# Unbiased estimator of a uniform distribution

For a random sample $$X_1,X_2,\ldots,X_n$$ from a $$\operatorname{Uniform}[0,\theta]$$ distribution, with probability density function $$f(x;\theta) = \begin{cases} 1/\theta, & 0 \le x \le \theta \\ 0, & \text{otherwise} \end{cases}$$

Let $$X_{\max} = \max(X_1,X_2,\ldots,X_n).$$ What is the value of k such that $$\hat \theta = kX_{\max}$$ is an unbiased estimator of $$\theta$$ ?

I'm not sure if there is more to this question, because my intuitive answer answer is just $$k=1$$. This is because if you order the sample like $$x_{(1)} \le x_{(2)} \le \cdots \le x_{(n)}$$ such that $$x_{(n)} = E[X_{\max}]$$. and the fact that the distribution is uniform, the estimator of $$\theta$$ should just be $$X_{\max}$$.

Unbiased estimator -> $$E\left[\widehat{\theta\,}\right] = kE[X_{\max}] = \theta$$

Is my logic wrong here?

Another answer has already pointed out why your intuition is flawed, so let us do some computations.

If $X$ is uniform, then: $$P(X_{max}<x)=P(X_i<x,\forall i)=\prod_i P(X_i<x)= \begin{cases} 1 & \text{if } x\ge \theta \\ \left(\frac{x}{\theta}\right)^n & \text{if } 0\le x\le \theta \\ 0 & \text{if } x\le 0 \end{cases}$$ so the density function of $X_{max}$ is: $$f_{max}(x;\theta)=\begin{cases} \frac{n}{\theta^n}x^{n-1} & \text{if } 0\le x\le \theta \\ 0 & \text{otherwise} \end{cases}$$ Then we can compute the average of $X_{max}$: $$E(X_{max})=\int_0^\theta x \frac{n}{\theta^n}x^{n-1} dx =\frac{n}{n+1} \theta$$ so $X_{max}$ is biased whereas $\frac{n+1}{n}X_{max}$ is an unbiased estimator of $\theta$.

Others have already provided excellent answers showing how $$X_{\mathrm{max}} = \mathrm{max}\{X_1,...,X_n\}$$ is a biased estimator and $$\frac{n+1}{n}X_{\mathrm{max}}$$ is an unbiased estimator. But $$\frac{n+1}{n}X_{\mathrm{max}}$$ is not the only unbiased estimator.

Let $$\hat{\theta} = 2\overline{X}_n$$. Then $$\hat{\theta}$$ is also an unbiased estimator.

\begin{align*} \mathrm{bias}(\hat{\theta}) &= \mathbb{E}(\hat{\theta}) - \theta\\ &= 2\mathbb{E}(\overline{X}_n) - \theta\\ &= \frac{2}{n}\sum_{i=1}^n \mathbb{E} (X_i) - \theta\\ &= \frac{2}{n}n\frac{\theta}{2} - \theta\\ &= 0 \end{align*}

To show that the sample maximum $x_{max} = \max_{i=1}^n\{x_i\}$ is an unbiased estimator of $\theta$ you would need to show that $E(x_{max}) = \theta.$ This is saying that the average value of the maximum of $n$ uniform variables on $[0,\theta]$ is $\theta.$

This cannot be right. $\theta$ is the maximum value that any of the uniform variables can have. $x_{max},$ the sample maximum might tend to be somewhat close to $\theta$, but it will always be less than or equal to it. (Actually, cause this is a continuous distribution, it will always be strictly less, but that's beside the point.) Thus the average value of the sample maximum will be somewhat less than $\theta$ and the sample maximum will be a biased estimator.

In order to compute the $k$ that gives you an unbiased estimator, you must demand $$E(\hat \theta) = kE(x_{max}) = \theta$$ so take $$k = \frac{\theta}{E(x_{max})}$$

The remaining work is to calculate $E(x_{max})$ as a function of $\theta$ and $n$.

Just to add further information concerning the variance:
Because: $$E(X_{max}^2)=\int_{0}^{\theta}x^2 x^{n-1}\frac{n}{\theta^n}dx=\frac{n\theta^2}{n+2}$$ So: $$V(X_{max})=E(X_{max}^2)-E^2(X_{max})= \frac{n\theta^2}{n+2}-(\frac{n \theta}{n+1})^2=\frac{n \theta^2}{(n+2)(n+1)^2} = V(X_{max})$$

• The variance of the average estimator is θ^2/3n and for this is θ^2/(n+2)(n). So does this means that is a better estimator for real θ (lower variance) for large n? Sep 16, 2023 at 22:42
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