random’ numbers, let Rk be thenumber of ‘run–ups’ of length k. It is possible to calculate theexpected values of Rk and compare with the observed values.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
c) E.g. for n = 5000:
k 1 2 3 4 5 ≥ 6
Expected Rk 833 1042 458 132 29 6
Observed Rk 824 1074 440 113 42 7
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
c) E.g. for n = 5000:
k 1 2 3 4 5 ≥ 6
Expected Rk 833 1042 458 132 29 6
Observed Rk 824 1074 440 113 42 7
If the observed frequencies differ greatly from the expectedvalues,then this would indicate dependence in the sequence, whichshouldn’t be there.
In the table for c) above, however, the discrepancies are notparticularly large, so there is no real evidence for non–randomness.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.2 (cont.)
c) E.g. for n = 5000:
k 1 2 3 4 5 ≥ 6
Expected Rk 833 1042 458 132 29 6
Observed Rk 824 1074 440 113 42 7
If the observed frequencies differ greatly from the expectedvalues,then this would indicate dependence in the sequence, whichshouldn’t be there.
In the table for c) above, however, the discrepancies are notparticularly large, so there is no real evidence for non–randomness.
Exercise: Think what would happen for a sequence like the one ina) above.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.3: Checking for dependence: scatterplots
Now consider a plot of the n + 1th number against the nth number.I.e. we plot the pairs (U1,U2), (U2,U3), (U3,U4), . . .
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Example 3.7.3: Checking for dependence: scatterplots
Now consider a plot of the n + 1th number against the nth number.I.e. we plot the pairs (U1,U2), (U2,U3), (U3,U4), . . .
For a good random number generator, these points should berandomly scattered in the unit square. We consider two ways inwhich a scatterplot can deviate from this . . .
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 1 of 2)
10
0 U
U
1
n+1
n
+
+
+
+
++ +
++++ +
+
++
++
++ +
++
+
+
+
+
+
+
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+++
++++
++
+
+
++
++
++
+
+
+ +
+
+
+
Here the points are ‘clustered’:— the mean separation betweenpoints is less than would be expected under randomness.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Solution (Slide 2 of 2)
10
0
++
+
+
U
U
1
n+1
n
+
+
++
+
+
+
++ +
+
+ +
+
+
+
+
+ +
+
+
+
++
+
+
Here the points are too evenly spread:— the mean separationbetween points is greater than would be expected underrandomness.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Summary
Summary
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Summary
Summary
All of the ideas above involve comparing some observed behaviourin our pseudo–random sequence with the expected behaviour ifrandomness and independence hold.
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
Simulating Continuous Random Behaviour (cont.)
Summary
Summary
All of the ideas above involve comparing some observed behaviourin our pseudo–random sequence with the expected behaviour ifrandomness and independence hold.
In order to check whether the observed behaviour is too far fromwhat we’d expect (or too close!), we would need to carry out anappropriate hypothesis test for goodness of fit. You will studythese in Stage 2!
Dr. David Walshaw
MAS1302:Computational Probability and Statistics:Week 7
