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Let $X_1,\ldots,X_n$ have the pdf given by $f(x;\theta)=(1/\sigma)e^{-(x-\mu)/\sigma}$. For the statistic $T$ given by $T=\sum_i (X_i - X_{n:1})$, I need to show that $T$ is sufficient for $\sigma$ when $\mu$ is known.

Previously, I have shown that $T'=\sum_i (X_i - \mu)$ is sufficient for $\sigma$ when $\mu$ is known. Is it enough for me to simply say that $T$ is sufficient because it is a one-to-one function of $T'$?

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Preliminary note: You have not stated the support of your density, so I am going to assume it is $x \geqslant \mu$, since this gives a valid density function. Under this interpretation, this means that you have a shifted exponential distribution with shift parameter $\mu$ and scale parameter $\sigma$.


Firstly, the function $T$ is not a one-to-one function of $T'$, so no, you can't just assert that. Secondly, the result you are trying to prove is in fact incorrect; the statistic $T$ is not sufficient for $\sigma$ in this case. To prove sufficiency you would generally apply the Fisher-Neyman factorisation theorem, which says that $T$ is sufficient if and only if you can write the likelihood function as:

$$L_\mathbf{x}(\theta) \propto g(\theta, T(\mathbf{x})),$$

where $\theta$ is the unknown parameter. (In other words, the likelihood function depends on the data vector $\mathbf{x}$ only through the statistic $T(\mathbf{x})$.) Denoting $m_n \equiv \min (x_1,...,x_n)$ we have:

$$\begin{equation} \begin{aligned} L_\mathbf{x}(\mu,\sigma) &= \prod_{i=1}^n f(x_i|\mu,\sigma) \\[6pt] &= \prod_{i=1}^n \frac{1}{\sigma} \cdot \exp \Big(- \frac{x_i-\mu}{\sigma} \Big) \cdot \mathbb{I}(x_i \geqslant \mu) \\[6pt] &= \frac{1}{\sigma^n} \cdot \exp \Big(- \frac{n(\bar{x} - \mu)}{\sigma} \Big) \cdot \mathbb{I}(m_n \geqslant \mu) \\[6pt] &= \frac{1}{\sigma^n} \cdot \exp \Big(- \frac{n(\bar{x} - m_n)}{\sigma} - \frac{n(m_n - \mu)}{\sigma} \Big) \cdot \mathbb{I}(m_n - \mu \geqslant 0) \\[6pt] &= \frac{1}{\sigma^n} \cdot \exp \Big(- \frac{n(\bar{x} - m_n)}{\sigma} \Big) \cdot \exp \Big(- \frac{n(m_n - \mu)}{\sigma} \Big) \cdot \mathbb{I}(m_n - \mu \geqslant 0). \\[6pt] \end{aligned} \end{equation}$$

From this decomposition, we can see that the statistic $(\bar{x} - m_n, m_n)$ is minimal sufficient for both parameters, and also minimal sufficient for $\sigma$ when $\mu$ is known. (We can also see that the statistic $m_n$ is minimal sufficient for $\mu$ when $\sigma$ is known.)

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  • $\begingroup$ The question doesn't ask whether $T$ is minimal sufficient. If you can express your statistic in terms of $T$--and it looks like you can when $\mu$ is known--then a fortiori $T$ is sufficient. $\endgroup$ – whuber Jun 14 '18 at 21:13
  • $\begingroup$ In this case $T = \bar{x} -m_n$, and from the above decomposition, this doesn't appear to me to be sufficient even when $\mu$ is known. If you disagree, can you show the proposed expression? $\endgroup$ – Ben Jun 14 '18 at 23:46

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