Question

binomial RV B(n,p) 2. Simulating a Binomial RV. One procedure for generating uses n EXi is binomial if realizations of a uniform random variable and exploits the fact that Y the Xi are Bernoulli RVs. Here is an alternative procedure that requires generating only a single (!) uniform variate: 1/p and B 1/(1 p) 0) Let 1) Set 0 U[0, 1] 2) Generate 3) If k n, go to step 5; else, k ++ au; if u B(u- p). Go to 3 4) If u<p, set 1 and 0 and u p, set 5) y k k1 Simulate 1000 Binomial variates (p 1/2, n 8) using both methods. Compare their his tograms with the true Binomial pmf. Explain why the second method works (Hint: consider uniform[0,1 conditioned on U<p.) the conditional distribution of U

Answer 1

We will use the following python code to simulate the binomial random variable import numpy as np import matplotlib. pyplot as plt from scipy. stats import binom def ManyUniform (n,p) 1np. random.uniform (0,1,n) у %3D 0 for in 1 if i <= p y 1 return y def OneUniform (n,p) а,b %3D 1/р, 1/(1-р) np.random. uniform (0,1) y =0 for k in range (1 , n+1): if u<p u= y 1 u a*u else ub(u-p) return y yl ManyUniform (8 ,0.5) for i in range (1000)] [OneUniform (8,0.5) for i in range (1000) у2 pl,bl =np. histogram (yl, bins = range (0,9) , density = True) p2, b2 = np. histogram (y2, bins = range (0,9), density = True) plt plot (range (0,8), pl label = 'Using-8_uniform-variate' plt.plot (range (0 ,8), p2, label = 'Using-1cuniform.variate') x = np.arange (0 ,8) plt plot (x, binom pmf(x, 8, 0.5), ms=8, label='binomial-pmf' plt. legend ( plt show This gives us the following output:

0.25 0.20 0.15 0.10 0.05 Using 8 uniform variate Using 1 uniform variate binomial pmf 0.00 0 1 2 5 6 7 CO

Observe that the conditional cdf of U given U < p is P(U u, U <p) P(U < p) Fu (uU <p) 0 if u<0 if 0 u<p Otherwise Therefore the conditional pdf of U would be 1 if 0 u< p fu (u/U < p) р Otherwise Hence aUU p ~ Uniform(0, 1), where a = 1 (It follows from Change of Variable theorem). Similarly B(U - p)|U p~ Uniform(0, 1). Hence we can use only one Uniform variable and keep using conditions on it to generate new Uniform random variables. Therefore this second method is as reliable as the first one