# Minimal sufficient statistics for 2-parameter exponential distribution

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?

• @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. Jan 26, 2021 at 20:29
• Changing the pair in any way modifies the likelihood function/ratio by more than a multiplicative constant, hence the pair is minimal. Jan 26, 2021 at 21:38
• @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. Jan 26, 2021 at 21:58
• Then apply the ratio argument you mention. Jan 27, 2021 at 7:20

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 $$(\alpha,\beta)$$.