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Let $Y_n$ be the nth order statistic of a random sample of size n from a distribution with pdf $f(x|\theta) = 1/\theta$, $0<x<\theta$, zero elsewhere. Take the loss function to be $L[\theta, \delta(y)] = [\theta - \delta(y_n)]^2$. Let $\theta$ be an observed value of the random variable $\Theta$, which ahs the prior pdf $h(\theta) = \beta\alpha^{\beta}/\theta^{\beta+1}$, $\alpha < \theta < \infty$, zero elsewhere, with $\alpha > 0$, $\beta > 0$. Find hte Bayes solution $\delta(y_n)$ for a point estimate of $\theta$.

I'm a little confused with finding the likelihood function. Since they are telling us that $Y_n$ is the nth statistic...does that mean that we only have one function in the likelihood function?

Thanks in advance

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  • $\begingroup$ One of your doubts: $Y_n=\max \{X_1,\dots,X_n\}$. Also, be careful with "Let $\theta$ be an observed value of the random variable $\Theta$." This sentence makes no sense, since we don't observe the values of parameters. Tip: write the indicators explicitly. Check this question: stats.stackexchange.com/questions/52531/… $\endgroup$
    – Zen
    Mar 28, 2013 at 16:05

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The nth order statistic means the nth largest. Hence we must use the likelihood given to determine the distribution function (or CDF) for the order statistic. Now if $Y_n\leq t$ then we also have $Y_{n-1}\leq t$ and also $Y_{n-2}\leq t$, etc. - as $Y_n$ is the maximum. The probability of this is given by:

$$Pr(Y_n\leq t)=Pr(Y_1\leq t,...,Y_n\leq t)=[F(t)]^n=\left(\frac{t}{\theta}\right)^n$$

Simple differentiation with respect to $t$ gives the likelihood function for $Y_n$

$$f(Y_n|\theta)=\frac{nY_n^{n-1}}{\theta^n}$$

I think this is all you were confused on, so I'll stop here. Let me know if you need more details

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  • $\begingroup$ Thanks a lot. I am trying to find $k(\theta|y_n) = \frac{g(y_n,\theta)}{g_1(y_n)} = \frac{L(y_n|\theta)h(\theta)}{g_1(y_n)}$. We know the likelihood $L(y_n|\theta)$ and $h(\theta)$. So now I have to find $g_1(y_1)$. But that is $g_1(y_n) = \int^{\infty}_{\infty} g(y_n, \theta)d\theta$. But when I compute that integral, I get infinity (because n is both in the numerator and the denominator)... $\endgroup$
    – Artus
    Mar 28, 2013 at 13:36
  • $\begingroup$ I think that (instead of doing what I wrote in the previous comment) I can use the fact that $k(\theta|y_n)$ is proportional to $g(y_n|\theta)h(\theta)$ and determine the pdf from there. Since $k(\theta|y_n)$ is proportional to (in this case) $1/(\theta^{n+\beta-1})$. It must be uniform, right? Since we are considering the square error loss function...we just find the mean, which is $\theta^{n+\beta-1}/2$, right? $\endgroup$
    – Artus
    Mar 28, 2013 at 13:43
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    $\begingroup$ Artus, you seem to be overlooking that the prior restricts $\theta$ to be $\alpha$ or larger, thus avoiding the infinities you have encountered. $\endgroup$
    – whuber
    Mar 28, 2013 at 17:02

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