Basic probability #
Probability is the branch of mathematics that studies randomness in a formal and rigorous way. Probability is also part of the foundation of statistics and data science, largely because it provides us with a way to link a sample to its population, and a way to make precise statements about uncertainty in data analysis. Probability also provides us with many tools for describing variation in a collection of measurements.
Recall that we view the sample of data that we collect and analyze as a selection from a larger population, and our goal in data analysis is to make statements about the population, not only about the sample. In a probability-based approach to statistics, the sample is viewed as a random “draw” from the population. In this context, we can view the population as something called a probability distribution, which is a precise description of all the possible observations that can be made, and how likely we are to observe each of them.
As a concrete example of a probability distribution, suppose we select a person of age 18 or older at random from a city, say Cairo, Egypt, and ask that person how many times they have been married. The possible outcomes are 0, 1, 2, … (in principle it is unbounded). There is a certain probability that the person responds with 0, another probability that the person responds with 1, and so on. These probabilities, taken together, constitute a probability distribution.
A random draw from a population can always be repeated, at least in theory. That is, if we are interested in sampling 10 people from the UM student population, and we randomly contact 10 people today to construct a sample, we can do this again tomorrow. The 10 people sampled tomorrow will almost certainly be different than the 10 people sampled today. However in some sense the two samples are equivalent because the same process was used to obtain them.
In situations where you have a sample, but you can’t imagine obtaining a replicate of your sample, then the sample that you have is likely not a random sample. For example, sometimes the sample is actually the entire population. Such a sample is known as a census. If you were to repeat a census, you would get back exactly the same results. In other cases, we have a convenience sample that was generated in such a haphazard way that there is no natural way to view it as a proper random draw from a population.
A probability distribution is formally defined by two entities, one of which, the sample space, is the collection of values that can be drawn. The sample space can be either finite of infinite. Starting with the finite case, the other entity that defines the probability distribution is its probability mass function, or pmf. The pmf is a function on the sample space that tells us the probability of observing each point when we take a draw.
For example if we have a box containing one red ball, two orange balls, and one green ball, then the sample space can be viewed as the set {red, orange, green}. If we reach into the box and draw a ball, the pmf can be denoted with the function \(p(\cdot)\) , where \(p({\rm red}) = 1/4\) , \(p({\rm orange}) = 1/2\) , and \(p({\rm green}) = 1/4\) . The probability for the orange point is twice that of the other colors since there are two orange balls in the box.
Based on the above example, we see that a probability distribution with a finite sample space can always be represented through a table:
x | red | orange | green |
---|---|---|---|
p(x) | 1/4 | 1/2 | 1/4 |
There are certain properties satisfied by all probability distributions. The most important of these are (i) every probability is a real number between 0 and 1, and (ii) the sum of probabilities over all points in the sample space is 1. We see this in the example above, since 1/4 + 1/2 + 1/4 = 1, and each of these probabilities falls between 0 and 1.
An event is a subset of the sample space. We say that the event “occurs” if the value we draw is in the event. For example, {orange, green} is an event that occurs if the ball that we draw is either orange or green. The probability of an event is the sum of all probabilities in the event, which is 3/4 in this example. An event can be complemented, by forming the set of all points that are not in the event. This yields a simple but important identity. If E is any event, the the probability that E does not occur is equal to 1 minus the probability that E does occur. Thus, the probability of not drawing an orange or a green ball is 1 - 3/4 = 1/4. We can write this fact mathematically as \(P({\rm not}\; E) = 1 - P(E)\) .
Another term for a probability distribution is a random variable. A random variable is usually denoted with a capital letter, such as \(X\) . When we write \(P(X=x)\) , this means “the probability that the random variable \(X\) takes on the value \(x\) ”. In this expression, \(x\) is a fixed constant and \(X\) represents a draw from a probability distribution, and therefore has no fixed value. If \(X\) follows the distribution for sampling colored balls discussed above, then we could write \(P(X={\rm red}) = 1/4\) . We can also write things like \(P(X \in \{{\rm orange, green}\}) = 3/4\) to denote the probability that the event {orange, green} occurs.
Summary statistics of quantitative probability distributions #
Quantitative distributions have numeric sample spaces. For example, if \(X\) represents a quantitative distribution, say the annual household income in the United States, then we can write \(P(X \le 30000)\) to denote the probability that a household’s income is 30,000 USD or less, or \(P(X > 100000)\) to denote the probability that the household’s income is greater than 100,000 USD.
Earlier in the course we studied summary statistics. Many of the summary statistics we discussed such as moments and quantiles are mainly used with quantitative data. Now that we have some basic definitions from probability, we can define analogous summary measures for probability distributions. We will work with the following example:
x | 1 | 2 | 4 | 7 |
---|---|---|---|---|
p(X=x) | 0.1 | 0.3 | 0.4 | 0.2 |
This is a quantitative distribution whose sample space contains the four points {1, 2, 4, 7}. Note that the probabilities in the table sum to 1. The expected value of this distribution is obtained by taking each point in the sample space, multiplying by its corresponding probability, and summing. In this case, we get 0.1·1 + 0.3·2 + 0.4·4 + 0.2·7 = 3.7, and this value may be denoted \(E[X]\) or \(EX\) , where \(E\) stands for “expectation”. Recall that the expected value is a moment, so this example shows how a moment can be defined for a probability distribution.
The expected value is the population analogue of the average of the data, and may also be called the population mean. The notion of a population analogue of a statistic is very important. The expected value represents the exact, true mean of a distribution. When we conduct a study, we collect data, but never observe the population. Thus, we would never know that 3.7 is the population mean. We can calculate the mean of our data (often called the sample mean to distinguish it from the population mean). For reasons to be discussed later, the sample mean will generally be close to the population mean, thus it is reasonable to argue that we have learned something about the population mean by calculating the sample mean. This is the process of generalization (generalizing from sample to population), which is central to all statistical data analyses.
Previously we have discussed the role of transformations in data analysis. Transformations can also be applied to populations. If we have a probability distribution for a random variable X, it is very easy to calculate the probability distribution for a transformation of X. For example, the probability distribution of \(Y=X^2\) is
y | 1 | 4 | 16 | 49 |
---|---|---|---|---|
p(Y=y) | 0.1 | 0.3 | 0.4 | 0.2 |
Note that we have applied the transformation (squaring) to each point in the sample space, and left the probabilities unchanged. In some cases, transforming a random variable may decrease the size of its sample space. For example, the distribution of the random variable \(Y=(X-3)^2\) is
x | 1 | 4 | 16 |
---|---|---|---|
p(X=x) | 0.7 | 0.1 | 0.2 |
A very important transformation is the centering transformation, which subtracts the population mean from each point in the sample space. The distribution of \(Z = X - EX\) , the centered version of X, is
z | -2.7 | -0.7 | 1.3 | 3.3 |
---|---|---|---|---|
p(Z=z) | 0.1 | 0.3 | 0.4 | 0.2 |
The values of \(X - EX\) are deviations from the mean, which play a very important role in many statistical analyses. Note that the expected value of the deviations from the mean is zero: EZ = 0.1·(-2.7) +0.3·(-0.7) +0.3·0.4 + 0.2·3.3 = 0.
We have previously discussed the variance and standard deviation as summary statistics of data, used to measure dispersion. The population analogue of the variance is \(E[(X - EX)^2]\) . This value tells us the expected squared deviation from the mean. It is a population-level measure of dispersion, since it will be large for distributions whose values often fall quite far from the expected value. Above we constructed the distribution of \(X - EX\) , now we can construct the distribution of \(Q=(X - EX)^2\) :
q | 7.29 | 0.49 | 1.69 | 10.89 |
---|---|---|---|---|
p(Q=q) | 0.1 | 0.3 | 0.4 | 0.2 |
Finally, we can calculate the variance of X, denoted Var[X], by taking the expected value of the above distribution: 0.1·7.29 + 0.3·0.49 + 0.4·1.69 + 0.2·10.89 = 3.73.
All of the above calculations were based on the probability mass function (pmf). An alternative representation of a probability distribution that is useful for quantitative data is the cumulative distribution function (CDF). The CDF tabulates all of the left tail probabilities, \(P(X \le x)\) . Note that we have earlier discussed the notion of the left tail proportion based on data. The left tail probabilities are the population analogues of the left tail proportions. For the example we have been using here, the CDF is
x | 1 | 2 | 4 | 7 |
---|---|---|---|---|
p(X<=x) | 0.1 | 0.4 | 0.8 | 1 |
If we sort the sample space of \(X\) , the left tail probabilities \(P(X\le x)\) are obtained by cumulatively summing the probabilities from the left. These cumulative probabilities will be non-negative, non-decreasing, and the final cumulative probability will always be equal to 1.
Another summary statistic we discussed earlier in the course is the quantile (or percentile). Certain quantiles can be obtained directly from the CDF. In the example above, the 0.1 quantile is 1, the 0.4 quantile is 2, the 0.8 quantile is 4, and the 1 quantile is 7. There are various conventions about how to define other quantiles for this population, but in practice, it is not very useful to consider quantiles for probability distributions with very small sample spaces such as these.
Many distributions are best thought of as having a sample space that is infinite, possibly consisting of all or part of the real number line. Suppose that \(X\) is a random variable representing such a distribution, then the left tail probability \(P(X \le x)\) , which is the CDF, has a natural definition and interpretation. Taking a probability p between 0 and 1, the p’th quantile is the point q such that \(P(X \le q) = p\) .
The probabilities of specific points \(P(X = x)\) do not have a natural definition for many random variables with infinite sample spaces, and while paradoxical, these probabilities may be equal to zero for every value of x. A proper treatment of random variables with infinite sample spaces requires calculus, and will not be developed further here.
Joint distributions #
So far we have only considered a single random variable. In practice, we often want to consider multiple random variables that are observed together. A probability distribution in this setting is called a joint distribution. As above, we will focus on the case where the sample spaces are finite. Suppose we have two random variables \(X\) and \(Y\) that we observe together. Suppose further that the sample space of \(X\) is {1, 3} and the sample space of \(Y\) is {2, 4, 5}. Then the joint distribution of \((X, Y)\) takes the form of a table:
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.3 | 0.3 | 0.2 |
3 | 0.1 | 0 | 0.1 |
This table contains six probabilities which sum to 1. Each cell in the table tells us the probability of observing the given values of \(X\) and \(Y\) jointly. For example, the probability of observing \(X=1\) and \(Y=4\) jointly is 0.3. Some combinations may have zero probability, for example the probability here of observing \(X=3\) and \(Y=4\) jointly is zero – this combination can never happen.
Given a joint probability distribution, such as the one above, it is possible to define marginal distributions for the rows (X) and the columns (Y) by summing the probabilities over the columns or rows, respectively. Thus, the marginal distribution of X is
x | 1 | 3 |
---|---|---|
P(X=x) | 0.8 | 0.2 |
The value 0.8 is obtained by calculating 0.3 + 0.3 + 0.2, the sum of the first row of the joint distribution, and the value of 0.2 is obtained by calculating 0.1 + 0 + 0.1 = 0.2, the sum of the second row of the joint distribution.
The marginal distribution of Y is
y | 2 | 4 | 5 |
---|---|---|---|
P(Y=y) | 0.4 | 0.3 | 0.3 |
which is obtained by summing the columns of the joint table. Note that the marginal distributions of X and Y are each probability distributions in their own right, since their probabilities sum to 1.
Independence #
One of the most fundamental notions in probability is that of independence. Suppose we observe two random variables \(X\) and \(Y\) jointly. Informally, \(X\) and \(Y\) are independent if knowing \(X\) does not tell us anything about the value of \(Y\) , and vice versa. In our example above, when \(X=4\) , \(Y\) must be equal to 1, since the \((X=4, Y=3)\) cell of the table has probability equal to zero. Thus, \(X\) and \(Y\) cannot be independent in that example.
When two variables are not independent, they are dependent. In the real world, variables such as income and education would be dependent, since people with more education tend to have higher income, and vice versa. It is very important to note that dependence does not imply causality. Although income and education are statistically dependent (i.e. not statistically independent), this does not mean that having more education causes people to have a higher income, or that having a higher income causes people to have more education. Either, both, or neither of these statements could be true.
Mathematically, we have perfect independence if and only if every joint probability is the product of the corresponding marginal probabilities. The following joint probability table has the same marginal probabilities as in the previous example. But unlike in the previous example, in this example, we can confirm that the random variables X and Y are independent. All six of the cell probabilities in the table below are equal to the product of their corresponding marginal probabilities, e.g. 0.32 = 0.8·0.4, and 0.24 = 0.8·0.3.
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.32 | 0.24 | 0.24 |
3 | 0.08 | 0.06 | 0.06 |
Conditional distributions #
Previously we have discussed the idea of stratifying data to better understand its properties. For example, when considering household incomes in the United States, we might stratify by state, or by demographic variables such as sex and age. A conditional distribution is a distribution obtained by starting with a joint distribution, and renormalizing it so that each row or each column sums to 1. For example, starting with this joint distribution,
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.3 | 0.3 | 0.2 |
3 | 0.1 | 0 | 0.1 |
we can divide each row by its total and obtain the conditional distributions of “ \(Y\) given \(X\) ”:
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 3/8 | 3/8 | 1/4 |
3 | 1/2 | 0 | 1/2 |
We would write these probabilities as, e.g. \(P(Y=4 | X=1) = 3/8\) . The symbol “|” is read “given”, or “conditioned on”. The meaning of this probability is that if we know that \(X\) is equal to 1, then the probability of observing \(Y\) to be equal to 4 would be 3/8. Note that each row of the above conditional probability table sums to 1, whereas the overall sum of all cells in the joint table is equal to 1.
We can also construct the conditional distributions of \(X\) given \(Y\) , by normalizing the columns:
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 3/4 | 1 | 2/3 |
3 | 1/4 | 0 | 1/3 |
These probabilities are denoted, e.g. \(P(X=1 | Y=2) = 3/4\) , and represent the probability of observing \(X\) to be equal to 1, when \(Y\) is known to be equal to 2.
The table considered above describes the joint distribution of two random variables \(X\) and \(Y\) that are not independent. Next let’s consider the table
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.32 | 0.24 | 0.24 |
3 | 0.08 | 0.06 | 0.06 |
for which the rows ( \(X\) ) and columns ( \(Y\) ) are independent. The conditional distributions of \(Y\) given \(X\) are
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.4 | 0.3 | 0.3 |
3 | 0.4 | 0.3 | 0.3 |
and the conditional distributions of \(X\) given \(Y\) are
X\Y | 2 | 4 | 5 |
---|---|---|---|
1 | 0.8 | 0.8 | 0.8 |
3 | 0.2 | 0.2 | 0.2 |
Note that \(P(X=x | Y=y)\) does not depend on \(y\) , and \(P(Y=y | X=x)\) does not depend on \(X\) . In other words, the conditional distribution of \(X\) given \(Y\) is the same regardless of what value of \(Y\) we condition on, and the conditional distribution of \(Y\) given \(X\) is the same regardless of what value of \(X\) we condition on. This happens if and only if \(X\) and \(Y\) are independent.
Conditional independence #
Many research questions hinge on the inter-relationship among three variables, which we will denote generically here by \(X\) , \(Y\) , and \(Z\) . In this setting, we may wish to combine the notions of conditioning and independence, leading to the concept of conditional independence. The basic idea of conditional independence is that two variables, say \(X\) and \(Y\) , may be dependent, but when we condition on a third variable \(Z\) , then \(X\) and \(Y\) may become independent.
As an example, consider the relationships among tea drinking, smoking, and cancer. Each of these variables can be considered to be a binary trait of a person (i.e. a person either drinks tea or does not, etc.). Since there are three variables, each with two levels, there are \(2^3=8\) combinations. The joint probabilities associated with these variables are presented in the following joint three-way table:
Tea | Smoke | Cancer | Probability |
---|---|---|---|
Y | Y | Y | 0.0033 |
Y | Y | N | 0.1080 |
Y | N | Y | 0.0049 |
Y | N | N | 0.4838 |
N | Y | Y | 0.0027 |
N | Y | N | 0.0860 |
N | N | Y | 0.0031 |
N | N | N | 0.3082 |
We can derive several marginal distributions from this joint three-way distribution. For example, the marginal distribution of tea drinking and cancer is
Cancer | No cancer | |
---|---|---|
Tea | 0.0082 | 0.5918 |
No tea | 0.0058 | 0.3942 |
Each value in this marginal tables is obtained by summing two values in the joint three-way table above. For example, the probability of drinking tea and not having cancer is the sum of the probabilities of drinking tea and not having cancer for smokers and for non-smokers, i.e. 0.5918 = 0.1080 + 0.4838.
There are actually three marginal two-way tables that can be constructed from the joint three-way table. Besides the joint distribution of tea drinking and cancer status presented here, there are also marginal distributions for tea drinking and smoking, and for smoking and cancer status, which is shown here:
Cancer | No cancer | |
---|---|---|
Smoke | 0.006 | 0.194 |
No smoke | 0.008 | 0.792 |
We can take this a step further, and obtain the marginal distribution for a single variable, say smoking. The marginal probability of smoking is the sum of the four cells in the joint three-way table corresponding to smokers: 0.0033 + 0.1080 + 0.0027 + 0.0860 = 0.2. This tells us that marginally (ignoring tea drinking and cancer), 20% of this population smokes. We could do the corresponding calculation for non-smokers, but since there are only two possible states for smoking, we immediately know that the probability of someone not smoking in this population is 1 - 0.2 = 0.8.
Are tea drinking and cancer independent or dependent? To assess this, we can convert the joint distribution between tea drinking and cancer, presented above, to a conditional distribution, conditioning on tea drinking. We do this by dividing the values in each row by the row total:
Cancer | No cancer | |
---|---|---|
Tea | 0.0137 | 0.9863 |
No tea | 0.0144 | 0.9856 |
Since the two rows of the above table are different, it follows that tea drinking and cancer are dependent. People who drink tea in this population have a slightly lower risk of having cancer than people who do not drink tea. However this may not be the full story, as we see next.
The idea of conditional independence is that when we hold one variable fixed, two other variables that were marginally dependent may become conditionally independent. If we condition on people not smoking, we get the following conditional distribution for tea drinking and cancer:
Cancer | No cancer | |
---|---|---|
Tea | 0.0061 | 0.6047 |
No tea | 0.0039 | 0.3853 |
The above distribution is obtained by taking the four probabilities in the joint three-way table corresponding to non-smokers, and dividing them by the marginal probability that someone does not smoke, which as determined above is 0.8. For example the conditional probability of a non-smoker being a tea-drinker without cancer is 0.4838 / 0.8 ~ 0.6048 (the last digit deviates due to rounding).
Are tea drinking and cancer dependent in this conditional distribution? To assess this, we condition the above table on tea drinking. This amounts to dividing each row by the row total, yielding
Cancer | No cancer | |
---|---|---|
Tea | 0.01 | 0.99 |
No tea | 0.01 | 0.99 |
This result shows us that among non-smokers, both tea drinkers and people who do not drink tea have the same risk of having cancer. If we were to do this calculation conditioning on people smoking rather than conditioning on people not smoking, then we would also find no dependence between tea drinking and cancer.
The explanation for what we are seeing here is that smoking is a confounder of the relationship between tea drinking and cancer. People who smoke are substantially more likely to get cancer than people who do not smoke, and people who smoke are somewhat less likely to drink tea than people who do not smoke. If we do not consider smoking, then a relationship appears between tea drinking and smoking, but when we then condition on smoking behavior, this relationship disappears. In other words, tea drinking and cancer are marginally dependent, but conditionally (given smoking) they are independent.