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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?

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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 $\theta$.

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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.

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

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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 $n$.

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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: $V(X_{max})=E(X_{max}^2)-E^2(X_{max})= \frac{n\theta^2}{n+2}-(\frac{n \theta}{n+1})^2=\frac{n \theta^2}{(n+2)(n+1)^2} = V(X_{max})$

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  • $\begingroup$ 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? $\endgroup$
    – pppp_prs
    Sep 16 at 22:42

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