# Sampling with replacement - what's the distribution of the histogram?

When sampling K numbers out of 1,...,N with replacement, and calculating the histogram of the sampled set - what's the distribution? Or if that's to hard, whats the expectation of each bin in the histogram?

Ex: Sampling K=5 out of N=3, we could get the set 2,2,2,3,3 in which case the histogram would be 0,3,2. The histogram has N elements, and I'm asking how each of them distributes.

• I think it depends with what probabilities you sample from the numbers $1,2,...,N$ and also you might need a much larger $K$ in order to understand how these numbers are distributed. Apr 13, 2021 at 19:59
• I assume a uniform sampling distribution. K and N are parameters of the histogram's distribution. Apr 13, 2021 at 23:16
• I think the answer is a Bernoulli distribution with probability 1/N and K repetitions. Apr 14, 2021 at 13:33
• But Bernoulli produces outcomes $0$ or $1$, I think what BruceET displays as the answer is really what you need. Apr 14, 2021 at 13:38

Let $$N = 5.$$ Then you are simulating the discrete uniform distribution on the integers 1 through $$5,$$ each with probability $$1/5.$$

With $$K = 1000$$ values sampled from among $$1$$ through $$5,$$ you ought to get an idea that the simulation process produces values that match this discrete uniform distribution. [I'm using R, following @Fiodor1234's Comment to use a larger $$K.]$$

set.seed(413)                # for reproducibility of results
x = sample(1:5, 1000, rep=T) # sampling 1000 values at random
summary(x);  length(x);  sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.000   2.000   3.000   3.017   4.000   5.000
[1] 1000     # sample size
[1] 1.3837   $sample standard deviation (aprx 1.414). table(x) x 1 2 3 4 5 180 219 194 218 189  Distribution: j = 1:5; pdf = rep(1/5, 5)$ PDF of discrete unif dist'n


$$P(X=1) = P(X=2) = P(X=3) = P(X=4) = P(X=5) = 1/5.$$ Thus $$E(X) = \frac{1}{5}\cdot 1 = \frac{1}{5}\cdot 2 + \cdots + \frac{1}{5}\cdot 5 = \frac{15}{5} = 3,$$ $$Var(X) = E[(X-3)^2] = \frac{1}{5}\left[(-2)^2 + (-1)^2 = 0^2 + 1^2 + 2^2\right] = \frac{10}{5} = 2,$$ and $$SD(X) =\sqrt{2} = 1.414.$$

sum((1/5)*j)
[1] 3                    # E(X)
sum((1/5)*(j-3)^2)
[1] 2                    # Var(X)


Make a histogram of this sample of 1000 values of $$X$$ and overlay red dots to show the exact distribution of $$X.$$

cutp = 0:5 + .5              # boundaries of histogram bins
hist(x, prob=T, br=cutp. col="skyblue2")
points(j, pdf, pch=19, col="red")