Inferential statistics

Inferential statistics or statistical induction comprises the use of statistics to make inferences concerning some unknown aspect (usually a parameter) of a population.

Two schools of inferential statistics are frequency probability using maximum likelihood estimation, and Bayesian inference. This is an example of the latter.

From a population containing N items of which I are special, a sample containing n items of which i are special can be chosen in

{I \choose i}

ways (see multiset and binomial coefficient).

Fixing (N,n,I), this expression is the unnormalized deduction distribution function of i.

Fixing (N,n,i) , this expression is the unnormalized induction distribution function of I.

The two most important parameters of a probability distribution are: the mean value and the standard deviation . The plus-minus sign, ± , is used to separate the mean from the deviation.

Deduction distribution formula

The mean value ± the standard deviation of the deduction distribution is used for estimating i knowing (N,n,I)

i \approx f(N,n,I)

f(N,n,I)=\frac{nI\pm\sqrt{\frac{nI(N-n)(N-I)}{N-1}}}{N}

where a(b±c)=ab±ac. Note that f defines two functions of three variables.

Example: The population contains two items one of which is special, and the sample contains one item. (N,n,I)=(2,1,1) gives

i\approx f(2,1,1)=\frac{1}{2}\pm\frac{1}{2}

confirming that the number of special items in the sample is either 0 or 1.

Induction distribution formula

The mean value ± the standard deviation of the induction distribution is used for estimating I knowing (N,n,i)

I \approx -1-f(-2-n,-2-N,-1-i)

where a+(b±c)=(a+b)±c.

Thus deduction is translated into induction by means of the involution

(N,n,I,i) \leftrightarrow (-2-n,-2-N,-1-i,-1-I).

Example: The population contains a single item and the sample is empty. (N,n,i)=(1,0,0) gives

I\approx -1-f(-2-0,-2-1,-1-0)=\frac{1}{2}\pm\frac{1}{2}

confirming that the number of special items in the population is either 0 or 1.

Note that the frequency probability solution to this problem is $I\approx \frac{Ni}{n}=\frac{0}{0}$ giving no meaning.

Binomial distribution formula

In the limiting case where N is a large number, the deduction distribution of i tends towards the binomial distribution with the probability

P=\frac{I}{N}

as a parameter,

i\approx nP\left (1\pm\sqrt{\frac{\frac{1}{P}-1}{n}}\right )

Example: The population is big, the probability $P=\frac{I}{N}=\frac{1}{2}$ , and the sample contains one item. n = 1 gives

i\approx \frac{1}{2}\pm\frac{1}{2}

confirming that the sample contains 0 or 1 special items, with equal probability.

Beta distribution formula

In the limiting case where N is a large number, the induction distribution of

P=\frac{I}{N}

tends towards the beta distribution

P\approx\frac{i+1\pm\sqrt{\frac{(i+1)(n-i+1)}{n+3}}}{n+2}.

The frequency probability solution to this problem is $P \approx \frac{i}{n}$ . The probability is estimated by the relative frequency.

Example: The population is big and the sample is empty. n = i = 0 gives

P \approx(50 \pm 29)\%

The frequency probability solution to this problem is $P \approx \frac{i}{n}=\frac{0}{0}$ , giving no meaning.

Poisson distribution formula

In the limiting case where

\frac{N}{n}

and

\ n

are large numbers, the deduction distribution of i tends towards the poisson distribution with the intensity

M=\frac{nI}{N}

as a parameter,

i \approx M \pm \sqrt{M}

Example: The population is big and the sample is big, and the intensity $M=\frac{nI}{N}=1$ gives

i\approx 1 \pm 1

Gamma distribution formula

In the limiting case where

\frac{N}{n}

and

\ n

are large numbers, the induction distribution of

M=\frac{nI}{N}

tends towards the gamma distribution with i as a parameter:

M \approx i+1 \pm \sqrt{i+1}.

Example: The population is big and the sample is big but contains no special items. i = 0 gives

M\approx 1 \pm 1

The frequency probability solution to this problem is $M\approx 0$ which is misleading. Even if you have not been wounded you may still be vulnerable.