Sufficiency (statistics)

In statistics, sufficiency is the property possessed by a statistic, with respect to a parameter, "when no other statistic which can be calculated from the same sample provides any additional information as to the value of the parameter"

This concept was due to Sir Ronald Fisher, and is equivalent to the most general statement of the above that, conditional on the value of a sufficient statistic, the distributions of samples drawn are independent of the underlying parameter(s) the statistic is sufficient for. Both the statistic and the underlying parameter can be vectors.

The concept has fallen out of favor in descriptive statistics because of the strong dependence on an assumption of the distributional form, but remains very important in theoretical work.

Mathematical definition
The concept is most general when defined as follows: a statistic T(X) is sufficient for underlying parameter &theta; precisely if the conditional probability distribution of the data X, given the statistic T(X), is independent of the parameter &theta;, i.e.


 * $$\Pr(X=x|T(X)=t,\theta) = \Pr(X=x|T(X)=t), \,$$

or in shorthand
 * $$\Pr(x|t,\theta) = \Pr(x|t).\,$$

Example
As an example, the sample mean is sufficient for the mean (&mu;) of a normal distribution with known variance. Once the sample mean is known, no further information about &mu; can be obtained from the sample itself.

Fisher-Neyman factorization theorem
Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. If the probability density function is &fnof;&theta;(x), then T is sufficient for &theta; if and only if functions g and h can be found such that


 * $$ f_\theta(x)=h(x) \, g_\theta(T(x)), \,\!$$

i.e. the density &fnof; can be factored into a product such that one factor, h, does not depend on &theta; and the other factor, which does depend on &theta;, depends on x only through T(x).

Interpretation
An implication of the theorem is that when using likelihood-based inference, two sets of data yielding the same value for the sufficient statistic T(X) will always yield the same inferences about &theta;. By the factorization criterion, the likelihood's dependence on &theta; is only in conjunction with T(X). As this is the same in both cases, the dependence on &theta; will be the same as well, leading to identical inferences.

Proof for the continuous case
Due to Hogg and Craig (ISBN 978-0023557224). Let X1, X2, ..., Xn, denote a random sample from a distribution having the pdf f(x,&theta;) for  &gamma; < &theta; < &delta;. Let Y = u(X1, X2, ..., Xn) be a statistic whose pdf is g(y;&theta;). Then Y = u(X1, X2, ..., Xn) is a sufficient statistic for &theta; if and only if, for some function H,


 * $$ \prod_{i=1}^{n} f(x_i; \theta) = g \left[u(x_1, x_2, \dots, x_n); \theta \right] H(x_1, x_2, \dots, x_n). \,\!$$

First, suppose that
 * $$ \prod_{i=1}^{n} f(x_i; \theta) = g \left[u(x_1, x_2, \dots, x_n); \theta \right] H(x_1, x_2, \dots, x_n). \,\!$$

We shall make the transformation yi = ui(x1, x2, ..., xn), for i = 1, ..., n, having inverse functions xi = wi(y1, y2, ..., yn), for i = 1, ..., n, and Jacobian J. Thus,



\prod_{i=1}^{n} f \left[ w_i(y_1, y_2, \dots, y_n); \theta \right] = |J| g(y; \theta) H \left[ w_1(y_1, y_2, \dots, y_n), \dots, w_n(y_1, y_2, \dots, y_n) \right]. $$

The left-hand member is the joint pdf g(y1, y2, ..., yn; &theta;) of Y1 = u1(X1, ..., Xn), ..., Yn = un(X1, ..., Xn). In the right-hand member, $$g(y_1,\dots,y_n;\theta)$$ is the pdf of $$Y_1$$, so that $$H[ w_1, \dots, w_n] |J|$$ is the quotient of $$g(y_1,\dots,y_n;\theta)$$ and $$g_1(y_1;\theta)$$; that is, it is the conditional pdf $$h(y_2, \dots, y_n | y_1; \theta)$$ of $$Y_2,\dots,Y_n$$ given $$Y_1=y_1$$.

But $$H(x_1,x_2,\dots,x_n)$$, and thus $$H\left[w_1(y_1,\dots,y_n), \dots, w_n(y_1, \dots, y_n))\right]$$, was given not to depend upon $$\theta$$. Since $$\theta$$ was not introduced in the transformation and accordingly not in the Jacobian $$J$$, it follows that $$h(y_2, \dots, y_n | y_1; \theta)$$ does not depend upon $$\theta$$ and that $$Y_1$$ is a sufficient statistics for $$\theta$$.

The converse is proven by taking:


 * $$g(y_1,\dots,y_n;\theta)=g_1(y_1; \theta) h(y_2, \dots, y_n | y_1),\,$$

where $$h(y_2, \dots, y_n | y_1)$$ does not depend upon $$\theta$$ because $$Y_2 ... Y_n$$ depend only upon $$X_1 ... X_n$$ which are independent on $$\Theta$$ when conditioned by $$Y_1$$, a sufficient statistics by hypothesis. Now divide both members by the absolute value of the non-vanishing Jacobian $$J$$, and replace $$y_1, \dots, y_n$$ by the functions $$u_1(x_1, \dots, x_n), \dots, u_n(x_1,\dots, x_n)$$ in $$x_1,\dots, x_n$$. This yields


 * $$\frac{g\left[ u_1(x_1, \dots, x_n), \dots, u_n(x_1, \dots, x_n); \theta \right]}{|J*|}=g_1\left[u_1(x_1,\dots,x_n); \theta\right] \frac{h(u_2, \dots, u_n | u_1)}{|J*|}$$

where $$J*$$ is the Jacobian with $$y_1,\dots,y_n$$ replaced by their value in terms $$x_1, \dots, x_n$$. The left-hand member is necessarily the joint pdf $$f(x_1;\theta)\cdots f(x_n;\theta)$$ of $$X_1,\dots,X_n$$. Since $$h(y_2,\dots,y_n|y_1)$$, and thus $$h(u_2,\dots,u_n|u_1)$$, does not depend upon $$\theta$$, then


 * $$H(x_1,\dots,x_2)=\frac{h(u_2,\dots,u_n|u_1)}{|J*|}$$

is a function that does not depend upon $$\theta$$.

Proof for the discrete case
We use the shorthand notation to denote the joint probability of $$(X, T(X))$$ by $$f_\theta(x,t)$$. Since $$T$$ is a function of $$X$$, we have $$f_\theta(x,t) = f_\theta(x)$$ and thus:


 * $$f_\theta(x) = f_\theta(x,t) = f_{\theta | t}(x) f_\theta(t) $$

with the last equality being true by the definition of conditional probability distributions. Thus $$f_\theta(x)=a(x) b_\theta(t)$$ with $$a(x) = f_{\theta | t}(x)$$ and $$b(x) = f_\theta(t)$$.

Reciprocally, if $$f_\theta(x)=a(x) b_\theta(t)$$, we have



\begin{align} f_\theta(t) & = \sum _{x : T(x) = t} f_\theta(x, t) \\ & = \sum _{x : T(x) = t} f_\theta(x) \\ & = \sum _{x : T(x) = t} a(x) b_\theta(t) \\ & = \left( \sum _{x : T(x) = t} a(x) \right) b_\theta(t) \end{align}$$

With the first equality by the definition of pdf for multiple variables, the second by the remark above, the third by hypothesis, and the fourth because the summation is not over $$t$$.

Thus, the conditional probability distribution is:

\begin{align} f_{\theta|t}(x) & = \frac{f_\theta(x, t)}{f_\theta(t)} \\ & = \frac{f_\theta(x)}{f_\theta(t)} \\ & = \frac{a(x) b_\theta(t)}{\left( \sum _{x : T(x) = t} a(x) \right) b_\theta(t)} \\ & = \frac{a(x)}{\sum _{x : T(x) = t} a(x)} \end{align}$$

With the first equality by definition of conditional probability density, the second by the remark above, the third by the equality proven above, and the fourth by simplification. This expression does not depend on $$\theta$$ and thus $$T$$ is a sufficient statistic.

Minimal sufficiency
A sufficient statistic is minimal sufficient if it can be represented as a function of any other sufficient statistic. In other words, S(X) is minimal sufficient if and only if
 * 1) S(X) is sufficient, and
 * 2) if T(X) is sufficient, then there exists a function f such that S(X) = f(T(X)).

Intuitively, a minimal sufficient statistic most efficiently captures all possible information about the parameter &theta;.

A useful characterization of minimal sufficiency is that when the density f&theta; exists, S(X) is minimal sufficient if and only if
 * $$\frac{f_\theta(x)}{f_\theta(y)}$$ is independent of &theta; :$$\Longleftrightarrow$$ S(x) = S(y)

This follows as a direct consequence from the Fisher's factorization theorem stated above.

A sufficient and complete statistic is necessarily minimal sufficient. A minimal sufficient statistic always exists; a complete statistic need not exist.

Bernoulli distribution
If X1, ...., Xn are independent Bernoulli-distributed random variables with expected value p, then the sum T(X) = X1 + ... + Xn is a sufficient statistic for p (here 'success' corresponds to $$X_i=1$$ and 'failure' to $$X_i=0$$; so T is the total number of successes)

This is seen by considering the joint probability distribution:


 * $$ \Pr(X=x)=P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n).$$

Because the observations are independent, this can be written as



p^{x_1}(1-p)^{1-x_1} p^{x_2}(1-p)^{1-x_2}\cdots p^{x_n}(1-p)^{1-x_n} \,\!$$

and, collecting powers of p and 1 &minus; p, gives



p^{\sum x_i}(1-p)^{n-\sum x_i}=p^{T(x)}(1-p)^{n-T(x)} \,\! $$

which satisfies the factorization criterion, with h(x)=1 being just a constant.

Note the crucial feature: the unknown parameter p interacts with the data x only via the statistic T(x) = &Sigma; xi.

Uniform distribution
If X1, ...., Xn are independent and uniformly distributed on the interval [0,&theta;], then T(X) = max(X1, ...., Xn ) is sufficient for &theta;.

To see this, consider the joint probability distribution:



\Pr(X=x)=P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n). $$

Because the observations are independent, this can be written as



\frac{\operatorname{H}(\theta-x_1)}{\theta}\cdot \frac{\operatorname{H}(\theta-x_2)}{\theta}\cdot\,\cdots\,\cdot \frac{\operatorname{H}(\theta-x_n)}{\theta} \,\! $$

where H(x) is the Heaviside step function. This may be written as



\frac{\operatorname{H}\left(\theta-\max_i \{\,x_i\,\}\right)}{\theta^n}\,\! $$

which can be viewed as a function of only &theta; and maxi(Xi) = T(X). This shows that the factorization criterion is satisfied, again where h(x)=1 is constant. Note that the parameter &theta; interacts with the data only through the data's maximum.

Poisson distribution
If X1, ...., Xn are independent and have a Poisson distribution with parameter &lambda;, then the sum T(X) = X1 + ... + Xn is a sufficient statistic for &lambda;.

To see this, consider the joint probability distribution:



\Pr(X=x)=P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n). $$

Because the observations are independent, this can be written as



{e^{-\lambda} \lambda^{x_1} \over x_1 !} \cdot {e^{-\lambda} \lambda^{x_2} \over x_2 !} \cdot\,\cdots\,\cdot {e^{-\lambda} \lambda^{x_n} \over x_n !} \,\! $$

which may be written as



e^{-n\lambda} \lambda^{(x_1+x_2+\cdots+x_n)} \cdot {1 \over x_1 ! x_2 !\cdots x_n ! } \,\! $$

which shows that the factorization criterion is satisfied, where h(x) is the reciprocal of the product of the factorials. Note the parameter &lambda; interacts with the data only through its sum T(X).

Rao-Blackwell theorem
Sufficiency finds a useful application in the Rao-Blackwell theorem. It states that if g(X) is any kind of estimator of &theta;, then typically the conditional expectation of g(X) given T(X) is a better estimator of &theta;, and is never worse. Sometimes one can very easily construct a very crude estimator g(X), and then evaluate that conditional expected value to get an estimator that is in various senses optimal.