is $\hat{\theta}$ unbiased

Question

consider a random sample of size n from a distribution with pdf $f(x;\theta)=\frac{1}{\theta}$ $0<x\leq \theta$ and zero otherwise. $0< \theta$

Now the first question was to find the MLE of $\hat{\theta}$ which I found to $X_{n:n}$ , now they want to find out if it is unbiased. My work so far: $$ \begin{align} E[\hat{\theta}] &=E[X_{n:n}] \\ &= E[n\frac{1}{\theta}[ln(\theta)]^{n-1}] \end{align} $$ now this is probably where i went wrong. isnt the cdf of $X_{n:n}$: $$ nf(x)[F(x)]^{n-1} $$ ?

since $X_{(n)} < \theta$ it is clearly not unbiased. No need to do any work. — mike, May 28 '13 at 14:24
every observation is in the range $(0, \theta)$, so the maximum is too. — mike, May 28 '13 at 15:58

response · Accepted Answer · 2013-05-28T14:35:42.707

1

The cdf of the maximum is given by $F(x)^n$. Thus, the pdf is $n f(x) F(x)^{n-1}$, In your case, we have for $0 < x \le \theta$:

$$f(x) = \frac{1}{\theta}$$

$$F(x) = \frac{x}{\theta}$$

Thus, the expected value of $\hat{\theta}$ is given by:

$$\int_0^{\theta}\frac{nx^{n-1}}{\theta^n}dx$$

That should hopefully help.

edited May 28 '13 at 14:35

answered May 28 '13 at 14:22

response

5,071

But I need the maximum right? $X_{n:n}$ denotes the maximum,, or not? Or should I not use the maximum of $X$ but the minimum of $\theta$? that sounds weird. – WiseStrawberry May 28 '13 at 14:24
You are correct. I have fixed my answer. – response May 28 '13 at 14:36
I did my cdf by integrating $\frac{1}{\theta}$ but on $\theta$ and not on $X$. Thanks! – WiseStrawberry May 28 '13 at 14:38

is $\hat{\theta}$ unbiased

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