Solution. The joint probability function is \[f_\theta (x_1, \dots , x_n) = \prod ^n_{i = 1}\frac {e^{-\theta }\theta ^{x_i}}{x_i!} = \frac {e^{-n\theta }\theta ^{\sum x_i}}{\prod x_i!},\qquad x_i = 0, 1, 2, \dots ,\] and \(T\sim \text {POI}(n\theta )\). The event \(\{X = x\}\) is contained in \(\{T = t\}\) when \(\sum x_i = t\), and is disjoint from it otherwise. So for \(\sum x_i = t\), \[P(X = x\mid T = t) = \frac {P(X = x)}{P(T = t)} = \frac {e^{-n\theta }\theta ^{t}}{\prod x_i!}\cdot \frac {t!}{e^{-n\theta }(n\theta )^t} = \frac {t!}{x_1!\,x_2!\cdots x_n!}\left (\frac 1n\right )^t ,\] and the probability is \(0\) otherwise. This does not involve \(\theta \), so \(T\) is sufficient. In fact, given \(T = t\), the sample has the multinomial distribution \(\text {MULT}\left (t; \frac 1n, \dots , \frac 1n\right )\): the \(t\) events are spread over the \(n\) observations at random.

Solution. Here \(T\sim N(n\theta , n\sigma ^2)\). Because \(X\) determines \(T\), the conditional density of \(X\) given \(T = t\) on the set \(\{\sum x_i = t\}\) is the ratio of the joint density of \(X\) to the density of \(T\): \[\frac {(2\pi \sigma ^2)^{-n/2}\exp \left (-\frac {1}{2\sigma ^2}\sum (x_i - \theta )^2\right )} {(2\pi n\sigma ^2)^{-1/2}\exp \left (-\frac {1}{2n\sigma ^2}(t - n\theta )^2\right )} .\] Expanding, \(\sum (x_i - \theta )^2 = \sum x_i^2 - 2\theta t + n\theta ^2\) and \(\frac 1n(t - n\theta )^2 = \frac {t^2}{n} - 2\theta t + n\theta ^2\). The terms in \(\theta \) cancel, leaving \[\frac {\sqrt n}{(2\pi \sigma ^2)^{(n-1)/2}}\exp \left (-\frac {1}{2\sigma ^2}\left [\sum x_i^2 - \frac {t^2}{n}\right ]\right ) = \frac {\sqrt n}{(2\pi \sigma ^2)^{(n-1)/2}}\exp \left (-\frac {1}{2\sigma ^2}\sum (x_i - \bar x)^2\right ),\] which does not depend on \(\theta \). So \(T\) is sufficient.

Proof. We give the proof for discrete \(X\); the continuous case is the same in spirit but needs measure theory.

Only if

Suppose \(T\) is sufficient. For any \(x\), put \(t = T(x)\). The event \(\{X = x\}\) is contained in \(\{T = t\}\), so \[f_\theta (x) = P_\theta (X = x) = P_\theta (X = x,\ T = t) = \underbrace {P(X = x\mid T = t)}_{h(x)}\ \underbrace {P_\theta (T = t)}_{g(t;\theta )} .\] The first factor does not depend on \(\theta \) by sufficiency and is a function of \(x\); the second depends on \(x\) only through \(t = T(x)\).

If

Suppose \(f_\theta (x) = g(T(x);\theta )h(x)\). For a value \(t\) with \(P_\theta (T = t)>0\), \[P_\theta (T = t) = \sum _{y : T(y) = t} g(t;\theta )h(y) = g(t;\theta )\sum _{y : T(y) = t} h(y) .\] Hence for \(x\) with \(T(x) = t\), \[P_\theta (X = x\mid T = t) = \frac {g(t;\theta )h(x)}{g(t;\theta )\sum _{y : T(y) = t} h(y)} = \frac {h(x)}{\sum _{y : T(y) = t} h(y)},\] and for \(x\) with \(T(x)\ne t\) the conditional probability is \(0\). Neither depends on \(\theta \), so \(T\) is sufficient. □

Solution. With \(\theta = (\mu , \sigma ^2)\), \begin {align*} f_\theta (x) &= (2\pi \sigma ^2)^{-n/2}\exp \left (-\frac {1}{2\sigma ^2}\sum (x_i - \mu )^2\right )\\ &= \underbrace {(\sigma ^2)^{-n/2}\exp \left (-\frac {\sum x_i^2 - 2\mu \sum x_i + n\mu ^2}{2\sigma ^2}\right )}_{g(T(x);\,\theta )}\ \underbrace {(2\pi )^{-n/2}}_{h(x)}, \end {align*}

so \(T = \left (\sum X_i, \sum X_i^2\right )\) is sufficient by theorem 1.3.4. Alternatively, writing \(x_i - \mu = (x_i - \bar x) + (\bar x - \mu )\) and using \(\sum (x_i - \bar x) = 0\), \[\sum (x_i - \mu )^2 = \sum (x_i - \bar x)^2 + n(\bar x - \mu )^2 = (n - 1)s^2 + n(\bar x - \mu )^2,\] so \[f_\theta (x) = (\sigma ^2)^{-n/2}\exp \left (-\frac {(n - 1)s^2 + n(\bar x - \mu )^2}{2\sigma ^2}\right )\cdot (2\pi )^{-n/2},\] and \((\bar X, S^2)\) is sufficient too. The two statistics are functions of each other, as corollary 1.3.8 below explains.

Solution. \[f_\theta (x_1, \dots , x_n) = \theta ^n\left (\prod x_i\right )^{\theta - 1}\exp \left (-\sum x_i^\theta \right ).\] The term \(\sum x_i^\theta \) cannot be written in terms of a fixed finite set of summaries of the data for all \(\theta \): it changes with \(\theta \) in a way that depends on every observation. It is, however, symmetric in the \(x_i\), so it is a function of the order statistics. Taking \(h = 1\) and \(g\) the whole expression, \(T(X) = (X_{(1)}, \dots , X_{(n)})\) is sufficient. Apart from forgetting the order in which the observations arrived, no reduction of the data is possible here.

Solution. Write \(I(A)\) for the indicator of the event \(A\), equal to \(1\) if \(A\) holds and \(0\) otherwise. Each density is \(\frac 1\theta I(0<x_i<\theta )\), and all \(n\) observations lie in \((0, \theta )\) exactly when the smallest is positive and the largest is below \(\theta \). So \[f_\theta (x) = \frac {1}{\theta ^n}\prod _{i=1}^n I(0<x_i<\theta ) = \underbrace {\frac {1}{\theta ^n}I\left (x_{(n)}<\theta \right )}_{g(x_{(n)};\,\theta )}\ \underbrace {I\left (x_{(1)}>0\right )}_{h(x)},\] and \(T = X_{(n)}\) is sufficient. When the support depends on the parameter, the indicator functions must be kept in the factorisation; dropping them gives the wrong answer.

Proof. By theorem 1.3.4, \(f_\theta (x) = g(T(x);\theta )h(x)\). Since \(r\) is one-to-one, \(T(x) = r^{-1}(T^*(x))\), so \[f_\theta (x) = g\left (r^{-1}(T^*(x));\theta \right )h(x) = g^*\left (T^*(x);\theta \right )h(x)\] with \(g^*(s;\theta ) = g(r^{-1}(s);\theta )\). By the theorem again, \(T^*\) is sufficient. □

Solution. \[f_\theta (x) = \prod _{i=1}^n\frac {x_i^{\alpha - 1}e^{-x_i/\beta }}{\beta ^\alpha \Gamma (\alpha )} = \left [\beta ^\alpha \Gamma (\alpha )\right ]^{-n}\left (\prod x_i\right )^{\alpha - 1}\exp \left (-\frac {1}{\beta }\sum x_i\right ),\qquad x_i>0 .\] With \(h = 1\) this is a function of \(\theta \) and of \(\left (\prod X_i, \sum X_i\right )\) only, so that pair is sufficient. By corollary 1.3.8 so are \(\left (\sum \ln X_i, \sum X_i\right )\) and the pair formed by the geometric and arithmetic means.