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If $Y_1,\ldots,Y_n$ independent each having pdf: $$ f(y\mid \beta,\theta, x)=\theta e^{-\theta(y-\beta x)},~~ y>\beta x$$ where $x_1,\ldots,x_n$ are given, the parameters $\beta$ and $\theta$ are unknown. I know the joint sufficent statistics for $\beta$ and $\theta$ are $\overline{Y}$ and $\min\{Y_i/X_i\}$. But can I say that the sufficient statistic for $\beta$ is $\min\{Y_i/X_i\}$?

I don't know why, but I feel strange calculating sufficient statistics for only part of the parameters.

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  • $\begingroup$ I'm guessing you mean $f(y_i\mid\beta, \theta,x_i) = \theta e^{-\theta(y_i-\beta x_i)}$ for $y_i>\beta x_i$. But your way of expressing that is unclear, and it contradicts your assertion that the $Y$s are i.i.d., since it implies they are not identically distributed. $\qquad$ $\endgroup$ Aug 16 '16 at 12:37
  • $\begingroup$ @MichaelHardy Yes, thanks for the correction! $\endgroup$
    – Zander
    Aug 16 '16 at 15:16
  • $\begingroup$ @ZhengWang How did you get those joint sufficient statistics? Through factorization the joint pdf is $$ \theta^n e^{-\theta (\sum_{i=1}^n y_i - \beta \sum_{i=1}^n x_i)} \prod_{i=1}^n \mathbb{1} \left[ \frac{y_i}{x_i} > \beta \right] = \theta^n e^{-\theta (\sum_{i=1}^n y_i - \beta \sum_{i=1}^n x_i)} \cdot \mathbb{1} \left[ \min_i \frac{y_i}{x_i} > \beta \right] $$ from here I see $\overline{Y}$ sufficient, but with $\min \limits_i \frac{y_i}{x_i} \neq \min \limits_i y_i$ present I don't know what to conclude. I don't think you can say $Y_{(1)}$ is part of the joint sufficiency $\endgroup$
    – user365239
    Dec 30 '16 at 3:44
  • $\begingroup$ If you assume $X_i$ are known, then comparing $ \prod_{i=1}^n \mathbb{1} [ y_i > \beta x_i]$ to $\mathbb{1}[ y_{(1)} > \beta x^*]$, where $x^* \in \{x_1,\dots,x_n\}$ is known and corresponds to the index of $y_{(1)}$ may do the job. Is this how you get $Y_{(1)}$ sufficient? $\endgroup$
    – user365239
    Dec 30 '16 at 3:59
  • $\begingroup$ @user365239 I think you are right. I checked my editing history. I wrote the later one as min Y_i as well, but changed it soon after. I think the former one would be the same mistake. Thank you for pointing it out! $\endgroup$
    – Zander
    Dec 30 '16 at 4:08
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Indeed if both parameters $(\theta, \beta)$ are unknown, calculating the MSS for only one of them is strange. But consider the case when only one of them is unknown, in such case the MSS for the unknown parameter may be of a lower dimension than the joint. In your shifted exponential distribution, for $\beta$, $X_{(1)}$ will be the MSS where $\theta$ is known and $\bar{Y}$ vice versa. So you can say that some $T(X)$ is MSS for only one parameter, but then you have to state (or assume) that the other one is known.

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