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Suppose $X_1,\cdots,X_n$ are $i.i.d$ from a distribution with p.d.f $$\delta_{(\theta,c)}(x)=\frac{1}{c}\mathbb{1}_{(x\in[\theta,\theta+c])},$$ where $\theta\in\mathbb{R}$ and $c\in\mathbb{R}^+$ unknown.

Find a minimal sufficient statistic for $(\theta,c)$.

From the range of $x_i$, i.e. $x_i\in[\theta,\theta+c]$, we can determine that $\theta\leq x_i$ and $\theta+c\geq x_i$, $\forall i\in\{1,\cdots,n\}$. This implies $$\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)},$$ where $x_{(1)}=\underset{i\in\{1,\cdots,n\}}{\min}x_i$ and $x_{(n)}=\underset{i\in\{1,\cdots,n\}}{\max}x_i$.

The plots shows the area of $\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)}.$ It look like that $(x_{(1)},x_{(n)})$ is a minimal sufficient statistic for $(\theta,c)$ because $(x_{(1)},x_{(n)})$ uniquely determines the shape of the log-likelihood function. Is this correct?

illustration

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2 Answers 2

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(Nobody answers so I post this answer which I am not sure is correct)

We show $(Y_{(1)},Y_{(n)})$ is a sufficient complete statistic, which implies $(Y_{(1)},Y_{(n)})$ is minimal sufficient. We first give the p.d.f of $(Y_{(1)},Y_{(n)})$ $$f(y_1, y_n,\theta,c)=\frac{n(n-1)}{c^n}(y_n-y_1)^{n-2},\quad\forall \theta\leq y_1\leq y_n\leq \theta+c\text{ and } (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$

Then, for any function $g(x_1,x_n)$ so that $\mathbb{E}_{\theta,c}\left[g(y_1,y_n)\right]=0,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$, we have $$0=\frac{n(n-1)}{c^n}\int_{\theta}^{\theta+c}\int_{\theta}^{y_n}g(y_1,y_n)(y_n-y_1)^{n-2} dy_1dy_n,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$ The integral area of $(y_1, y_n)$ is a triangle with vertices $(\theta,\theta)$, $(\theta,\theta+c)$ and $(\theta+c,\theta+c)$. With varying $\theta\in\mathbb{R}$ $c\in\mathbb{R}^+$, these triangles generate the Beral $\sigma$-algebra of $\mathcal{B}=\{(x,z)\in\mathbb{R}^2:x\leq z\}$. Thus $$0=\frac{n(n-1)}{c^n}\int_{A}g(y_1,y_n)(y_n-y_1)^{n-2} d(y_1,y_n),\text{ for any Borel set }A\subset\mathcal{B}.$$ This means $$g(y_1,y_n)(y_n-y_1)^{n-2}\equiv0,a.e.\iff g\equiv 0,a.e.$$ Thus we conclude that $(Y_{(1)},Y_{(n)})$ is a complete statistic for $(\theta,c).$

Since $(Y_{(1)},Y_{(n)})$ is sufficient by factorization theorem, we conclude that $(Y_{(1)},Y_{(n)})$ is minimal sufficient.

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  • $\begingroup$ Could you clarify why $(Y_{(1)}, Y_{(n)})$ being complete sufficient implies that it is minimally sufficient? This might be just my lack of knowledge, but I don't see how one follows from the other. $\endgroup$ Jan 7, 2021 at 16:35
  • $\begingroup$ Certainly. It is Bahadur's theorem. $\endgroup$
    – Tan
    Jan 7, 2021 at 17:29
  • $\begingroup$ I believe this is correct -- the only thing I'm not 100% sure on is the claim that the triangles generate the Borel algebra. Could you expand on that? Also, maybe add something about Bahadur's theorem in the answer (I could add it if you don't mind). $\endgroup$ Jan 23, 2021 at 0:02
  • $\begingroup$ I do not have rigorous proof but see here. It's great if you could add Bahadur's theorem. I simply remember it $T$ sufficient + complete $\implies$ $T$ minimal sufficient. $\endgroup$
    – Tan
    Jan 23, 2021 at 5:58
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This is an attempt at an alternative proof that $(x_{(1)}, x_{(n)})$ is a minimal sufficient statistic for $(\theta, c)$ using a direct approach.

Consider the likelihood ratio $$ R(x,y) = \frac{p(x | \theta, c)}{p(y | \theta, c)} $$ where $x = (x_{1}, x_{2}, \dots x_{n}), y = (y_{1}, y_{2}, \dots y_{n})$. The ''factorization'' theorem of minimal sufficient statistics states that if a statistic $T(x)$ has the following property P then it is a minimally sufficient statistic for $\theta$:

P: $R(x, y)$ does not depend on $\theta$ if and only if $T(x) = T(y)$.

I claim that $T(x) = (x_{(1)}, x_{(n)})$ has this property and therefore is minimally sufficient.

To see this, first consider what $p(x | \theta, c)$ is. I claim that $$p(x | \theta, c) = c^{-n} I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c].$$ The indicator $I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c]$ just determines if $x$ is a valid value in the range; certainly $p(x | \theta, c) = 0$ if $x$ is not within $[\theta, \theta+c]$. Otherwise, since the distribution is uniform we have that $p(x | \theta, c)$ should just be $c^{-n}$ because the density function of each $x_{i}$ is just $c^{-1}$.

Therefore, we have that $$ R(x, y) = \frac{p(x | \theta, c)}{p(y | \theta, c)} = \frac{I[x_{(1)} \geq \theta, x_{(n)} \leq \theta+c]}{I[y_{(1)} \geq \theta, y_{(n)} \leq \theta+c]} $$ since the $c^{-n}$ values cancel out.

At this point, I should admit that the factorization theorem is not technically be applicable here, but I'd guess that it is possible to extend that theorem to this situation. Intuitively, we can see that the ratio of $p(x | \theta, c)$ to $p(y | \theta, c)$ is independent of $(\theta, c)$ if and only if $T(x) = T(y)$, despite the technicality that $R(x, y) = \frac{0}{0}$ when both $x$ and $y$ are not in the interval. That is, if we pretend that $\frac{0}{0} = 1$, then $R(x, y) = 1$ whenever $T(x) = T(y)$, whereas it might either be 0 or $+\infty$ when $T(x) \not = T(y)$.

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  • $\begingroup$ Reading your answer. The p.d.f of $(X_{1},X_{(n)})$ should be $p(x_{(1)},x_{(n)})=n(n-1)\frac{(x_{(n)}-x_{(1)})^{n-2}}{c^n}$, I think. $\endgroup$
    – Tan
    Jan 9, 2021 at 20:56

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