How do we conclude that a statistic is sufficient but not minimal sufficient? #2

Question

This is related to a question I recently asked.

I want to show that the statistic $\left(\sum_{i = 1}^n Y_i, \sum_{i = 1}^n Y_i^2 \right)$ is sufficient for $\mu$ but not minimal sufficient where $(Y_1, \dots, Y_n)$ is a random sample from $N(\mu, \mu)$ for $\mu > 0$.

The textbook All of Statistics: A Concise Course in Statistical Inference by Larry Wasserman gives the following definition and theorem of minimal sufficiency:

9.32 Definition. Write $x^n \leftrightarrow y^n$ if $f(x^n; \theta) = cf(y^n; \theta)$ for some constant $c$ that might depend on $x^n$ and $y^n$ but not $\theta$. A statistic $T(x^n)$ is sufficient if $T(x^n) \leftrightarrow T(y^n)$ implies that $x^n \leftrightarrow y^n$.

9.35 Definition. A statistic $T$ is minimal sufficient if (i) it is sufficient; and (ii) it is a function of every other sufficient statistic.

9.36 Theorem. $T$ is minimal sufficient if the following is true: $$T(x^n) = T(y^n) \ \text{if and only if} \ x^n \leftrightarrow y^n.$$

Using theorem 9.36 as a guide, how do we show in practice that a statistic is not minimal sufficient?

For the statistic $T(\mathbf{Y}) = \left(\sum_{i = 1}^n Y_i, \sum_{i = 1}^n Y_i^2 \right)$, let's assume that $T(\mathbf{Y}) = T(\mathbf{X})$. I calculated the likelihood $$\begin{align} L(\mu; \mathbf{y}) &= (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}} \end{align},$$ and so we also have $$\begin{align} L(\mu; \mathbf{x}) &= (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n x_i^2 - 2\mu \sum_{i = 1}^n x_i + n\mu^2 \right) \right\}} \end{align}.$$ Taking the ratio of these, as in definition 9.32, we get $$\begin{align} \dfrac{L(\mu; \mathbf{y})}{L(\mu; \mathbf{x})} &= \dfrac{(2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}}}{(2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n x_i^2 - 2\mu \sum_{i = 1}^n x_i + n\mu^2 \right) \right\}}} \\ &= \dfrac{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}}}{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n x_i^2 - 2\mu \sum_{i = 1}^n x_i + n\mu^2 \right) \right\}}} \\ &= \dfrac{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i \right) \right\}}}{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n x_i^2 - 2\mu \sum_{i = 1}^n x_i \right) \right\}}} \end{align}$$

Is this the type of calculation that we need to do? I don't really understand what I'm doing here, partly because I don't really understand how $c$ in definition 9.32 is supposed to work in practice.

Answer 1

To apply¹ Theorem 9.36, the relevant implication for minimality is that $x^n \leftrightarrow y^n$ implies $T(x^n) = T(y^n)$ (as the reverse implication always holds for any sufficient statistic $T(\mathbf y)$).

Hence if one starts from the property $x^n \leftrightarrow y^n$ holding for a given pair $(\mathbf y,\mathbf x)$, it means that $$L(\mu;\mathbf y)=c(\mathbf y,\mathbf x)L(\mu;\mathbf x)\qquad \forall\mu\in\mathbb R^+$$ where $c(\mathbf y,\mathbf x)$ is a function of $(\mathbf y,\mathbf x)$ that does not depend on $\mu$. As correctly developped in the question \begin{align} \dfrac{L(\mu;\mathbf y)}{L(\mu;\mathbf x)} &= \dfrac{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i \right) \right\}}}{\exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n x_i^2 - 2\mu \sum_{i = 1}^n x_i \right) \right\}}}\\ \end{align} Regrouping terms together \begin{align} \dfrac{L(\mu;\mathbf y)}{L(\mu;\mathbf x)} &=\dfrac{\exp{\left\{ -\dfrac{T_1(\mathbf y)}{2 \mu} + T_2(\mathbf y) \right\}}}{\exp{\left\{ -\dfrac{T_1(\mathbf x)}{2 \mu} + T_2(\mathbf x) \right\}}} =\dfrac{\exp{\left\{ \dfrac{T_1(\mathbf x)-T_1(\mathbf y)}{2 \mu} \right\}}}{\exp{\left\{ T_2(\mathbf x) - T_2(\mathbf y) \right\}}} \end{align} where $T(\mathbf y)=(T_1(\mathbf y),T_2(\mathbf y))$. This ratio being constant in $\mu$ by assumption, this implies $$T_1(\mathbf y)=T_1(\mathbf x)$$ rather than the intended $$T(\mathbf y)=T(\mathbf x)$$ Therefore $T(\mathbf y)$ is not minimal by virtue of Theorem 9.6.

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¹A direct argument made in my comment is that, since $T_1(\mathbf y)$ is a minimal statistic, the pair $(T_1(\mathbf y),T_2(\mathbf y))$ cannot be a minimal statistic.