Answer

It is not appropriate in terms of algorithm. Below is an explanation:

To use continuous version of inverse transform, note that the cdf of discrete uniform is \(F(x) = x/m\) for \(x=1,\ldots,m\). Set \(u = F(x)\), we have \[ x = um \implies F^{-1}(u) = um? \] There are at least two problems here:
1) As \(F(x)\) is only defined at \(x=1,...,m\), the inverse function \(F^{-1}(u)\) is actually not well-defined. As you can try in simulation, \(um\) may not be an integer
2) Even if you round the “inverse” properly, they may fail on boundary cases. For instance, if \(0 \le u \le 1\), \(\lceil um \rceil\) (i.e. round up to nearest integer) will fail at \(u=0\) since \(\lceil 0 \rceil = 0\) may not be in the support of \(X\)

Hence you should use discrete version of inverse transform for generating for discrete uniform random variable.

However, if you round the the “inverse” properly, the implementation is fine in R practically because the documentation states that “runif(1,0,1)” will not return 0 or 1. This avoids the boundary cases but I do not recommend it as it depends on specification of uniform random number in the language.

m = 100
n = 10000
U = runif(n) #use same set of U to compare plot
X = Y = vector(length = n)
# Standard way
for (i in 1:n)
{
  for (j in 1:m)
  {
    if (U[i] > (j-1)/m && U[i] <= j/m)
    {
      X[i] = j
      break
    }
  }
}
hist(X)

# Non-standard way
for (i in 1:n)
{
  Y[i] = ceiling(U[i]*m)
}
hist(Y) #same shape as X's plot

RMSC5102 2020S A Question on Discrete Uniform

LEUNG Man Fung, Heman

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