Probability and Relative Frequency

Abstract

The concept of probability seems to have been inexplicable since its invention in the seventeenth century. In its use in science, probability is closely related with relative frequency. So the task seems to be interpreting that relation. In this paper, we start with predicted relative frequency and show that its structure is the same as that of probability. I propose to call that the ‘prediction interpretation’ of probability. The consequences of that definition are discussed. The “ladder”-structure of the probability calculus is analyzed. The expectation of the relative frequency is shown to be equal to the predicted relative frequency. Probability is shown to be the most general empirically testable prediction.

Appendix: Derivation of the Predicted Variance

The predicted variance $\sigma ^{2}$ of the relative frequency is

$$\begin{aligned} \sigma ^2&:=\,E\left( {\left( \frac{k}{N}-p\right) ^2}\right) =E\left( {\left( {\frac{k}{N}}\right) ^2-2\frac{k}{N}\cdot p+p^2}\right) \\&=\,E\left( \left( {\frac{k}{N}}\right) ^2\right) -2\cdot E\left( {\frac{k}{N}}\right) \cdot p+p^2=E\left( \left( {\frac{k}{N}}\right) ^2\right) -p^2, \end{aligned}$$

since (cf. Eq. 7) $E( {\frac{k}{N}})=p.$ Using Eq. (6) we get,

$$\begin{aligned}&E\left( \left( {\frac{k}{N}}\right) ^2\right) =\sum \limits _{k=0}^N {\left( {\frac{k}{N}}\right) ^2\cdot p(A_k^N )}\\&\quad =\,\sum \limits _{k=0}^N {\left( {\frac{k}{N}}\right) ^2\cdot ( {_k^N })\cdot p^k\cdot (1-p)^{N-k}}\\&\quad =\,\sum \limits _{k=1}^N {\frac{k}{N}\cdot \frac{k}{N}\cdot \frac{N!}{k!(N-k)!}\cdot p^k\cdot (1-p)^{N-k}}\\&\quad =\,\frac{1}{N}\sum \limits _{k=1}^N {k\cdot \frac{(N-1)!}{(k-1)!(N-k)!}\cdot p^k\cdot (1-p)^{N-k}} . \end{aligned}$$

Replacing the factor k by the sum of k–1 and 1, we get the sum of two terms, $T_{1 }$and $T_{2}$:

$$\begin{aligned} E\left( \left( {\frac{k}{N}}\right) ^2\right)&=\frac{1}{N}\sum \limits _{k=1}^N {(k-1)\cdot \frac{(N-1)!}{(k-1)!(N-k)!}\cdot p^k\cdot (1-p)^{N-k}}\\&\quad +\,\frac{1}{N}\sum \limits _{k=1}^N {\frac{(N-1)!}{(k-1)!(N-k)!}\cdot p^k\cdot (1-p)^{N-k}} \\&=T_1 +T_2 . \end{aligned}$$

In $T_1$ the summand with k = 1 is zero. Thus:

$$\begin{aligned} T_1 =\frac{p^2\cdot (N-1)}{N}\cdot \sum \limits _{k=2}^N {\frac{(N-2)!}{(k-2)!(N-k)!}\cdot p^{k-2}\cdot (1-p)^{N-k}} . \end{aligned}$$

Substituting in the sum, l for k–2 and M for N–2 we get a binomial form:

$$\begin{aligned} T_1&=\,\frac{p^2\cdot (N-1)}{N}\cdot \sum \limits _{l=0}^M {\frac{(M)!}{l!(M-l)!}\cdot p^l\cdot (1-p)^{M-l}}\\&=\,\frac{p^2\cdot (N-1)}{N}\cdot (p+(1-p))^M=\frac{p^2\cdot (N-1)}{N}. \end{aligned}$$

Similarly we transform $T_{2}$, substituting l for k–1 and M for N–1:

$$\begin{aligned} T_2&=\,\frac{p}{N}\cdot \sum \limits _{k=1}^N {\frac{(N-1)!}{(k-1)!(N-k)!}\cdot p^{k-1}\cdot (1-p)^{N-k}}\\&=\,\frac{p}{N}\cdot \sum \limits _{l=0}^M {\frac{(M)!}{l!(M-l)!}\cdot p^l\cdot (1-p)^{M-l}=} \frac{p}{N}. \end{aligned}$$

Thus

$$\begin{aligned} E\left( \left( {\frac{k}{N}}\right) ^2\right) =\frac{p^2\cdot (N-1)}{N}+\frac{p}{N}=p^2-\frac{p^2}{N}+\frac{p}{N}=p^2+\frac{p\cdot (1-p)}{N}. \end{aligned}$$

So we get the variance

$$\begin{aligned} \sigma ^2=E\left( \left( {\frac{k}{N}}\right) ^2\right) -p^2=\frac{p\cdot (1-p)}{N}. \end{aligned}$$