Home Articles 2021/03/04

Is it easier to tell what color a ball is or if it is red?

This is a draft post which will hopefully be expanded on in the future.

Introduction

While analyzing some data for a research project, I noticed effect A seemed a lot harder to get a significant result for compared to a closely related effect B. I wanted to know if it was actually the case that A was harder to “prove” than B, or if we should be suspicious of our significant B effects since their A counterparts weren't significant. Intuitively, it seemed like A could be harder to prove than B, but I wanted to be able to make a more solid argument than “seems right to me!” Unfortunately, my statistics education and a few web searches did not offer much advice on how to compare significance levels for different tests. This article describes how I went about trying to answer this question and the conclusions I reached.

To make this question more accessible, I simplified it to: “Is it easier to tell what color a ball is or if it is red?” As it turns out, this is slightly too simplified, but it's simple enough to pose to people online and get good speculative answers and ideas that I had not considered. Some set about describing how a color detector might work; it seems the answer depends on the technology used for the detector. Others took an information-theoretic approach: whether a ball is red is 1 bit of information, but what color it is is more than 1 bit if you have more than two color possibilities. Philosophy has a plethora of interesting questions to ask about what it means to “be a color”; these are well beyond my ability to answer. We can also discuss this in terms of how reliable the tests are, how many are performed, and what conditions they're performed under. A hypothesis testing perspective suggests we test a hypothesis for each color, or perhaps with clever hypothesis design we can test \(\log_2(\text{# of colors})\) hypotheses.

All these ideas, as well as a late-night trip to Wikipedia, led me to a solution that's convincing enough, at least for me. We'll start with a more explicit statistical “experiment” that describes the data we want to analyze. Next, we'll ask two questions of this data and explore how we might answer them. This leads us to an answer for our original question, “how do we say whether effect A is harder to prove than effect B?” Finally, we'll draw connections to some statistical measures of serial dependence in time-series data.

The Experiment

Say we have balls, one for each of \(C\) colors, e.g., a red ball, a green ball, and a violet ball. We do an experiment where we randomly pick a ball and put it in front of a color detector, then note down both the color of the ball and the color reported by the detector. If we let the random variable \(X\) represent the color of the ball and \(Y\) the reported color, this experiment gives a series of observations \((X_i, Y_i), 0 \leq i \lt T\) of a bivariate distribution. Our experimenter picks balls blindfolded, so the observations \(X_i\) are independent and identically distributed.

If this seems too abstract for you, we can switch the scenario up: suppose we have cards with letters written on them and are using some image recognition software to identify the letter. This scenario allows for the situation where we have multiple cards with 'A' written on them in different handwriting. It also permits the image recognition software to be nondeterministic, so repeat detections of the same card may produce different results!

Question 1: Is the color detector right?

Suppose we want to know whether the color detector is right; that is, if it's given a red ball, does it report red? And so on for each color. If \(X\) and \(Y\) are independent — that is, the color detector is picking answers at random with no regard to the actual color of the ball — then \(P(X = c, Y = c) = P(X = c)P(Y = c)\) for all colors \(c \in C\). If the color detector is “truthy”, then \(P(X = c, Y = c) \gt P(X = c)P(Y = c)\), and if it's actively being deceptive, then \(P(X = c, Y = c) \lt P(X = c)P(Y = c)\).

However, we don't actually know any of these values; we just have the observations \((X_i, Y_i)\) to work from. How do we decide if our estimates for \(P(X = c)\), \(P(Y = c)\), and \(P(X = c, Y = c)\) indicate that we can believe the detector right (or wrong)?

To measure how right the detector is, we define the random variable \(Z_T\) to be the number of times the detector gave the right answer in \(T\) observations. In other words, \(Z_T = \sum_{i = 0}^T \mathbf{I}(X_i, Y_i)\) where \(\mathbf{I}\) is the indicator function that returns 1 if its inputs are equal and 0 otherwise. \(Z_T\) follows a Bernoulli distribution1 with \(T\) trials and a probability of success \(p = {1 \over T} \sum_{i = 0}^T \mathbf{I}(X_i,Y_i)\).

A relevant fact is that a Bernoulli distribution \(\mathcal{B}(T,p)\) with sufficiently large \(T\) and \(p\) not near 0 or 1 can be approximated by a normal distribution \(\mathcal{N}(Tp, Tp(1-p))\). In the case that our \(Y_i\) are independent and identically distributed — the case that the detector is making answers up as it pleases — \(p = p_{i.i.d.} = \sum_{c \in C} p_{X,c} p_{Y,c}\) where \(p_{V,c} = {1 \over T} \sum_{i = 0}^T \mathbf{I}(V_i, c)\) is the probability of observing color \(c\) in observations \(V_i\). Thus, we can use significance testing to determine if \(Z_T\) follows a \(\mathcal{N}(Tp_{i.i.d.}, Tp_{i.i.d.}(1-p_{i.i.d.}))\) distribution. One testing approach is to calculate the lower and upper critical values directly from the normal distribution above. Alternatively, we can normalize \(Z_T\) into \(\overline{Z}_T\) such that \(\overline{Z}_T\) follows the “standard normal” distribution, \(\mathcal{N}(0,1)\), when \(X\) and \(Y\) are independent.2 Then \(\overline{Z}_T^2\) follows a chi-squared distribution3 with one degree of freedom, \(\chi^2_1\), and the independence of \(X\) and \(Y\) can be tested using a \(\chi^2\) significance test.

Question 2: Could the detector be right?

This is all well and good, but suppose we let Wittgenstein or Quine loose in the lab before the experiment and they explained referential inscrutability to the color detector. Now the color detector still works or doesn't work, but it also might have mixed up its labels for colors so that, for example, it calls “green” what we'd call “red”. Can we tell if it's just making stuff up or if there is some method to its madness — a mapping from its outputs to the “right” answers? In short, can we tell whether it could be right?

We can even relax this question a little by letting the detector, say, report “green” for both “red” and “yellow” balls. This question is equivalent to asking whether there is a mapping \(\phi : C \to C\) such that \(P(X = c, Y = \phi(c)) \gt P(X = c)P(Y = c)\). In other words, can we figure out a way of assigning our color labels to the answers given by the color detector? Note that our first question, “is the detector right?”, is a special case of this broader question where we assume \(\phi(c) = c\) — that is, no relabeling happens.

As an aside, from a formal logic perspective, we can see that question 2 is more complex to ask. Question 1 asks if \(\forall i, X_i = Y_i\). Question 2 asks if \(\exists \phi, \forall i, X_i = \phi(Y_i)\). Since there's an extra quantifier, we might anticipate that answering question 2 would require a longer proof or more powerful proof techniques.

We can split this up into \(|C|^2\) Bernoulli trials, each counting the number of times the detector gave the reply \(r \in C\) for a ball of color \(c\). Define \(Z_T^{c,r}\) to be the number of times \((c,r)\) appears in \(T\) observations of \((X,Y)\). This variable follows a Bernoulli distribution with \(T\) trials and \(p^{c,r} = {1 \over T}\sum_{i = 0}^T\mathbf{I}(X_i,c)\mathbf{I}(Y_i,r)\) — the probability that \(X_i = c\) and \(Y_i = r\). If \(X\) and \(Y\) are independent, we have \(p^{c,r} = p_{i.i.d.}^{c,r} = p_{X,c}p_{Y,r}\).

As an example, let \(C = \{R,G,V\}\) (“red”, “green”, and “violet”) be our color space. We can arrange the Bernoulli trial probabilities for the independent case into a matrix as shown: \begin{equation*} \begin{bmatrix} p_{X,R}p_{Y,R} & p_{X,R}p_{Y,G} & p_{X,R}p_{Y,V} \\ p_{X,G}p_{Y,R} & p_{X,G}p_{Y,G} & p_{X,G}p_{Y,V} \\ p_{X,V}p_{Y,R} & p_{X,V}p_{Y,G} & p_{X,V}p_{Y,V} \\ \end{bmatrix} \end{equation*} Notice that each row sums to \(p_{X,c}\) and each column sums to \(p_{Y,c}\), so while we have \(|C|^2\) entries, only \((|C|-1)^2\) of them are independent. Let \(C^*\) be the set of colors \(C\) with one removed; in this example, we might let \(C^* = \{R,G\}\). Thus we have \((|C|-1)^2\) independent Bernoulli trials \(Z_T^{c,r}\) where \(c,r \in C^*\).

Normalize the \(Z_T^{c,r}\) into \(\overline{Z}_T^{c,r}\) so that they follow the standard normal distribution when \(X\) and \(Y\) are independent. Let \(Q = \sum_{c,r \in C^*} \left(\overline{Z}_T^{c,r}\right)^2\) be the sum of these independent standard normal random variables. Then, in the case \(X\) and \(Y\) are independent, \(Q\) follows a \(\chi^2\) distribution with \((|C|-1)^2\) degrees of freedom.

Summary

To summarize: if we want to know whether the detector is right, the significance test critical values are taken from the \(\chi^2_1\) distribution. If we want to know whether there is a way it could be right, the critical values are taken from the \(\chi^2_{(|C|-1)^2}\) distribution.

As shown in Figure 1, the more degrees of freedom in the distribution, the longer the \(\chi^2\) CDF takes to converge to 1. To visually find the critical value for a significance test, pick a significance level \(\alpha\), then draw a horizontal line on the CDF at \(1 - \alpha\). The critical value is the \(x\) value where the CDF crosses that horizontal line. Test statistic values greater than the critical value are considered to demonstrate a statistically significant effect. In more familiar terms, the \(p\) value for a given test statistic value \(x\) is \(1 - F_k(x)\); if that \(p\) value is larger than \(1-\alpha\), the result is significant.

The critical values are lower for the first question than the second assuming we have at least 3 colors to pick between. For the above example, compare the CDF for \(k = 1\) (Q1) to \(k = 4\) (Q2).

Which is to say: it takes a less extreme effect to determine that a detector is right. Determining whether it might be right or if it's just talking nonsense requires a stronger effect.4


Figure 1: \(\chi^2\) cumulative distribution function for 1–9 degrees of freedom.
Source: Wikipedia

Serial Dependence

All this play in the statistical ball pit is not just for fun, but was motivated by questions about statistics that you won't find so easily in a Chuck-e-Cheese. My research has led me into the realm of Categorical Time Series Analysis. A time series is a sequence of observations; for instance, the outside temperature measured every hour. These are used all over the place to study weather, heart rhythms, stock market performance, and so on. Our data happens to be categorical, meaning that it takes discrete values with no inherent ordering. Categorical models appear in a variety of places; for instance, studies of sleep phases and bird calls. While real-valued data can be averaged, sorted from least to greatest, and run through all kinds of functions, the operations that are allowed for categorical data are much more restricted. Since we can't compute the average, median, or expected value of a categorical distribution, the statistical techniques for it are a bit different from their real-valued counterparts.

We'll focus on measures of categorical serial dependence, which is analogous to autocorrelation for real-valued time series. These measures ask, “does knowing part of a time series help predict the rest?” or, “how repetitive is this time series?” We'll ask these for a particular time offset, called lag: given an observation at time \(t-k\), what can we say about the observation at time \(t\)? Depending on the answer, we observe one of several effects:

Unsigned serial dependence is a measure of how repetitive a dataset is: if we always see \(R\) followed by \(G\), this gives some amount of unsigned serial dependence at lag 1. Signed serial dependence is a measure of how self-similar a dataset is: if we always see \(R\) followed by \(R\), we have some signed serial dependence.

Notice that wherever we have signed serial dependence, we also have unsigned serial dependence. However, the reverse is not true. Unsigned serial dependence is harder to prove precisely because it encompasses more possible relationships among the data.

We can relate the questions asked of our colored balls data to serial dependence measures at lag \(k\). In these measures, the categorical labels correspond to “colors” and our random variables are \(X_{t-k} = X\) and \(X_t = Y\). Denote the marginal probabilities of each category by \(\pi_c = P(X_t = c)\); when estimating from samples, \(\pi_c = p_{X,c} = p_{Y,c}\). Lastly, we define the lagged bivariate probability \(p_{cr} = P(X_{t-k} = c, X_t = r) = P(X = c, Y = r)\); when estimated, \(p_{cr} = p^{c,r}\).

Signed serial dependence corresponds to the first question: \(X_t\) and \(X_{t-k}\) have to match, just like \(X\) and \(Y\) have to match in the colored balls experiment. In statistical terms, we get perfect signed serial dependence under these conditions:

Common measures for signed serial dependence are Cohen's \(\kappa\) and several modifications thereof, denoted \(\kappa^*\) and \(\kappa^\star\). All these measures have been shown to be normally distributed when estimated from i.i.d. time series data and thus can be tested using a \(\chi^2_1\) distribution, just as with our Bernoulli trial test for question 1.

Unsigned serial dependence corresponds to the second question: “perfect insight” corresponds to picking a mapping \(\phi\) that allows us to predict \(X_t\) from \(X_{t-k}\). In statistical terms, we get perfect unsigned serial dependence when we can pick \(r = \phi(c)\) so that \(p_{cr} = \pi_c\) or \(P(X = c, Y = r) = P(X = c)\) for all \(c\). A common measure of unsigned serial dependence is Cramer's (\v\), which follows a \(\chi^2_{d^2}\) distribution when estimated from i.i.d. time series data having \(d+1\) categories, just as with our Bernoulli trial test for question 2.

The Bernoulli trial statistics developed here don't exactly follow the properties one would desire of serial dependence measures, which is why we use Cramer's \(v\), Cohen's \(\kappa\), \(\kappa^*\), and \(\kappa^\star\) instead. However, they offer an intuitive sense for how one might design such a test. It is significantly easier to see why they follow \(\chi^2\) distributions when estimated from i.i.d. data than it is to follow the proofs for the above test statistics. See Further Reading for far more detail on serial dependence measures.

Epilogue

I am not a statistician, as you can probably tell. This argument features a few handwaved bits which I would be very interested in shoring up. If you have thoughts on how to improve this argument or if you spot a hole in it, I'd like to hear! Also, if there's articles written about related topics, please send them my way. Write me an email, or if you got this through social media, replies there are good too.

Thanks to Canageek, parenthetical, cpsdqs, cephie, mithrandir, hex, and crystalmoon for their input!

Further Reading

[1]
C. H. Weiß and R. Göb, “Measuring serial dependence in categorical time series,” AStA, vol. 92, no. 1, pp. 71–89, Feb. 2008, doi: 10.1007/s10182-008-0055-4.
[2]
C. H. Weiß, “Serial dependence of NDARMA processes,” Computational Statistics & Data Analysis, vol. 68, pp. 213–238, Dec. 2013, doi: 10.1016/j.csda.2013.07.009.
[3]
C. H. Weiß, “Measures of Dispersion and Serial Dependence in Categorical Time Series,” Econometrics, vol. 7, no. 2, Art. no. 2, Jun. 2019, doi: 10.3390/econometrics7020017.

1. If \(X \sim \mathcal{B}(n,p)\), the probability \(P(X = k)\) is the chance of getting \(k\) successes in \(n\) independent trials with probability of success \(p\).

2. If \(X \sim \mathcal{N}(\mu,\sigma^2)\), we can normalize this to \(\overline{X} = \frac{X - \mu}{\sigma} \sim \mathcal{N}(0,1)\).

3. Given random variables \(X_i \sim \mathcal{N}(0,1)\), the random variable \(Q = \sum_{i = 0}^k X_i^2\) follows the distribution \(\chi^2_k\).

4. There is some handwaving here; namely, we assume the effect of scaling to standard normal distributions is approximately equivalent for both questions. Whether this is true is somewhat difficult to show; fortunately, it has little effect on the overall conclusion, which concerns measures which empirically produce similar-scale values when evaluated on the same dataset.