2
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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 $\theta$.

$\endgroup$
2
$\begingroup$

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 $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.