Difference drawing random numbers from distributions R [migrated] - r

Question

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I am comparing these two forms of drawing random numbers from a beta and a Gaussian distribution. What are their differences? Why are they different?
The first way (_1) simulates from a Uniform(0,1) and then applies the inverse CDF of the Beta (Normal) distribution on those uniform draws to get draws from the Beta (Normal) distribution.
While the second way (_2) uses the default function to generate random numbers from the distribution.
Beta Distribution
set.seed(1)
beta_1 <- qbeta(runif(1000,0,1), 2, 5)
set.seed(1)
beta_2 <- rbeta(1000, 2,5)
> summary(beta_1); summary(beta_2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.009481 0.164551 0.257283 0.286655 0.387597 0.895144
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.006497 0.158083 0.261649 0.284843 0.396099 0.841760
Here every number is different.
Normal distribution
set.seed(1)
norm_1 <- qnorm(runif(1000, 0,1), 0, 0.1)
set.seed(1)
norm_2 <- rnorm(1000, 0, 0.1)
> summary(norm_1); summary(norm_2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-0.3008048 -0.0649125 -0.0041975 0.0009382 0.0664868 0.3810274
Min. 1st Qu. Median Mean 3rd Qu. Max.
-0.300805 -0.069737 -0.003532 -0.001165 0.068843 0.381028
Here the numbers are almost the same except in the mean and median
Shouldn't all be equal? Because I am generating random numbers from distributions with the same parameters

Answer 1

I think your question boils down to the assumption about the random number generator. If rnorm used the same RNG as runif under the hood, then your expectation would hold. It does not use the same RNG. The normal distribution RNG and uniform RNG are separate. See ?RNGkind. Without that exact match, you are left with the statistical tests below:
Is the mean of norm_1 different from the mean of norm_2?
t.test(x = norm_1, y = norm_2)
p-value > 0.05 indicates there is insufficient evidence to reject the null hypothesis that the means are equal at the 0.05 type I error level
Are the distributions different?
ks.test(x = norm_1, y = norm_2)
p-value > 0.05 indicates there is insufficient evidence to reject the null hypothesis that the distributions are equal at the 0.05 type I error level

Answer 2

I tried to sample a Bernoulli parameter at home using two different ways.
I flip a coin and assign 1 to heads and 0 to tails
I roll a six sided dice and assign the result 1 to a roll of the 3 highest numbers and the result 0 to a roll of the 3 lowest numbers.
I did this only twenty times instead of thousand times but the principle is the same. I got the following results:
result 0
result 1
Method 1
11
9
Method 2
8
12
Q: Why did I not get the same result for both methods?
A: Well, it is of course because they are samples and are supposed to be variable everytime.
If I would be able to reset some random seed to remove the variability, then this still doesn't matter because they are different methods.
Why is there no use of inverse transform sampling?
The normal distribution actually does use inverse transform sampling. The following command returns the same value of 0.3735462
set.seed(1)
rnorm(1,1,1)
set.seed(1)
qnorm(runif(1),1,1)
Also the rbeta uses inverse transform sampling and the following returns the same 0.7344913 and 0.2655087, which are only different by the relationship Y = 1-X (so internally there is some inversion)
alpha = 1
beta = 1
set.seed(1)
rbeta(1,alpha,beta)
set.seed(1)
qbeta(runif(1),alpha,beta)
The beta function becomes different when when $\alpha$ and $\beta$ are not both equal to one. This is because the inverse sampling is not very efficient and the the rbeta function will do some algorithm that creates the sample in a different way. Below is a code with the algorithm for the case that $min(\alpha,\beta) \leq 1$.
See for more about the algorithm: Hung, Ying-Chao, Narayanaswamy Balakrishnan, and Yi-Te Lin. "Evaluation of beta generation algorithms." Communications in Statistics-Simulation and Computation 38.4 (2009): 750-770.
You can see a few points that are calculated differently. The algorithm has a few steps where it starts redrawing random numbers, and it does this because redrawing numbers is easier than computing the inverse transform for a difficult case.
alpha = 0.9
beta = 0.9
#### Cheng's BC algorithm
### used if min(alpha,beta)<=1
### initialize
set.seed(1)
p = min(alpha,beta)
q = max(alpha,beta)
a = p+q
b = p^-1
delta = 1+q-p
k1 = delta*(0.0138889+0.0416667*p)/(q*b-0.777778)
k2 = 0.25 + (0.5+0.25/delta)*p
sample = function() {
### Perform steps of algorithm in a loop
step = 1
while(step<6) {
if (step == 1) {
U1 = runif(1)
U2 = runif(1)
if (U1 < 0.5)
{step = 2}
else
{step = 3}
}
if (step == 2) {
Y = U1*U2
Z = U1*Y
if (0.25*U2 + Z-Y >= k1) {
step = 1
} else {
step = 5
}
}
if (step == 3) {
Z = U1^2*U2
if (Z > 0.25) {
step = 4
} else {
V = b*log(U1/(1-U1))
W = q*exp(V)
step = 6
}
}
if (step == 4) {
if (Z < k2) {
step = 5
} else {
step = 1
}
}
if (step == 5) {
V = b*log(U1/(1-U1))
W = q*exp(V)
if (a*(log(a/(p+W))+V) - 1.3862944 < log(Z)) {
step = 1
} else {
step = 6
}
}
}
if (q == alpha) {
X = W/(p+W)
} else {
X = p/(p+W)
}
return(X)
}
sample()
n = 20
beta_orig = sapply(1:n,function(x) {
set.seed(x)
rbeta(1,alpha,beta)
})
beta_quantile = sapply(1:n,function(x) {
set.seed(x)
qbeta(runif(1),alpha,beta)
})
beta_BC = sapply(1:n,function(x) {
set.seed(x)
sample()
})
plot(beta_orig,beta_BC, pch = 1, xlim = c(0,1), ylim = c(0,1))
points(beta_orig,beta_quantile, col = 2, pch = 3)
legend(0.3,1, c("rbeta compared to inverse transform sampling", "rbeta compared to manual"), pch=c(3,1), col = c(2,1), cex = 0.85)
Some weird effect
In the code above I was resetting the random seed for each computation. The inverse transform is only the same for the first number. When you compute multiple numbers then only the first number is the same.
The following code
set.seed(1)
rnorm(6,1,1)
set.seed(1)
qnorm(runif(6),1,1)
set.seed(2)
rnorm(6,1,1)
set.seed(2)
qnorm(runif(6),1,1)
returns
[1] 0.3735462 1.1836433 0.1643714 2.5952808 1.3295078 0.1795316
[1] 0.3735462 0.6737666 1.1836433 2.3297993 0.1643714 2.2724293
[1] 0.1030855 1.1848492 2.5878453 -0.1303757 0.9197482 1.1324203
[1] 0.10308546 1.53124079 1.18484918 0.03810797 2.58784531 2.58463150
What you see here is that rnorm function skips a number. The reason is because it samples two random numbers to create more precision.
See these lines in the source ode of the norm_rand() function that R uses https://svn.r-project.org/R/trunk/src/nmath/snorm.c
define BIG 134217728 /* 2^27 */
/* unif_rand() alone is not of high enough precision */
u1 = unif_rand();
u1 = (int)(BIG*u1) + unif_rand();
return qnorm5(u1/BIG, 0.0, 1.0, 1, 0);