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Can someone help me solve this out, please? Thanks a lot.

Find the Bayes estimate of $\theta$ based on a single observation of $5$ from a distribution that is uniform on the interval 0 to $\theta$. Use square-error loss and a prior distribution of $\theta$ which has p.d.f. $p(\theta) = \theta \cdot \mathrm{e}^{-\theta}$ where $0<\theta <\infty$.

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    That's almost correct, but what would happen if $\theta$ was smaller than 5? Could you ever draw a 5 from such a uniform distribution? Also, instead of just looking for the "peak" as I wrote, it sounds like you need the mean squared error. First work on getting to the point where you are confidence that you understand $p(\theta|5)$ then we can use the definition of mean squared error to finish off the problem.2012-04-05

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The prior density is $p(\theta)=\theta e^{-\theta}$ for $\theta>0$, and the likelihood function is $ L(\theta) = \begin{cases} \frac 1 \theta & \text{for }\theta\ge5, \\ \\ \\ 0 & \text{otherwise}. \end{cases} $ Multiplying them, you get $ g(\theta)=p(\theta) L(\theta) = \begin{cases} e^{-\theta} & \text{for }\theta\ge5, \\ \\ \\ 0 & \text{otherwise}. \end{cases} $ Then we have $ \int_5^\infty g(\theta)\;d\theta = e^{-5},\text{ so }\int_5^\infty e^5 g(\theta)\;d\theta = 1. $ Therefore $f(\theta)= \begin{cases} e^{5-\theta} & \text{if }\theta>5 \\ \\ \\ 0 & \text{if }\theta<5 \end{cases}$ is the posterior probability density function. With squared-error loss, the Bayes estimate is just the posterior expected value $ \int_5^\infty \theta f(\theta)\;d\theta = 6. $

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    Its mean is not $1$ because the values of random variables with this distribution are always in the interval $[5,\infty)$. (And notice that it wouldn't make sense to use $1$ as an estimate of something that must obviously be at least $5$. The observation must be between $0$ and $\theta$, and the observation was $5$, so $\theta\ge5$.)2012-04-05