Continuous Random Variables

\[\newcommand{\P}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\V}{\mathbb{V}}\]

In this set of notes, we introduce the fundamentals of continuous random variables and their distributions. At a high level, continuous RVs are much like their discrete counterparts - we can compute expectations and variances, compute probabilities of events, see how they behave under transformations. They differ in one key way however: they do not have a probability mass function. Continuous RVs are instead characterized by a probability density function. If you take an “expectations first” view of probability theory, this is not a ground-shaking change; if you instead take a “distributions first” view, the changes are a bit more obvious. Regardless, we begin by recalling our discussion of “events” in probability.

Events and Sigma Algebras

Recall that in our earliest discussion of probability, we studiously avoided asking the probability of an outcome; we instead insisted on only assigning probabilities to events. At the time, I argued that events are easier to generalize: it is easier to have one uniform set of rules for events than to have two interlocking sets of rules for outcomes and events separately.¹ This is certainly true, but it’s now time to see the other advantage of event-thinking.

Formally, a probability space consists of three things:

$\Omega$, a set of possible outcomes
$\mathcal{E}$, a set of allowed events
$\P: \mathcal{E} \to [0, 1]$, a measure function that assigns a numerical value to every event $E \in \mathcal{E}$

The set of events $\mathcal{E}$ is required to have certain properties:

If $E \in \mathcal{E}$, the complement is as well: $E^c \in \mathcal{E}$.
$\emptyset, \Omega \in \mathcal{E}$
If $E_1, E_2, \dots$ are (a countable list of) elements of $\mathcal{E}$, so is their union: $\bigcup_{i=1}^{\infty} E_i \in \mathcal{E}$

Many - if not all - of the apparent paradoxes of probability theory are resolved by the set $\mathcal{E}$. $\mathcal{E}$ essentially defines the set of allowed questions we can ask of our probability measure. If a set is “too weird”, we simply exclude it from $\mathcal{E}$ and the problems go away. (Life is much easier in math-land.)

Recalling that $\mathcal{E}$ defines the “allowed questions”, let’s review our three rules:

If we are allowed to ask the probability of $E$, we must also be allowed to ask the probability of not $E$ ($E^c$).
We have to be able to ask whether anything happens ($\Omega \in \mathcal{E}$) or whether nothing happens ($\emptyset \in \mathcal{E}$). (Recall that we take $\P(\emptyset) = 0$ and $\P(\Omega) = 1$.)
If we are allowed to ask whether $E_1, E_2, \dots$ happen separately, we are also allowed to ask whether any of them, i.e., their union, happens.

The fundamental curiosity of continuous random variables is that we take out the singleton events $\{X = x\}$ from the set $\mathcal{E}$. That is, questions like “what is the probability that $X$ equals 3” are no longer valid in the land of continuous random variables.

Continuous Random Variables

Exact Equality is Ill-Posed

Suppose we seek to develop a probabilistic model for a ‘natural’ quantity, e.g., my height. You might seek to calculate the probability that I am 5’10”. But upon a bit of reflection, that’s actually not a well-posed question. It depends how accurately you are measuring:

If you are giving an estimated height to a public safety officer, 5’10” might be a very exact measurement.
If you are taking my height for my annual physical, 5’10” might seem to be ‘correct.’
If you pull out a tape measure, I may be closer to 5’10.625”. And
If you use lab grade equipment to measure me down to the final atom, there’s not even a real answer - my hair is constantly moving back and forth.²

With this perspective, let’s reconsider the question “What is the probability that Michael is exactly 5’10”?“. Depending on your philosophical inclinations, the answer is either 0 or the question is ill-defined. It is customary to introduce continuous random variables from the former view (”probability 0”); I prefer the latter (“invalid question”). Regardless, we need a better building block for probabilities of ‘natural’ events.

Cumulative Distribution Functions

In our discussion of discrete random variables, we emphasized the role of the probability mass function. But there was another, equivalent, function that characterizes a distribution: the cumulative distribution function (CDF).

\[ F_X(x) = \P(X \leq x) \]

For discrete distributions, we simply defined $F_X(\cdot)$ in terms of the PMF. But we can also take the CDF as our starting point. If we start from a CDF, the PMF of a discrete random variable is given by

\[ P(X = x) = F_X(x) - F_X(x-1) \quad \text{ for discrete } X \]

What does a CDF look like for a continuous random variable? In essence, nothing changes! We can still ask the question $\P(X \leq x)$ - formally, the set $(-\infty, x] \in \mathcal{E}$ - regardless of the continuity or discreteness of $X$. Given some reference height, we simply ask whether I am taller than it or not. We might mutter about needing more precise measuring tools to be sure - such is the concern of experimentalists - but at least the question is well-posed.

Continuous Random Variables

We are now ready to define continuous random variables. A random variable $X$ is continuous if its CDF $F_X(\cdot)$ satisfies the following:

$\lim_{x \to -\infty} F_X(x) = 0$
$\lim_{x \to +\infty} F_X(x) = 1$
$F_X(\cdot)$ is non-decreasing. ($x \leq y \implies F_X(x)\leq F_X(y)$)
$F_X(\cdot)$ is continuous and almost everywhere differentiable.

These rules may be more continuous than what we have seen so far, but they are mostly restatements of our basic probability properties.

$F(-\infty) = 0$. What is the chance that $X$ is less than $-\infty$? Zero of course!
$F(+\infty) = 1$. What is the chance that $X$ is less than $+\infty$? 100%.³
$F_X(\cdot)$ is non-decreasing. Of course! Since $\{X \leq x\} \subseteq \{X \leq y\}$, we have to have $\P\{X \leq x\} \leq \P\{X \leq y\} \implies F_X(x) \leq F_X(y)$.

The final assumption - about continuity and differentiability - is a bit more technical, but it lets us avoid weird pathologies. In brief, we’re just enforcing the rule that there are no “jumps” in $F_X(\cdot)$. Such “jumps” are indicative of discrete, not continuous, behavior.⁴

We now have a cleaner - and perhaps simpler - categorization of random variables. All random variables are defined by their CDF. If the CDF of a random variable $X$ is a “step-function”, $X$ is a discrete random variable; if the CDF of a random variable $X$ is a continuous function, $X$ is a continuous random variable.

Probability Densities

Now suppose we are interested in some continuous random variable $X$. Because $X$ is continuous, $F_X(\cdot)$ is continuous and has a derivative (almost everywhere). This derivative turns out to be useful enough to get its own name:

\[\frac{\text{d}F_X}{\text{d}x}(x) = F_X'(x) \text { is the \emph{probability density function} of $X$}\]

The probability density function (PDF) is the closest analogue a continuous RV has to a PMF.

Before we dive into the use of PDFs, let’s see what properties we can derive.

First, we note that, because $F_X(\cdot)$ is non-decreasing, its derivative $f_X(\cdot)$ is non-negative. So both PDFs and PMFs are non-negative.
Next, we apply the fundamental theorem of calculus to $F_X(\cdot)$ and $f_X(\cdot)$.

\[ F_X(a) = \int_{-\infty}^a F'_X(x)\,\text{d}x = \int_{-\infty}^a f_X(x)\,\text{d}x = \P(X \leq a)\]

That is, we can compute the CDF by integrating the PDF.
By subtraction, we can also use the PDF to compute probabilities of intervals.

\[\begin{align*} \P(a \leq X \leq b) &= \P(X \leq b) - \P(X \leq a) \\ &= \int_{-\infty}^b f_X(x)\,\text{d}x - \int_{-\infty}^a f_X(x)\,\text{d}x \\ &= \int_{a}^b f_X(x)\,\text{d}x \end{align*}\]

Of course, if we take $a = b$ and attempt to compute $\P(a \leq X \leq a) = \P(X = a)$, we get $\int_a^a f_X(x)\,\text{d}x$. By convention, this is taken to be 0, consistent with our argument that “exact equality” must have probability 0 if we insist on putting a probability on it.⁵

Expectations and Moments

If you stare at the discussion above for a bit, you might notice an interesting way to rewrite our key integrals:

\[\P(a \leq X \leq b) = \int_a^b f_X(x)\, \text{d}x = \int_{-\infty}^{\infty} 1_{x \in [a, b]}(x)\, f_X(x)\,\text{d}x\] That is, rather than using bounds on our integral, we can use an indicator function. Because the indicator function is 0 outside of the range of interest, this implicitly focuses the integral only on the interval $[a, b]$ without requiring us to keep track of bounds. We also recall that the probability of any event can be written as the expectation of the relevant indicator function:

\[ \E[1_{X \in [a, b]}] = \int_{-\infty}^{\infty} 1_{x \in [a, b]}(x)\,f_X(x)\,\text{d}x\]

From this, it’s not too big a leap to argue that expectations in general can be written as:

\[ \E[h(X)] = \int_{-\infty}^{\infty} h(x) f(x)\,\text{d}x \]

Indeed, we take the case of $h(x) = x$ to define the “plain” expectation:

\[ \E[X] = \int_{-\infty}^{\infty} x\, f(x)\,\text{d}x \]

Stepping back, we see that - as far as expectations are concerned - the only shift from discrete to continuous is to replace the sum with an integral. If we recall the standard motivation of the integral as a “infinite sum of infinitely small things”, this doesn’t feel too implausible.

Tip

For practice, convince yourself that all our formulas for expectations and variances hold in the continuous case as well.

Conditional Distributions of Continuous Random Variables

The “conditioning” rules for continuous random variables parallel their discrete counterparts:

\[ f_{Y | X}(y, x) = \frac{f_{Y, X}(y, x)}{f_X(x)} \]

Because of this, the rules of conditional expectations and variances translate to the continuous case with only the standard “sum to integral” substitution.

Mixture Distributions and Classification

A particularly important family of discrete-continuous distributions is the set of mixtures. Given a Bernoulli RV $Z$ and two continuous RVs $X_1, X_2$, the derived RV

\[ X = Z X_1 + (1-Z) X_2 \]

can be seen to be continuous as well, despite its ‘discrete input’ Z. (Why?)

Such distributions arise commonly in the modeling of classification problems.

We can derive the CDF of $X$ quite easily:

\[\begin{align*} F_X(x) &= \P(X \leq x) \\ &= \P(Z X_1 + (1-Z) X_2 \leq x) \\ &= \P(Z X_1 + (1-Z) X_2 \leq x | Z = 0)\P(Z = 0) + \P(Z X_1 + (1-Z) X_2 \leq x | Z = 1)\P(Z = 1)\\ &= \P(X_2 \leq x)\P(Z = 0) + \P(X_1\leq x)\P(Z = 1)\\ &= F_{X_2}(x)(1-p) + F_{X_1}(x)p \end{align*}\]

It’s not hard to show that this defines a valid CDF. Differentiating with respect to $x$, we can also see that the PDF is given by:

\[ f_X(x) = f_{X_1}(x) * p + f_{X_2}(x) * (1-p) \]

Frequently, we will observe $X$ and want to know whether $Z = 0$ or $Z = 1$? If the supports of $X_1, X_2$ are overlapping, we cannot be certain, but we can use Bayes’ rule:

\[\begin{align*} \P(Z = 0 | X = x) &= \frac{\P(X = x | Z = 0)\P(Z = 0)}{\P(X = x)} \end{align*}\]

Here, we can mix PMFs and PDFs at will within our Bayes’ rule calculation.⁶ This gives us:

\[\begin{align*} \P(Z = 0 | X = x) &= \frac{\P(X = x | Z = 0)\P(Z = 0)}{\P(X = x)} \\ &= \frac{f_X(X = x | Z = 0)\P(Z = 0)}{f_X(X = x)} \\ &= \frac{f_{X_1}(x) * p}{f_{X_1}(x) * p + f_{X_2}(x) * (1-p)} \end{align*}\]

This type of rule gives rise to a generative classifier in Machine Learning. I have also recently used this type of logic to give probabilities of certain climate drivers giving observed data.

Important Continuous Distributions

Unlike discrete random variables - which can mainly be built out of increasingly arcane combinations of Bernoulli random variables -

Continuous Uniform Distribution

Footnotes

We let ourselves be a bit sloppier and talk about outcomes for random variables, $\P(X = x)$. If we had been properly rigorous throughout, we should have spoken only of the “atomic” event $\P(\{X = x\})$, but it’s customary to equate a single outcome with the event containing only that outcome.↩︎
You might recognize this discussion as having a “how many angels can dance on the head of a pin?” flavor. You would not be wrong to note such a similarity.↩︎
In this course, we take random variables to be real-valued and disallow $\pm\infty$ as valid values. It’s not too much harder to allow extended real-valued variables, but it’s generally not worth the effort.↩︎
It is of course possible to have mixed discrete-continuous random variables. These are most commonly needed in survival analysis. Consider, e.g., how long a patient survives in a medical trial of 5 years. All amounts between 0 and 5 are valid (continuous) but there’s also a “chunk” of probability at “5 or more”. Here the underlying measurement (future lifetime) is arguably continuous, but our observations thereof have a mixed discrete-continuous flavor. In this course, we’ll mainly stay away from this sort of mixed discrete-continuous structure.↩︎
This is perhaps as good a place to note that we essentially always choose $\mathcal{E}$ to contain all intervals of the form $(-\infty, x]$ so that $F_X(\cdot)$ can always be computed. The smallest $\mathcal{E}$ that allows this is the so-called “Borel $\sigma$-algebra on $\mathbb{R}$ and it is the nigh-universal default in the study of random variables. Now that you have read this footnote, you can immediately forget it, because you will never encounter a fundamentally different choice of $\mathcal{E}$ in this course.↩︎
It’s actually a bit hard to show that this works formally, but we don’t worry about such things in this course.↩︎