Suppose $X_1, \ldots, X_n$ is a random sample with pdf $$f_{X_i}(x_i \mid \alpha, \beta) = \beta^{-1} \exp \left(\frac{-(x_i-\alpha)}{\beta}\right) I(x_i \geq \alpha)$$ for all $i = 1, 2, \ldots, n; \ n \geq 2$, with $\beta > 0$. I want to find a minimal sufficient statistic for $(\alpha, \beta)$. I am having trouble working out this problem and can't find a lot of information about this particular distribution so I thought I would ask here.

I do not think that this distribution belongs to an exponential family, but I do think it belongs to a location-scale family. So my approach was to get the PDF into a form where the Neyman-Pearson Factorization Theorem can be applied. We can write

$$\begin{aligned}[t] f_\mathbf{X}(\mathbf{x} \mid \alpha, \beta) &= \prod_{i=1}^n f_{X_i}(x_i \mid \alpha, \beta) \\ &= \beta^{-n}\exp\left( -\frac{1}{\beta} \left(\sum_{i=1}^n x_i - \alpha n\right) \right) I(x_1 \geq \alpha, \ldots, x_n \geq \alpha). \end{aligned}$$

Please let me know if the error is due to my algebra (I omitted the simplification steps here since they are a bit long).

Now I am unsure how to proceed to form $T(\mathbf{x})$, the sufficient statistic. I thought that perhaps I could say

$$I(x_1 \geq \alpha, \ldots, x_n \geq \alpha) = I(x_{(1)} \geq \alpha)$$

and then take $T(\mathbf{x}) = \left( \sum_{i=1}^n x_i, \ x_{(1)} \right) $ and if my step with the indicator function is allowed, then $T$ is sufficient by the Neyman-Pearson Factorization Theorem. But, I am not sure if this is minimal sufficient. When I take the ratio of the pdfs and write

$$\frac{f_{\mathbf{X}}(\mathbf{x} \mid \alpha, \beta)}{f_{\mathbf{Y}}(\mathbf{y} \mid \alpha, \beta)} = \frac{I(x_{(1)} \geq \alpha)}{I(y_{(1)} \geq \alpha)} \exp \left(-\frac{1}{\beta} \left(\sum_{i=1}^n x_i - \sum_{i=1}^n y_i \right) \right)$$

I know that if a sufficient statistic is minimal sufficient if given two samples, $\mathbf{x}, \ \mathbf{y}$, the ratio $f(\mathbf{x}\mid \theta) / f(\mathbf{y} \mid \theta)$ is "constant as a function of theta" if and only if $T(\mathbf{x}) = T(\mathbf{y})$.

How can I determine if this is true for my ratio? Clearly if the sums of the two samples are equal, the ratio is constant as a function of beta, so the sum is minimal sufficient for beta.

But I am unsure whether $x_{(1)}$ is sufficient for alpha based on the ratio argument. How can I use the ratio to determine this? If $x_{(1)}$ is not minimal sufficient for alpha, how can I modify my approach?

  • $\begingroup$ @Xi'an thank you, I have fixed the typo. I thought that was the case but then I am not sure how to find a minimal sufficient statistic. $\endgroup$
    – wzbillings
    Jan 26, 2021 at 20:29
  • $\begingroup$ Changing the pair in any way modifies the likelihood function/ratio by more than a multiplicative constant, hence the pair is minimal. $\endgroup$
    – Xi'an
    Jan 26, 2021 at 21:38
  • $\begingroup$ @Xi'an Unfortunately I am not allowed to use a likelihood argument as my course has not covered likelihood yet. I have updated my question to better reflect what I am asking. $\endgroup$
    – wzbillings
    Jan 26, 2021 at 21:58
  • $\begingroup$ Then apply the ratio argument you mention. $\endgroup$
    – Xi'an
    Jan 27, 2021 at 7:20

1 Answer 1


First this family of distributions is an Exponential $\mathcal E(\beta^{-1})$ translated by $\alpha$. Since the support of the density depends on $\alpha$, the family is not an exponential family.

Second, the ratio $$\frac{f_{\mathbf{X}}(\mathbf{x} \mid \alpha, \beta)}{f_{\mathbf{Y}}(\mathbf{y} \mid \alpha, \beta)} = \frac{\mathbb I(x_{(1)} \geq \alpha)}{\mathbb I(y_{(1)} \geq \alpha)} \exp \left(-\frac{1}{\beta} \left(\sum_{i=1}^n x_i - \sum_{i=1}^n y_i \right) \right)$$ factorises in a function of $\alpha$ only $$\frac{\mathbb I(x_{(1)} \geq \alpha)}{\mathbb I(y_{(1)} \geq \alpha)}\tag{1}$$ and a function of $\beta$ only $$\exp \left(-\frac{1}{\beta} \left(\sum_{i=1}^n x_i - \sum_{i=1}^n y_i \right) \right)\tag{2}$$

  1. If $x_{(1)}\ne y_{(1)}$, (1) is not constant in $\alpha$ but takes the values $0$, $1$ and $\infty$.
  2. If $\bar x_n\ne\bar y_n$, (2) is not constant in $\beta$.

Hence $(x_{(1)},\bar x)$ is a minimal sufficient statistic for $(\alpha,\beta)$.

  • 1
    $\begingroup$ Thank you for the explanation, I think I was having trouble understanding what the theorem for the constant ratio actually means. But thanks to this I think I understand now. $\endgroup$
    – wzbillings
    Jan 28, 2021 at 20:40

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