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. [**D**0.05 = 0.565]

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Alson Garbuja
on 14 Apr, 2022
0

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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