I am working through some textbook problems and came across a problem I am having difficulty with. The problem asks to give the bias of a point estimate, namely for a given set of data $X = \{x_1, x_2,\ldots,x_n\}$ drawn from a discrete $\text{Uniform}\{0,1,\ldots, k\}$, what is the bias of the estimator $\hat\theta = \max(X)$?

I'm pretty sure I have a representation for $E[\hat\theta]$, namely:

$$E[\hat\theta] = \sum_{i=0}^{n-1} \left(\frac{k-1}{k}\right)^{ni} \left(1-\left(\frac{k-1}{k}\right)^n\right)(k-i)\\ \hspace{.91cm} = \left(1-\left(\frac{k-1}{k}\right)^n\right)\sum_{i=0}^{n-1} \left(\frac{k-1}{k}\right)^{ni} (k-i)$$

The first term represents the probability that the numbers larger than $k-i$ have not been drawn, while $1 - (\frac{k-1}{k})^n$ represents the probability of any number in the range being drawn. Thus, those terms together represent the probability that $k-i$ is the largest number in the sample. However, to measure the bias, I need to simplify the summation as bias is $E[\hat\theta] - k$, but I am a bit unsure how to simplify the summation.

  • $\begingroup$ Could you clarify what you mean by "Uniform$(0,k)$"? Is is a discrete uniform distribution on the integers $\{0,1,\ldots, k\}$ or a continuous uniform distribution on the interval $[0,k]$? $\endgroup$
    – whuber
    Jun 8, 2019 at 23:17
  • $\begingroup$ It is discrete, sorry about that. $\endgroup$ Jun 9, 2019 at 5:14
  • $\begingroup$ Please edit the information into your question. $\endgroup$
    – Glen_b
    Jun 9, 2019 at 6:27

1 Answer 1


Let $X_1,\dots,X_n$ be a random sample from a discrete uniform distribution $\text{U}\{0,1,\dots,\kappa\}$. Let $\hat{\kappa}=\max\{X_1,\dots,X_n\}$. Then (Why? Work out the details), $$ \text{E}[\hat{\kappa}] = \sum_{m=0}^{\kappa-1} \Pr\{\hat{\kappa}>m\} = \sum_{m=0}^{\kappa-1} \left( 1 - \left(\frac{m+1}{\kappa+1}\right)^n\right) = \kappa - \frac{1}{(\kappa+1)^n} \sum_{m=1}^\kappa m^n, $$ in which $\sum_{m=1}^\kappa m^n$ can be written as a polynomial in $\kappa$ (if you want an explicit formula which involves Bernoulli numbers). Subtract $\kappa$ from $\text{E}[\hat{\kappa}]$ to get the bias of $\hat{\kappa}$.

kappa <- 10
n <- 15
N <- 10^5
x <- matrix(sample(0:kappa, N*n, replace = TRUE), ncol = n)
kappa_hat <- apply(x, 1, max)
(bias_MC <- mean(kappa_hat) - kappa)
(-1/(kappa+1)^n) * sum((1:kappa)^n)

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