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Let $X_1,X_2,\dots,X_n$ be a random sample from a distribution with p.d.f.: $$f(x, \theta)= e^{-(x-\theta)}; \theta<x<\infty;-\infty<\theta<\infty$$ obtain the sufficient statistic for parameter $\theta$. I obtain this likelihood function $$L=\exp\left\{-\sum_1^n x_I\right\}e^{n\theta}.$$ I want to use factorization theorem (only know this rule to find sufficient statistic), but not able to factorize it into desired form as the $e^{n\theta}$ does not have any function of $x$ with it.

What should I do and how can I find the sufficient statistic in other such questions. I also know how to find maximum likelihood estimator but that also resulted in $\theta=0$ (pointless).

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  • $\begingroup$ e^{-\lambda x} shows $e^{-\lambda x}$; \exp(-\lambda x) shows $\exp(-\lambda x)$; \infty shows $\infty$; more formatting tips here. $\endgroup$ – Em. Jan 11 '16 at 17:14
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Write the joint distribution for an IID sample $\boldsymbol x = (x_1, x_2, \ldots, x_n)$ drawn from this distribution: $$f(\boldsymbol x \mid \theta) = \prod_{i=1}^n f(x_i \mid \theta) = \prod_{i=1}^n e^{-(x_i - \theta)} \mathbb{1} (x_i > \theta).$$ Note here I have used the indicator function $$\mathbb{1}(x_i > \theta)$$ to express the idea that each observation $x_i$ must be greater than $\theta$, otherwise the density is zero. It is not hard to see from this that $$\prod_{i=1}^n \mathbb 1(x_i > \theta) = \mathbb{1}(x_{(1)} > \theta),$$ since $x_{(1)} = \min_i x_i$ is the first order statistic of the sample, and each $x_i$ exceeds $\theta$ if and only if the smallest of the $x_i$s exceeds $\theta$. We may therefore write $$f(\boldsymbol x \mid \theta) = e^{-n(\bar x - \theta)} \mathbb{1}(x_{(1)} > \theta).$$ Note I used the fact that $$n \bar x = x_1 + x_2 + \cdots + x_n.$$ From this, I think you should be able to complete the problem: what is the appropriate factorization, and what is a sufficient statistic for $\theta$?

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  • $\begingroup$ It is already given that $x>\theta$ but I didn't get the point that if $x<\theta $ then the joint density will be 0 may be this will be the case if any random observation $x=\theta$ among the n sample observations. $\endgroup$ – Onix Jan 12 '16 at 4:47
  • $\begingroup$ I also obtained the same joint distribution but I am not able to understand the appropriate factorization, and also the indicator function what is it? Sorry for being this much lame, I am studying on my own and may have missed those things somewhere, thanks for yur help $\endgroup$ – Onix Jan 12 '16 at 4:56

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