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I am currently trying to show that the statistic $\sum\limits_{y = 1}^n Y_i^2$ is minimal sufficient for $\mu$ 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 definitions and theorems:

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: $$T(x^n) = T(y^n) \ \text{if and only if} \ x^n \leftrightarrow y^n.$$

9.40 Theorem (Factorization Theorem). $T$ is sufficient if and only if there are functions $g(t, \theta)$ and $h(x)$ such that $f(x^n; \theta) = g(t(x^n), \theta)h(x^n)$.

I first calculate the likelihood

$$\begin{align} L(\mu; \mathbf{y}) &= \prod_{i = 1}^n L(\mu; y_i) \\ &= \prod_{i = 1}^n \dfrac{1}{\sqrt{2\pi \mu}} \exp{\left\{ -\dfrac{1}{2 \mu}(y_i - \mu)^2 \right\}} \\ &= \dfrac{1}{(2 \pi \mu)^{n/2}} \exp{\left\{ -\dfrac{1}{2 \mu} \sum_{i = 1}^n (y_i - \mu)^2 \right\}} \\ &= (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \sum_{i = 1}^n (y_i - \mu)^2 \right\}} \\ &= (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\sum_{i = 1}^n y_i \mu + \sum_{i = 1}^n \mu^2 \right) \right\}} \\ &= (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}} \end{align}$$

Using the factorization theorem, I get

$$\begin{align}&T(\mathbf{Y}) = \sum_{i = 1}^n Y_i^2, \\ &g(T(\mathbf{y}), \mu) = (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}},\\ &h(\mathbf{y}) = 1, \end{align}$$

so we get

$$\begin{align} L(\mu; \mathbf{y}) = g(T(\mathbf{y}), \mu) \times h(\mathbf{y}) = (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}} \end{align}$$

I then conclude that the statistic $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i^2$ is sufficient for $\mu$.

I now have the likelihood ratio

$$\begin{align} \dfrac{L(\mu; \mathbf{y_1})}{L(\mu; \mathbf{y_2})} &= \exp{\left\{ \dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_{2i}^2 - \sum_{i = 1}^n y_{1i}^2 + 2\mu \left( \sum_{i = 1}^n y_{1i} - \sum_{i = 1}^n y_{2i} \right) \right) \right\}} \\ &= \exp{\left\{ \dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_{2i}^2 - \sum_{i = 1}^n y_{1i}^2 \right) + \left( \sum_{i = 1}^n y_{1i} - \sum_{i = 1}^n y_{2i} \right) \right\}} \end{align}$$

This likelihood ratio does not depend on the parameter $\mu$ iff $T(\mathbf{y}_2) = \sum_{i = 1}^n y_{2i}^2 = \sum_{i = 1}^n y_{1i}^2 = T(\mathbf{y}_1)$: $$\begin{align} \exp{\left\{ \dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_{2i}^2 - \sum_{i = 1}^n y_{1i}^2 \right) + \left( \sum_{i = 1}^n y_{1i} - \sum_{i = 1}^n y_{2i} \right) \right\}} = \exp{\left\{ \left( \sum_{i = 1}^n y_{1i} - \sum_{i = 1}^n y_{2i} \right) \right\}} \end{align}$$

I then conclude that the statistic $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i^2$ is minimal sufficient.

However, I'm not sure whether I did this all correctly in accordance with the above definitions and theorems. The problem is that some of these are "iff", meaning that they require you to show the two-way implication (otherwise known as 'equivalence'), and I'm not sure that I did this correctly. The user microhaus posted this related answer, but I'm not sure if I've applied it correctly. First, I'm unsure whether my use of the factorization theorem was correct to conclude that the statistic $T(\mathbf{Y}) = \sum_{i = 1}^n Y_i^2$ is sufficient for $\mu$. And second, I'm unsure how one is supposed to show that a statistic is "a function of every other sufficient statistic".

So is my work here correct? Did I use the factorization theorem correctly to conclude that the statistic is sufficient? How does one show that a statistic is "a function of every other sufficient statistic"? If there are any problems here, how can the reasoning be improved?

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In AoS, there are definitions and there are theorems. Theorems are helpful in identifying properties. When achieving the equality $$L(\mu;\mathbf y) = (2\pi \mu)^{-n/2} \exp{\left\{ -\dfrac{1}{2 \mu} \left( \sum_{i = 1}^n y_i^2 - 2\mu \sum_{i = 1}^n y_i + n\mu^2 \right) \right\}}$$ the exponential can be broken into parts \begin{align} L(\mu;\mathbf y) =& (2\pi \mu)^{-n/2} \exp \left\{ -\dfrac{\sum_{i = 1}^n y_i^2}{2 \mu} \right\} \exp\left\{\sum_{i = 1}^n y_i\right\} \exp\left\{- n\mu/2 \right\}\\ =& \underbrace{(2\pi \mu)^{-n/2} \exp\left\{- n\mu/2 \right\}^{(a)}}_\text{function of $\mu$ only} \underbrace{\exp \left\{ -\dfrac{\sum_{i = 1}^n y_i^2}{2 \mu} \right\}^{(b)}}_\text{function of $(\mu,T(\mathbf y))$} \underbrace{\exp \left\{ \sum_{i = 1}^n y_i\right\} ^{(c)}}_\text{function of $y$ only} \end{align}

Theorem 9.40 thus shows that $T(\mathbf y)$ is sufficient since $L(\mu;\mathbf y)$ factorises into a function of $(\mu,T(\mathbf y))$ [first two terms, (a) and (b)] times a function of $y$ only [last term (c)].

Theorem 9.36 shows that $T(\mathbf y)$ is minimal* sufficient since, assuming that ["only if" part]$$T(\mathbf y)=T(\mathbf x)$$ this implies that only (c) differs between $\mathbf y$ and $\mathbf x$ [this is true for any sufficient statistic] and, conversely ["if" part], if the equality $$L(\mu;\mathbf y)=c(\mathbf y,\mathbf x)L(\mu;\mathbf x)\qquad \forall\mu\in\mathbb R^+$$holds, then $$\exp \left\{ -\dfrac{T(\mathbf y)}{2 \mu} \right\}\exp\left\{\sum_{i = 1}^n y_i\right\}=c(\mathbf y,\mathbf x)\exp \left\{ -\dfrac{T(\mathbf x)}{2 \mu} \right\}\exp\left\{\sum_{i = 1}^n x_i\right\} \qquad \forall\mu\in\mathbb R^+$$ implies that $$\exp \left\{ \dfrac{T(\mathbf x)}{2 \mu}-\dfrac{T(\mathbf y)}{2 \mu} \right\}= c(\mathbf y,\mathbf x)\exp \left\{ \sum_{i = 1}^n x_i-\sum_{i = 1}^n y_i\right\}$$ i.e., the lhs is a constant function of $\mu$. Therefore, $$T(\mathbf y)=T(\mathbf x)$$ and the necessary and sufficient condition is established.

By applying Theorem 9.36 to the question, one does not need to further establish that $T(\mathbf y)$ is a function of every sufficient statistic for this model, since Theorem 9.36 provides a sufficient condition for minimal sufficiency to hold.

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  • $\begingroup$ Thanks for the great answer! Can you please explain how $$\exp \left\{ -\dfrac{T(\mathbf y)}{2 \mu} \right\}\exp\left\{\sum_{i = 1}^n y_i\right\}=c(\mathbf y,\mathbf x)\exp \left\{ -\dfrac{T(\mathbf x)}{2 \mu} \right\}\exp\left\{\sum_{i = 1}^n x_i\right\}\qquad \forall\mu\in\mathbb R^+$$ implies that $$\exp \left\{ \dfrac{T(\mathbf x)}{2 \mu}-\dfrac{T(\mathbf y)}{2 \mu} \right\}$$ is a constant function of $\mu$? I understand that we can get $\exp \left\{ \dfrac{T(\mathbf x)}{2 \mu}-\dfrac{T(\mathbf y)}{2 \mu} \right\}$ through division of the exponential on the RHS. $\endgroup$ Commented Apr 20, 2021 at 8:56
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    $\begingroup$ You just need to move the$$\exp \left\{ -\dfrac{T(\mathbf x)}{2 \mu} \right\}$$term from the rhs to the lhs and the$$\exp\left\{\sum_{i = 1}^n y_i\right\}$$term from the lhs to the rhs. $\endgroup$
    – Xi'an
    Commented Apr 20, 2021 at 10:10
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    $\begingroup$ Wow, this is great. Thanks! $\endgroup$ Commented Apr 21, 2021 at 18:41

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