Question

1) Consider a linear congruential random number generator with parameters a = 35, c = 20 and m = 100.

a- Generate 5 random numbers by using this method. Use 84.

b- By using inverse transform method, generate 2 random variate for an exponential distribution with parameter λ = 0.5. Use the first two random numbers you generated in part a.

Answer 1

1) generating random numbers using linear congreuential number generator

a= 35

c= 20

m=100

X₀= 84, to generate 5 random numbers

Xi+!= (a X_i+c)* mod m

Ri= Xi/m

X1= ( 35 x 84 + 20) mod 100= 60, R1= X1/m= 60/100= 0.6

X2= (35 x 60 + 20) mod 100= 20,, R2=X2/m= 20/100= 0.2

X3= ( 35x20+20) mod 100= 20, R3=X2/m= 20/100= 0.2

X4= (35x20+20) mod 100= 20, R4=X2/m= 20/100= 0.2

X5= (35x20+20) mod100= 20, R5=X2/m= 20/100= 0.2

5 random numbers are = 0.6, 0.2,0.2,0.2,0.2

b) inverse sampling method

to find random variates first we need inverse cdf for it

cdf of exponential = 1-e^-px, p= lambda

let u = 1-e^-px, , u-1= -e^-px, 1-u= e^-px,, -px= log(1-u), x=- log(1-u)/p

so our inv-cdf = - log(1-u)/p

using two random numbers from above we have

Xi= - log(1-u)/p

X1= -log(1-0.6)/0.5= 1.8

X2=-log(1-0.2)/0.5= 0.4

two random variates are 1.8, 0.4