Good–Turing frequency estimation

The method

The Good–Turing estimator is largely independent of the distribution of species frequencies.

Notation

• Assuming that ${\displaystyle X}$ distinct species have been observed, enumerated ${\displaystyle 1,\dots ,X.}$
• Then the frequency vector, ${\displaystyle \overline{R},}$ has elements ${\displaystyle R_x}$ that give the number of individuals that have been observed for species ${\displaystyle x.}$
• The frequency of frequencies vector, ${\displaystyle (N_r{)}_{r=0,1,\dots },}$ shows how many times the frequency r occurs in the vector ${\displaystyle \overline{R}}$ (i.e. among the elements ${\displaystyle R_x}$):

${\displaystyle N_r=|\left\{x\mid R_x=r\right\}|.}$

For example, ${\displaystyle N_1}$ is the number of species for which only one individual was observed. Note that the total number of objects observed, ${\displaystyle N,}$ can be found from

${\displaystyle N=\sum _{r=1}^{\mathrm{\infty }}r{N}_r=\sum _{x=1}^X{R}_x.}$

Calculation

The first step in the calculation is to estimate the probability that a future observed individual (or the next observed individual) is a member of a thus far unseen species. This estimate is:

${\displaystyle p_0=\frac{N_1}{N}.}$

The next step is to estimate the probability that the next observed individual is from a species which has been seen ${\displaystyle r}$ times. For a single species this estimate is:

${\displaystyle p_r=\frac{(r+1)S(N_{r+1})}{NS(N_r)}.}$

Here, the notation ${\displaystyle S(\cdot )}$ means the smoothed, or adjusted value of the frequency shown in parentheses. An overview of how to perform this smoothing follows in the next section (see also empirical Bayes method).

To estimate the probability that the next observed individual is from any species from this group (i.e., the group of species seen ${\displaystyle r}$ times) one can use the following formula:

${\displaystyle \frac{(r+1)S(N_{r+1})}{N}.}$

Smoothing

For smoothing the erratic values in ${\displaystyle N_r}$ for large r, we would like to make a plot of ${\displaystyle \log N_r}$ versus ${\displaystyle \log r}$ but this is problematic because for large ${\displaystyle r}$ many ${\displaystyle N_r}$ will be zero. Instead a revised quantity, ${\displaystyle \log Z_r,}$ is plotted versus ${\displaystyle \log r,}$ where ${\displaystyle Z_r}$ is defined as

${\displaystyle Z_r=\frac{N_r}{\frac{1}{2}(t-q)},}$

and where q, r, and t are three consecutive subscripts with non-zero counts ${\displaystyle N_q,N_r,N_t}$. For the special case when r is 1, take q to be 0. In the opposite special case, when ${\displaystyle r=r_{\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{t}}}$ is the index of the last non-zero count, replace the divisor ${\displaystyle \frac{1}{2}(t-q)}$ with ${\displaystyle r_{\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{t}}-q,}$ so ${\displaystyle Z_{r_{\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{t}}}=\frac{N_{r_{\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{t}}}}{r_{\mathsf{l}\mathsf{a}\mathsf{s}\mathsf{t}}-q}.}$

A simple linear regression is then fitted to the log–log plot.

For small values of ${\displaystyle r}$ it is reasonable to set ${\displaystyle S(N_r)=N_r}$ – that is, no smoothing is performed.

For large values of r, values of ${\displaystyle S(N_r)}$ are read off the regression line. An automatic procedure (not described here) can be used to specify at what point the switch from no smoothing to linear smoothing should take place. Code for the method is available in the public domain.

Derivation

Many different derivations of the above formula for ${\displaystyle p_r}$ have been given.

One of the simplest ways to motivate the formula is by assuming the next item will behave similarly to the previous item. The overall idea of the estimator is that currently we are seeing never-seen items at a certain frequency, seen-once items at a certain frequency, seen-twice items at a certain frequency, and so on. Our goal is to estimate just how likely each of these categories is, for the next item we will see. Put another way, we want to know the current rate at which seen-twice items are becoming seen-thrice items, and so on. Since we don't assume anything about the underlying probability distribution, it does sound a bit mysterious at first. But it is extremely easy to calculate these probabilities empirically for the previous item we saw, even assuming we don't remember exactly which item that was: Take all the items we have seen so far (including multiplicities) — the last item we saw was a random one of these, all equally likely. Specifically, the chance that we saw an item for the ${\displaystyle (r+1)}$th time is simply the chance that it was one of the items that we have now seen ${\displaystyle r+1}$ times, namely ${\displaystyle \frac{(r+1)N_{r+1}}{N}}$. In other words, our chance of seeing an item that had been seen r times before was ${\displaystyle \frac{(r+1)N_{r+1}}{N}}$. So now we simply assume that this chance will be about the same for the next item we see. This immediately gives us the formula above for ${\displaystyle p_0}$, by setting ${\displaystyle r=0}$. And for ${\displaystyle r>0}$, to get the probability that a particular one of the ${\displaystyle N_r}$ items is going to be the next one seen, we need to divide this probability (of seeing some item that has been seen r times) among the ${\displaystyle N_r}$ possibilities for which particular item that could be. This gives us the formula ${\displaystyle p_r=\frac{(r+1)N_{r+1}}{N{N}_r}}$. Of course, your actual data will probably be a bit noisy, so you will want to smooth the values first to get a better estimate of how quickly the category counts are growing, and this gives the formula as shown above. This approach is in the same spirit as deriving the standard Bernoulli estimator by simply asking what the two probabilities were for the previous coin flip (after scrambling the trials seen so far), given only the current result counts, while assuming nothing about the underlying distribution.