From Probability Concepts and Random Number Generation chapter in PU/ Simulation and Modelling
Asked by arjun adhikari on 13 Apr, 2022
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Generate random numbers using linear congruential method with Xi = 27, a = 17, c = 43, and m = 100. And test their uniformity with Kolmogorov Smirnov test with 5% level of significance. [D0.05 = 0.565]
We have
Xi = 27, a = 17, c = 43, m = 100
Then using LCM,
X(i+1) = (a*Xi + c) mod m
X1 = (17*27 + 43) mod 100 = 2
X2 = (17*2 + 43) mod 100 = 77
X3 = (17*77 + 43) mod 100 = 52
X4 = (17*52 + 43) mod 100 = 27
X5 = (17*27 + 43) mod 100 = 2
Now,
R1 = X1/m = 2/100 = 0.02
R2 = X2/m = 77/100 = 0.77
R3 = X3/m = 52/100 = 0.52
R4 = X4/m = 27/100 = 0.27
R5 = X5/m = 2/100 = 0.02
Now using KS-test
Ri | 0.02 | 0.77 | 0.52 | 0.27 | 0.02 |
(i/n - Ri) | 0.18 | -0.37 | 0.08 | 0.53 | 0.98 |
(Ri - (i-1)/n) | 0.02 | 0.57 | 0.12 | -0.33 | -0.78 |
Then
D+ = MAX(i/n - Ri) = 0.98
D- = MAX(Ri - (i-1)/n) = 0.57
D = MAX(D+, D-) = 0.98
Since D > Da hence the random numbers generated are not uniform
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