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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.

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  • $\begingroup$ This is essentially asking the same question. In the previous question you show that $T^0(\mathbf y)=\sum_{i=1}^n y_i^2$ is minimal sufficient. This implies that (a) $T^1(\mathbf y)=(T^0(\mathbf y),\sum_{i=1}^n y_i)$ is sufficient [since it contains $T^0$] and (b) it is not minimal [since it is not a function of $T^0$]. $\endgroup$
    – Xi'an
    Apr 20, 2021 at 7:51
  • $\begingroup$ Could you erase and rewrite the previous comment as the sentence about $T^0$ not being a function of $T^0$ does not make sense. My point is that $\sum_i y_i$ cannot be expressed as a function of $\sum_i y_i^2$. If I know that $\sum_i y_i^2=3.1415$, I cannot find the numerical value of $\sum_i y_i$ (even when $n=1$). $\endgroup$
    – Xi'an
    Apr 20, 2021 at 10:19
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    $\begingroup$ @Xi'an and The Pointer. Having re-read my answer to the linked question, I have realised that I could be more explicit in showing why $U = (\sum Y_i, \sum Y^2_i)$ is sufficient but not minimal sufficient, I will edit accordingly. $\endgroup$
    – microhaus
    Apr 20, 2021 at 13:42

1 Answer 1

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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.


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.

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    $\begingroup$ Thank you for supplying a much more direct way of getting the solution than the one I originally supplied here. Reading your solution was instructive for me also. $\endgroup$
    – microhaus
    Apr 20, 2021 at 15:40
  • $\begingroup$ @microhaus: Thanks for the support, I have not first realised The Pointer had somehow posted the question three times over the past days. $\endgroup$
    – Xi'an
    Apr 20, 2021 at 16:32
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    $\begingroup$ I guess they are being thorough, and sometimes it helps to see multiple ways of getting the same answer. For example, the part of your solution which states "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 $T(\mathbf{y})$)." is a point of emphasis that is new to me. $\endgroup$
    – microhaus
    Apr 20, 2021 at 16:55
  • $\begingroup$ @Xi'an The questions are on the same subject, but they are asking different things (different aspects of the same subject). $\endgroup$ Apr 21, 2021 at 22:27
  • $\begingroup$ "To apply 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 $T(\mathbf y)$)." Good point to know. This would have saved me some time had I known it earlier. $\endgroup$ Apr 22, 2021 at 1:53

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