Conditional probability

There can be rich relationships between two or more variables that are useful to understand. For example a car insurance company will consider information about a person’s driving history to assess the risk that they will be responsible for an accident. These types of relationships are the realm of conditional probabilities.

Exploring probabilities with a contingency table

The photo_classify data set represents a classifier a sample of 1822 photos from a photo sharing website. Data scientists have been working to improve a classifier for whether the photo is about fashion or not, and these 1822 photos represent a test for their classifier. Each photo gets two classifications: the first is called mach_learn and gives a classification from a machine learning (ML) system of either pred_fashion or pred_not. Each of these 1822 photos have also been classified carefully by a team of people, which we take to be the source of truth; this variable is called truth and takes values fashion and not. Figure 3.11 summarizes the results.

Marginal and joint probabilities

Figure 3.11 includes row and column totals for each variable separately in the photo_classify data set. These totals represent marginal probabilities for the sample, which are the probabilities based on a single variable without regard to any other variables. For instance, a probability based solely on the mach_learn variable is a marginal probability: $$\begin{aligned} P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) = \frac{219{}}{1822{}} = 0.12 \end{aligned}$$ A probability of outcomes for two or more variables or processes is called a joint probability: $$\begin{aligned} P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion} and \texttt{truth} is \texttt{fashion}}) = \frac{197{}}{1822{}} = 0.11 \end{aligned}$$ It is common to substitute a comma for “and” in a joint probability, although using either the word “and” or a comma is acceptable:

$P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion}, \texttt{truth} is \texttt{fashion}})$
means the same thing as
$P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion} and \texttt{truth} is \texttt{fashion}})$

We use table proportions to summarize joint probabilities for the photo_classify sample. These proportions are computed by dividing each count in Figure 3.11 by the table’s total, 1822, to obtain the proportions in Figure 3.13. The joint probability distribution of the mach_learn and truth variables is shown in Figure 3.14.

We can compute marginal probabilities using joint probabilities in simple cases. For example, the probability a randomly selected photo from the data set is about fashion is found by summing the outcomes where truth takes value fashion: $$\begin{aligned} P(\text{\texttt{truth} is \texttt{fashion}{}}) &= P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion} and \texttt{truth} is \texttt{fashion}{}}) \\ & \qquad + P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}not} and \texttt{truth} is \texttt{fashion}{}}) \\ &= 0.1081 + 0.0615 \\ &= 0.1696 \end{aligned}$$

Defining conditional probability

The ML classifier predicts whether a photo is about fashion, even if it is not perfect. We would like to better understand how to use information from a variable like mach_learn to improve our probability estimation of a second variable, which in this example is truth.

The probability that a random photo from the data set is about fashion is about 0.17. If we knew the machine learning classifier predicted the photo was about fashion, could we get a better estimate of the probability the photo is actually about fashion? Absolutely. To do so, we limit our view to only those 219 cases where the ML classifier predicted that the photo was about fashion and look at the fraction where the photo was actually about fashion: $$\begin{aligned} P(\text{\texttt{truth} is \texttt{fashion} given \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) = \frac{197{}}{219{}} = 0.900 \end{aligned}$$ We call this a conditional probability because we computed the probability under a condition: the ML classifier prediction said the photo was about fashion.

There are two parts to a conditional probability, the outcome of interest and the condition. It is useful to think of the condition as information we know to be true, and this information usually can be described as a known outcome or event. We generally separate the text inside our probability notation into the outcome of interest and the condition with a vertical bar: $$\begin{aligned} && P(\text{\texttt{truth} is \texttt{fashion} given \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) \\ && \quad = P(\text{\texttt{truth} is \texttt{fashion}\ }| \text{\ \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) = \frac{197{}}{219{}} = 0.900 \end{aligned}$$ The vertical bar “$|$” is read as given.

In the last equation, we computed the probability a photo was about fashion based on the condition that the ML algorithm predicted it was about fashion as a fraction: $$\begin{aligned} & P(\text{\texttt{truth} is \texttt{fashion}\ }| \text{\ \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) \\ &\quad = \frac{\text{\# cases where \texttt{truth} is \texttt{fashion} and \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}} {\text{\# cases where \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}} \\ &\quad = \frac{197{}}{219{}} = 0.900 \end{aligned}$$ We considered only those cases that met the condition, mach_learn is pred_fashion, and then we computed the ratio of those cases that satisfied our outcome of interest, photo was actually about fashion.

Frequently, marginal and joint probabilities are provided instead of count data. For example, disease rates are commonly listed in percentages rather than in a count format. We would like to be able to compute conditional probabilities even when no counts are available, and we use the last equation as a template to understand this technique.

We considered only those cases that satisfied the condition, where the ML algorithm predicted fashion. Of these cases, the conditional probability was the fraction representing the outcome of interest, that the photo was about fashion. Suppose we were provided only the information in Figure 3.13, i.e. only probability data. Then if we took a sample of 1000 photos, we would anticipate about 12.0% or $0.120\times 1000 = 120$ would be predicted to be about fashion (mach_learn is pred_fashion). Similarly, we would expect about 10.8% or $0.108\times 1000 = 108$ to meet both the information criteria and represent our outcome of interest. Then the conditional probability can be computed as $$\begin{aligned} &P(\text{\texttt{truth} is \texttt{fashion}}\ |\ \text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) \\ &= \frac{\text{\# (\texttt{truth} is \texttt{fashion} and \texttt{mach\_{}learn} is \texttt{pred\_{}fashion})}} {\text{\# (\texttt{mach\_{}learn} is \texttt{pred\_{}fashion})}} \\ &= \frac{108}{120} = \frac{0.108}{0.120} = 0.90 \end{aligned}$$ Here we are examining exactly the fraction of two probabilities, 0.108 and 0.120, which we can write as $$\begin{aligned} P(\text{\texttt{truth} is \texttt{fashion} and \texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}) \quad\text{and}\quad P(\text{\texttt{mach\_{}learn} is \texttt{pred\_{}fashion}}). \end{aligned}$$ The fraction of these probabilities is an example of the general formula for conditional probability.

Smallpox in Boston, 1721

The smallpox data set provides a sample of 6,224 individuals from the year 1721 who were exposed to smallpox in Boston. Doctors at the time believed that inoculation, which involves exposing a person to the disease in a controlled form, could reduce the likelihood of death.

Each case represents one person with two variables: inoculated and result. The variable inoculated takes two levels: yes or no, indicating whether the person was inoculated or not. The variable result has outcomes lived or died. These data are summarized in Tables 3.15 and 3.16.

General multiplication rule

Section 3.1.7 introduced the Multiplication Rule for independent processes. Here we provide the General Multiplication Rule for events that might not be independent.

General Multiplication Rule is simply a rearrangement of the conditional probability equation.

Independence considerations in conditional probability

If two events are independent, then knowing the outcome of one should provide no information about the other. We can show this is mathematically true using conditional probabilities.

We can show in Guided Practice 3.38(c) that the conditioning information has no influence by using the Multiplication Rule for independence processes: $$\begin{aligned} P(Y=\text{\texttt{1}}\ |\ X=\text{\texttt{1}}) &= \frac{P(Y=\text{\texttt{1} and }X=\text{\texttt{1}})} {P(X=\text{\texttt{1}})} \\ &= \frac{P(Y=\text{\texttt{1}}) \times P(X=\text{\texttt{1}})} {P(X=\text{\texttt{1}})} \\ &= P(Y=\text{\texttt{1}}) \\ \end{aligned}$$

Tree diagrams

Tree diagrams are a tool to organize outcomes and probabilities around the structure of the data. They are most useful when two or more processes occur in a sequence and each process is conditioned on its predecessors.

The smallpox data fit this description. We see the population as split by inoculation: yes and no. Following this split, survival rates were observed for each group. This structure is reflected in the tree diagram shown in Figure 3.17. The first branch for inoculation is said to be the primary branch while the other branches are secondary.

Tree diagrams are annotated with marginal and conditional probabilities, as shown in Figure 3.17. This tree diagram splits the smallpox data by inoculation into the yes and no groups with respective marginal probabilities 0.0392 and 0.9608. The secondary branches are conditioned on the first, so we assign conditional probabilities to these branches. For example, the top branch in Figure 3.17 is the probability that result = lived conditioned on the information that inoculated = yes. We may (and usually do) construct joint probabilities at the end of each branch in our tree by multiplying the numbers we come across as we move from left to right. These joint probabilities are computed using the General Multiplication Rule: $$\begin{aligned} & P(\text{\texttt{inoculated} = \texttt{yes} and \texttt{result} = \texttt{lived}}) \\ &\quad = P(\text{\texttt{inoculated} = \texttt{yes}})\times P(\text{\texttt{result} = \texttt{lived}}| \text{\texttt{inoculated} = \texttt{yes}}) \\ &\quad = 0.0392\times 0.9754=0.0382 \end{aligned}$$

Bayes’ Theorem

In many instances, we are given a conditional probability of the form $$\begin{aligned} P(\text{statement about variable 1 } | \text{ statement about variable 2}) \end{aligned}$$ but we would really like to know the inverted conditional probability: $$\begin{aligned} P(\text{statement about variable 2 } | \text{ statement about variable 1}) \end{aligned}$$ Tree diagrams can be used to find the second conditional probability when given the first. However, sometimes it is not possible to draw the scenario in a tree diagram. In these cases, we can apply a very useful and general formula: Bayes’ Theorem.

We first take a critical look at an example of inverting conditional probabilities where we still apply a tree diagram.

Example 3.42 highlights why doctors often run more tests regardless of a first positive test result. When a medical condition is rare, a single positive test isn’t generally definitive.

Consider again the last equation of Example 3.42. Using the tree diagram, we can see that the numerator (the top of the fraction) is equal to the following product: $$\begin{aligned} P(\text{has BC and mammogram}^+\text{}) = P(\text{mammogram}^+\text{ } | \text{ has BC})P(\text{has BC}) \end{aligned}$$ The denominator – the probability the screening was positive – is equal to the sum of probabilities for each positive screening scenario: $$\begin{aligned} P(\text{mammogram}^+\text{}) &= P(\text{mammogram}^+\text{ and no BC}) + P(\text{mammogram}^+\text{ and has BC}) \end{aligned}$$ In the example, each of the probabilities on the right side was broken down into a product of a conditional probability and marginal probability using the tree diagram. $$\begin{aligned} P(\text{mammogram}^+\text{}) &= P(\text{mammogram}^+\text{ and no BC}) + P(\text{mammogram}^+\text{ and has BC}) \\ &= P(\text{mammogram}^+\text{ } | \text{ no BC})P(\text{no BC}) \\ &\qquad\qquad + P(\text{mammogram}^+\text{ } | \text{ has BC})P(\text{has BC}) \end{aligned}$$ We can see an application of Bayes’ Theorem by substituting the resulting probability expressions into the numerator and denominator of the original conditional probability. $$\begin{aligned} & P(\text{has BC } | \text{ mammogram}^+\text{}) \\ & \qquad= \frac{P(\text{mammogram}^+\text{ } | \text{ has BC})P(\text{has BC})} {P(\text{mammogram}^+\text{ } | \text{ no BC})P(\text{no BC}) + P(\text{mammogram}^+\text{ } | \text{ has BC})P(\text{has BC})} \end{aligned}$$

Bayes’ Theorem is a generalization of what we have done using tree diagrams. The numerator identifies the probability of getting both $A_1$ and $B$. The denominator is the marginal probability of getting $B$. This bottom component of the fraction appears long and complicated since we have to add up probabilities from all of the different ways to get $B$. We always completed this step when using tree diagrams. However, we usually did it in a separate step so it didn’t seem as complex.

apply Bayes’ Theorem correctly, there are two preparatory steps:

1. First identify the marginal probabilities of each possible outcome of the first variable: $P(A_1)$, $P(A_2)$, ..., $P(A_k)$.

2. Then identify the probability of the outcome $B$, conditioned on each possible scenario for the first variable: $P(B | A_1)$, $P(B | A_2)$, ..., $P(B | A_k)$.

Once each of these probabilities are identified, they can be applied directly within the formula. Bayes’ Theorem tends to be a good option when there are so many scenarios that drawing a tree diagram would be complex.

The last several exercises offered a way to update our belief about whether there is a sporting event, academic event, or no event going on at the school based on the information that the parking lot was full. This strategy of updating beliefs using Bayes’ Theorem is actually the foundation of an entire section of statistics called Bayesian statistics. While Bayesian statistics is very important and useful, we will not have time to cover much more of it in this book.

Exercises