With the advent of machine learning into our IT landscapes, a previously rather academic conflict of the statistical community surfaces in blogs and other forums of discussion every other week. It is the question of frequentism vs. Bayesianism. This debate, often one that is as emotional as the famous editor-wars, is in fact a very fundamental one that touches the foundations of statistics and probability theory. In that sense, it isn’t your usual bike shedding discussion, even if it is sometimes lead as one. Metaphorically, it is a custody trial to determine who may claim the interpretational sovereignty over nothing less than the Theory of Probability.

Frequentism and Bayesianism are both established approaches to statistics. Their differences start, with their core definitions. frequentism treats probabilities as ratios of frequency-counts collected from an infinite number of trials; and frequentist practitioners will tell you that a finite number of trials will also suffice (as in: from 100 coin flips, 50 times we will obtain head, thus the probability for head is $$\wp(H=1)=0.5$$).

For Bayesianists, probabilities are degrees of belief; also Bayesianists use Bayes’ theorem for inference. A Bayesianist would take the probability $$\wp(H=1)=0.5$$

in the coin-flip example above to mean something like "It is credible, that head and non-head (tail) are results of a coin flip, without one option being more likely than the other’. A programmer can think of the Bayesian interpretation of probabilities as an extension of Boolean algebra: true (1, firm-belief) and false (0, firm-disbelief) are complemented with a spectrum of values $$0 \ldots 1$$.

These brief characterizations are already enough to understand much of the criticism either method faces:

• Bayesian probabilities are criticised as "subjective" or as not a genuine measurement parameter (degree of belief).

• frequentist probabilities are said to be limited to infinitely repeatable trials, and thus not applicable to any real world data set, with a finite number of measurements.

This criticism is too simplistic, however. And to those, who strongly associate with one camp, there are probably many embarassing commonalities: Both approaches often lead to very similar results. In this post I will show you how you can solve a problem with both methods and compare the results.

## Estimating The Probability for a Bernoulli-Trial

We all expect coins to be fairly balanced. I.e. if we flip a coin, we expect to roughly obtain head half of the times, and tail the other half of the times. Yet there are many processes with two outcomes, where we don’t know the individual probabilities beforehand. For example, a researcher might be interested in the immunization rate of a population.

Our researcher determines the immunization rate of $$N=40$$ people. The measured results could be a series of numbers (1 for immunized and 0 for not-immunized) like: [1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1].

The questions we are bound to solve are

• What are the immunization rates (what is the probability for a person to be immunized)?

• How reliable (and under what circumstances) would that inferred probability be?

## Frequentist Approaches

I divided this section into three parts

1. first we apply a common-sense approach to the problem

2. then, we see that our first approach is in fact the solution of the maximum-likelihood approach

3. we apply the bootstrap method to get more than just the maximum-likelihood estimate of the immunization rate.

### Common-sense (naïve) Treatment

A very simple approach to this problem is to just count the number of immunized and the number of people screened. for the above list, we have $$k = 34$$ immunized people of a total of $$N = 40$$ people, which leads to an immunization probability of $$0.85$$ .

If the researcher had only screened the first 20 people, the result would have looked a bit different, $$0.95$$ . If we had only looked at the probabilities from people 20-40, we would have gottten a probability lower than $$0.85$$ . Thus, we have a method that gives us immunization rates, yet it heavily depends on the sample size. Also, we don’t have a means to quantify how certain we are about these enumbers.

### Maximum Likelihood Estimate

The likelihood-function [1]:

$L(\mu \mid N, k)$

is the probability $$k$$ immunized subjects of $$N$$ subjects in total, under the condition of a parameter $$\mu$$ , which we’ll write down as $$\wp(N, k \mid \mu)$$ . We identify, that this

$\wp(N, k \mid \mu)$
$L(\mu) = \wp(N, k \mid \mu) = \text{Binomial}(k\mid N, \mu) = \binom N k \mu^k(1-\mu)^{N-k}$

We now want to find the $$\mu$$ that maximizes the likelihood $$L(\mu)$$. We could of course work out the equations by hand. I use sympy here, which will do all the tiresome calculations for us:

In [1]: N, k, mu = symbols("N, k, mu")
In [2]: likelihood = binomial(N, k) * mu**k * (1-mu)**(N-k)

Sympy will nicely render the likelihood term

In [3]: likelihood
Out[3]:
k         N - k ⎛N⎞
μ ⋅(-μ + 1)     ⋅⎜ ⎟
⎝k⎠

Now let’s see if sympy can come up with the derivative with respect to $$\mu$$:

In [4]: diff(likelihood, mu)
Out[4]:
k         N - k ⎛N⎞    k                  N - k ⎛N⎞
k⋅μ ⋅(-μ + 1)     ⋅⎜ ⎟   μ ⋅(-N + k)⋅(-μ + 1)     ⋅⎜ ⎟
⎝k⎠                             ⎝k⎠
────────────────────── + ─────────────────────────────
μ                          -μ + 1

Finally, we are only interested in the value of $$\mu$$, for which the derivative is zero.

In [5]: solve(diff(likelihood, mu), mu)
Out[5]:
⎡k⎤
⎢─⎥
⎣N⎦

Result: We obtain $$\mu = \frac{k}{N}$$ as as the maximum likelihood estimate, which basically is what we expected as the naïve result.

### The Bootstrap Method

We can use the fact, that different subsets of our data yield different results, to get a better picture of the reliability / variance of our probabilities. Just like above, we will take a look at subsets of the total data set. This time we will take a systematic approach, we will

• construct a new data sets from the recorded trials by randomized sampling with replacement. The new data set has the same size as the original one, but might contain some data points multiple times, and other data-points will be missing.

• calculate rates on the newly contructed data-set.

• Repeat this many times to get as many rates as possible (best: calculate it for all possible combinations, but mind the combinatorial explosion)

• plot the histogram of these rates (each subsample yielding one data point).

#### Distribution of Rates in Subsample

The histogram below was made from a sample of 100 measurements, drawing many (100000) subsamples of 100 measurements, from which the rates were calculated.

np.random.seed(1)
N = 100

x = scipy.stats.bernoulli(0.86).rvs(size=N)
m = np.fromiter((np.mean(np.random.choice(x, N))
for _ in range(100000)),
dtype=float)

plt.hist(m, np.linspace(0, 1, 50))
plt.xlabel("Rate")
plt.ylabel("Occurences")

Depending on the subsample, we get different results for the calculated rate. All calculated rates from a distribution. We must assume, that the single rate calculated in the maximum likelihood approach above is just as noisy as the rates calculated in the bootstrap approach, because we assume independence of the individual events. So the bootstrapped rate distribution gives us an idea on how credible and accurate the maximum-likelihood rate is.

## Bayesian Approach

 Note If you aren’t so much interested in Bayes' theorem, you can just scroll down to the heading Incremental updates and enjoy the graphs.

### Bayes Theorem

Bayes’ theorem is the central hub of Bayesian methods. It is, however, not a postulated assumption, just happening to work, but a direct consequence of conditional probabilities. If you look at the probability for two propositions[2] $$A$$ and $$B$$ — $$\wp(A, B)$$ — you can express this joint probability by conditional probabilities:

$\wp(A, B) = \wp(A\mid B) \wp(B) = \wp(B \mid A) \wp (A)$

If we take A for the probability of the street to be wet, and B for the probability of rainfall in the last hour, then $$\wp(A, B)$$ is the probabilty for rainfall and a wet street.

$$\wp(A\mid B)$$ is the conditional probability for a wet street, given that it has rained, $$\wp(B \mid A)$$ the conditional probability for rainfall, given that the street is wet; and $$\wp(B)$$ is the probability of rainfall, $$\wp(A)$$ is the probability of a wet street.

We can rearrange the above equation dividing through $$\wp(B)$$ on both sides and get an equation that expresses the conditional probability $$\wp(A\mid B)$$ by the inverse probability $$\wp(B\mid A)$$.

$\wp(A\mid B) = \frac{\wp(B \mid A)}{\wp(B)} \wp(A)$

Until now, I think, frequentists and Bayesianists can agree. The disagreement starts on when and how to use this equation. Bayesianists infer (conclude)

$\wp(A\mid B)$

on the left side of the equation from the right hand side. Whereas critics of Bayesianism consider this to be a dangerous endeavour.

#### Bayes’ Theorem and model estimation

Bayesian statistics is concerned with data $$D$$ and a model $$M$$ (a fit parameter, a parameter of a distribution, a quantity that should be inferred). By substituting $$A\to M, B \to D$$, we get Bayes’ theorem with Bayesian semantics:

$\wp(M\mid D) = \frac{\wp(D \mid M)}{\wp(D)} \wp(M)$

The probabilities involved are:

• $$\wp(M\mid D)$$ is the probability for the model, given data. Obtaining this probability distribution, we have the inference (Inference means nothing but: Estimating a model from/given data). We also call this quantity the posterior.

• $$\wp(D\mid M)$$ is the probability for data, given the model. This is the likelihood we have already seen above [3].

• $$\wp(M)$$ is the prior, a probability (distribution), that is independent of $$D$$.

• $$\wp(D)$$ is the probability for data [4]. For this treatment here, we can think of it as a normalization constant (which can be very costly to compute).

The most important aspect of this theorem is: We can express the probability for a model, given data by some term, that involves the probability for data, given the model.

Why is this important? Because it is often much easier to give an expression for the, than to come up with the posterior $$\wp(M\mid D)$$ directly.

### Bayesian Estimation of Immunization Rates

The immunization rate $$\mu$$ that we are looking for is a probability (in the frequentist sense). Nevertheless, it is in the Bayesian sense also model parameter $$M = \mu$$ , whose probability distribution given data $$\wp(\mu\mid D)$$ can be inferred using Bayes’ theorem. We would like to find the distribution for this parameter $$\wp(\mu\mid D)$$ , so that we can get an expectancy value $$\mathcal{E}[\mu \mid D$$] and the variance.

Bayes rule gives us this distribution. First, we will ignore the denominator, which is more of a normalization parameter, without loss of generality, as $$\wp(\mu\mid D) \propto \wp(D \mid \mu) \wp(\mu).$$

What could $$\wp(D\mid \mu)$$ be? During the trial, we have $$k$$ observations of immunized subjects, out of $$N$$ observations in total, so $$\wp(D\mid \mu)= \wp(k \mid N, \mu)$$. Does the k-out-of-N sound familiar? It is what the Binomial distribution describes

$\wp(N, k \mid \mu) \sim \text{Binomial}(k\mid N, \mu).$

Now we need a suitable expression for $$\wp(\mu)$$ . It needs to be a probability distribution that just contains the model alone (no data). This is a tough choice (and one of the main sources for distrust of the Bayesian methods, probably). Luckily, the Bayesian literature tells us, that the Beta-distribution is a suitable choice for this,

$\wp(M) \sim \text{Beta}(\mu\mid \alpha_0, \beta_0)$

and $$\alpha_0$$ and $$\beta_0$$ are the a priori parameters. Choosing $$\alpha_0=\beta_0=1$$ for example would assume a uniform prior distribution between $$0 \leq \mu \leq 1$$. I use $$\alpha_0=\beta_0=\frac{1}{2}$$ [5].

Due to some neat properties of binomial and beta distribution, what follows is that Bayes’ theorem in this instance simplifies to a very simple rule. Starting with a Beta-prior distribution, we can obtain the posterior distribution by just adding our observed $$k$$ and $$N-k$$ data points to the parameters of the beta distribution.

$\wp(\mu \mid N,k) = \text{Beta}(\mu\mid \alpha_0 + k, \beta_0 + (N-k))$

### Data

Using the above beta-prior model with our collected data, we can obtain posterior distributions for the immunization rate. The following plot shows such distributions, adding more data with each subplot.

In the plot, we can follow along, how wit hmore data, the distribution gets narrower (i.e. with more data, we are more certain). The first subplot labelled $$N= 0$$ is the prior distribution.

The infered value for the immunization rate, $$\mu$$, is the expectancy value of these distributions (represented by the vertical black lines in the plot). For comparison, the red, dotted line is the true immunization rate, that we know, because we put it into the random number generator that provides us with the data set.

We can see in the plots, that gradually our estimate changes, as the data fluctuates, although with many measurements it seems to stabilize. The learn curve is a good way to visualize this.

## Learn curve for Bayes

Pretending that we only perform one trial at a time, we can plot a learning curve that shows how our method performs and how the inference quality improves with more data:

## Learn curve for Boostrap

We have plotted a learning curve for the Bayesian approach, how does the bootstrap method compare? Let’s add a learn curve for the bootstrap method to the plot:

What we can see here is, that with enough data points, both methods give very similar results, to the point where they seem equivalent. However, the Bayesian approach is better when infering from fewer data points. Why is that?

• Bootstrap relies on constructing ad-hoc data-sets from the original samples. With few data points, it suffers from the same bias as the original sample.

In our data set, the first few samples are 1` consistently, and thus the bootstrap approach must yield a rate of 1 and 0-sized error bars.

• The Bayesian approach uses a prior. This prevents the method from focussing too strongly on the first few events. Recording the first event, we do have error bars that are fairly large (which fits nicely with our expectation of being uncertain about the true value of our parameter).

The subtle, systematic difference that remains between the Bayesian and the bootstrap method for larger sample sizes (the Bayesian rate is consistently smaller than the bootstrapped rate) is the influence of the prior, that never completely vanishes (although it is negligible given the statistical fluctuations).

## Conclusion

There are many conclusions that one can draw from this simple example. Of course not everything that we can learn from this example generalize to all questions about Bayes and frequentism. So I’ll limit my conclusion here to the one very simple advice: I learned a lot more about statistical methods and algorithms by constantly looking at how Bayesianists and frequentists approach and derive them. Most statistical topics are treated in Bayesian and frequentist literature.

## Notes

Thank you Christopher and Daniel for your feedback on this blog post.

4. It doesn’t have a much of a canonical name. Calling it the evidence is popular. Since $$\wp(D) = \sum \wp(D \mid M_i) \wp(M_i)$$, I like “marginalized likelihood”.
5. The prior distribution $$\text{Beta}\left(\mu \mid \frac{1}{2}, \frac{1}{2}\right)$$ is called the Jeffreys prior. Choice of priors would be the material for a series of blog posts.