Lecture notes. Written from the material of MIT 6.041SC, licensed CC BY-NC-SA 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

5. Random Variables, PMFs, and Expectation¶

What this covers¶

This chapter opens the study of random variables, the central object for the rest of the course. It answers three questions: what a random variable is, formally; how its possible values and their likelihoods are recorded by a probability mass function (PMF); and how a random variable is summarized by a single typical value (its expectation) and a measure of how spread out it is (its variance). It builds directly on the sample-space, event and counting machinery of the earlier lectures — in particular the coin-tossing and binomial-coefficient counting argument — and treats only the discrete case; continuous random variables are taken up in a later lecture.

Random variables¶

A random variable is a function from the sample space $\Omega$ to the real numbers. It assigns a number to every possible outcome of an experiment. That is the whole definition — despite the name, a random variable is neither random nor a variable in the algebraic sense; it is an ordinary function, and all the randomness lives in which outcome $\omega \in \Omega$ occurs.

Standard notation keeps the function and its values apart: a capital letter such as $X$ denotes the random variable — the function itself — and the corresponding lowercase letter $x$ denotes a particular numerical value it can take. Concretely: pick a random student from the class (the sample space $\Omega$) and measure their height $H$ in inches. $H$ is the function; a specific student's height, say $60$ or $71$, is a value $h$ that $H$ takes. It helps to think of $H$ as a subroutine: it takes a student as input, puts them on a scale, and returns a number as output. Fix a student and the number is fixed; before that, all you have is the function.

A single experiment can carry several random variables at once. The same random student also has a weight, so weight $W$ is a second function on the same $\Omega$ — unrelated in form to $H$, but defined on the same underlying outcomes.

A function of a random variable is again a random variable. If $\bar H = cH$ converts height in inches to height in centimeters, for some fixed conversion constant $c$, then $\bar H$ is itself a function $\Omega \to \mathbb{R}$: once the outcome (the student) is fixed, $H$ is determined, and so is $\bar H$. This point recurs constantly through the course: for any real function $g$, $g(X)$ is a new random variable on the same sample space.

Random variables can be discrete (finitely or countably many possible values — height rounded to the nearest inch) or continuous (a continuum of possible values — height measured with unlimited precision). The course covers discrete random variables first, because the ideas are the same but the bookkeeping is simpler, and returns to continuous random variables later once the concepts here — PMF, expectation, variance — are secure.

The probability mass function¶

The probability mass function (PMF) of a discrete random variable $X$ — also called its probability law or probability distribution — records how likely each value is:

\[p_X(x) = P(X = x) = P\big(\{\omega \in \Omega : X(\omega) = x\}\big).\]

That is, $p_X(x)$ is the total probability of all outcomes that make $X$ equal to $x$. Two properties follow immediately from the probability axioms:

\[p_X(x) \ge 0 \ \text{ for every } x, \qquad \sum_x p_X(x) = 1,\]

the second because every outcome maps to some value of $x$, and the events $\{X = x\}$ for different $x$ partition $\Omega$.

Computing a PMF is always the same three-step procedure: for each candidate value $x$, collect every outcome $\omega$ with $X(\omega) = x$, add up their probabilities, and repeat for every $x$.

Example: the geometric PMF¶

Flip a coin, independently, with $P(H) = p > 0$ on each toss, until the first head appears, and let $X$ be the toss number on which that first head occurs. The event $\{X = k\}$ happens in exactly one way: the first $k - 1$ tosses are tails and the $k$-th is heads. By independence,

\[p_X(k) = P(\underbrace{TT\cdots T}_{k-1}H) = (1-p)^{k-1}p, \qquad k = 1, 2, \dots\]

This is the geometric PMF. Plotted as a bar graph it starts at $p$ when $k = 1$, and each successive bar is smaller than the last by a factor of $1-p$ — the shape of a geometric sequence, hence the name, and $X$ is called a geometric random variable. This case was easy because only one outcome produced each value of $X$; in general many outcomes share a value, and all of their probabilities must be added together.

Example: the minimum of two die rolls¶

Roll a fair tetrahedral die (faces $1$–$4$) twice, independently. Let $F$ and $S$ be the first and second roll — themselves random variables — and let $X = \min(F, S)$. There are $16$ equally likely outcomes $(F, S)$, each with probability $1/16$. To find $p_X(2)$, collect every outcome where the smaller of the two rolls is $2$:

The 16 equally likely outcomes of two independent tetrahedral die rolls. The five shaded cells are exactly the outcomes where $\min(F,S)=2$, so $p_X(2) = 5/16$.

There are five such outcomes — $(2,2), (2,3), (2,4), (3,2), (4,2)$ — so

\[p_X(2) = \frac{5}{16}.\]

The rest of the PMF, $p_X(1)$, $p_X(3)$, $p_X(4)$, is found the same way.

Example: the binomial PMF¶

Toss a coin $n$ times, independently, with $P(H) = p$ on each toss, and let $X$ be the total number of heads — the same counting problem as an earlier lecture, now written in PMF notation. For $n = 4$, the event $\{X = 2\}$ is the union of the six equally-probable arrangements of two heads and two tails ($HHTT, HTHT, HTTH, THHT, THTH, TTHH$), each with probability $p^2(1-p)^2$, so

\[p_X(2) = 6\,p^2(1-p)^2 = \binom{4}{2}p^2(1-p)^2.\]

The $6$ is exactly $\binom{4}{2}$, the number of ways to place $2$ heads among $4$ tosses, and in general

\[p_X(k) = \binom{n}{k}p^k(1-p)^{n-k}, \qquad k = 0, 1, \dots, n.\]

This is the binomial PMF. Plotted for large $n$, it takes on a bell-shaped curve concentrated around the middle values of $k$, with very few heads or very many heads both unlikely — a first glimpse of a fact (the emergence of that bell shape) the course proves properly much later; for now it is only worth noticing.

Expectation¶

The expectation (or expected value, or mean) of a discrete random variable is

\[E[X] = \sum_x x\, p_X(x).\]

There are two useful ways to read this number. First, as a long-run average: if probabilities are interpreted as the frequencies with which values occur over many independent repetitions of the experiment, $E[X]$ is the average payoff per repetition. Second, and often faster for a quick answer, as a center of gravity: place a mass $p_X(x)$ at position $x$ on the real line; $E[X]$ is the point where this system of masses balances.

The expectation is the point where the probability-weighted values balance, exactly like the center of gravity of masses $1/6$, $1/2$ and $1/3$ placed at $1$, $2$ and $4$.

For example, suppose a game pays $\$1$ with probability $1/6$, $\$2$ with probability $1/2$, and $\$4$ with probability $1/3$. Then

\[E[X] = 1\cdot\frac16 + 2\cdot\frac12 + 4\cdot\frac13 = 2.5,\]

matching the point where the diagram above balances.

The center-of-gravity picture also gives quick answers by symmetry. Let $X$ be uniform on $\{0, 1, \dots, n\}$, i.e. $p_X(x) = 1/(n+1)$ for each of these $n+1$ values. Rather than summing $\sum_{k=0}^n k/(n+1)$ term by term, notice the PMF is symmetric about the midpoint of its range, so the balance point — the expectation — must be that midpoint:

\[E[X] = \frac{n}{2}.\]

Whenever a PMF is symmetric about some point, that point is the expectation, with no calculation needed.

The expected value rule¶

Often what is wanted is not $E[X]$ itself but $E[Y]$ for some $Y = g(X)$ derived from it — say $Y = X^2$, or a payoff that is some function of the number of heads. One route is the hard way: find the PMF of $Y$ first, by grouping outcomes according to the value of $y$ they produce, then apply the definition,

\[E[Y] = \sum_y y\, p_Y(y).\]

The easy way skips finding $p_Y$ entirely:

\[E[g(X)] = \sum_x g(x)\, p_X(x).\]

Both formulas add up exactly the same quantity — the probability of each outcome times the payoff it produces — and differ only in how the terms are grouped: the first groups outcomes by the value of $y$ they share before multiplying by $y$; the second runs over values of $x$ directly, multiplying the probability of each $x$ by $g(x)$ without first bundling together outcomes that happen to produce the same $y$. Because it feels so natural, this fact is often assumed to hold "by definition" — it does not; it is a genuine claim that needs a proof (one of the exercises below), which is why it carries a name: the law of the unconscious statistician. It is the single most-used shortcut for computing expectations of derived quantities, because it lets you work entirely with the PMF of $X$ you already have, never with the PMF of $Y$.

A caution worth stating before using the rule at all: in general,

\[E[g(X)] \ne g(E[X]).\]

Averaging and applying a function do not commute, and the course will produce many examples where they disagree. The one broad exception is when $g$ is linear. For constants $\alpha, \beta$:

\[E[\alpha] = \alpha, \qquad E[\alpha X] = \alpha E[X], \qquad E[\alpha X + \beta] = \alpha E[X] + \beta.\]

The first says that a "random variable" which always equals $\alpha$ — a degenerate case with no actual randomness — has expectation $\alpha$: the only term in the sum is $\alpha$ itself, with probability $1$. The second follows from the expected value rule applied to $g(x) = \alpha x$:

\[E[\alpha X] = \sum_x (\alpha x)\, p_X(x) = \alpha \sum_x x\, p_X(x) = \alpha E[X],\]

and combining the two gives the third. Concretely: if $H$ is height in inches and $\bar H = cH$ is height in centimeters, then $E[\bar H] = c\, E[H]$ — converting every student's height by the same factor converts the average height by the same factor, exactly as intuition demands. This is the one case where "reasoning on the average" is legitimate; the general caution above is the rule, and linearity is the exception.

Variance¶

Two more expectations are worth naming. The second moment of $X$ is $E[X^2] = \sum_x x^2\, p_X(x)$, an instance of the expected value rule with $g(x) = x^2$. More useful is the expectation of the squared distance from the mean,

\[\operatorname{var}(X) = E\big[(X - E[X])^2\big] = \sum_x (x - E[X])^2\, p_X(x),\]

called the variance of $X$. The quantity $(X - E[X])^2$ inside the outer expectation is itself a random variable — a number is subtracted from $X$, and the result is squared — so this too is just an application of the expected value rule, now to $g(x) = (x - E[X])^2$. Variance measures how spread out the PMF is around its mean, with the squaring giving extra weight to values far from the mean: a small variance means outcomes cluster tightly around $E[X]$; a large variance means the bars of the PMF reach far to either side.

A shortcut formula, easier to compute with, is

\[\operatorname{var}(X) = E[X^2] - (E[X])^2\]

(its short derivation is one of the exercises below). Two basic properties:

\[\operatorname{var}(X) \ge 0\]

— immediate, since it is an average of non-negative quantities (squares) — and

\[\operatorname{var}(\alpha X + \beta) = \alpha^2 \operatorname{var}(X).\]

The plausibility of the second is worth seeing even before deriving it formally. Adding a constant $\beta$ shifts every value of $X$, but it shifts the mean by exactly the same amount, so the distance of any value from the mean — and hence the spread — is unchanged: $\operatorname{var}(X + \beta) = \operatorname{var}(X)$. Multiplying by $\alpha$ instead stretches both the values and the mean by $\alpha$, so a deviation that was $x - E[X]$ becomes $\alpha(x - E[X])$; squaring that deviation multiplies it by $\alpha^2$, which is where the square on $\alpha$ comes from. Combining the two: scaling by $\alpha$ and shifting by $\beta$ multiplies the variance by $\alpha^2$, and the shift has no effect on it at all.

Exercises¶

These are Recitation 5's problems, which work with exactly the material above.

(a) Derive the expected value rule (the law of the unconscious statistician): for $Y = g(X)$, show that $E[Y] = \sum_x g(x)\, p_X(x)$. (b) For $Y = aX + b$ with constants $a, b$, derive $E[Y] = aE[X] + b$ and $\operatorname{var}(Y) = a^2 \operatorname{var}(X)$. (c) Derive the shortcut formula $\operatorname{var}(X) = E[X^2] - (E[X])^2$.
A marksman takes $10$ shots at a target, hitting with probability $0.2$ on each shot, independently of all other shots. Let $X$ be the number of hits. (a) Find and sketch the PMF of $X$. (b) What is the probability that he scores no hits? (c) What is the probability that he scores more hits than misses? (d) Find $E[X]$ and $\operatorname{var}(X)$. (e) He must pay $\$3$ to enter the range and receives $\$2$ for every hit. Let $Y$ be his net profit. Find $E[Y]$ and $\operatorname{var}(Y)$. (f) Instead, suppose entry is free and his profit equals the square of his number of hits, $Z = X^2$. Find $E[Z]$.
Four buses carry $40$, $33$, $25$ and $50$ job-seeking students, respectively ($148$ students in total), to a job convention. One of the $148$ students is chosen at random; let $X$ be the number of students on that student's bus. Separately, one of the four bus drivers is chosen at random; let $Y$ be the number of students on that driver's bus. (a) Before calculating anything: which of $E[X]$ and $E[Y]$ do you expect to be larger? Explain your reasoning. (b) Compute $E[X]$ and $E[Y]$.
St. Petersburg paradox. You toss a fair coin repeatedly and independently until it first comes up tails. If the first tails occurs on toss $n$, you are paid $2^n$ dollars. What is the expected amount you receive? How much would you actually be willing to pay to play this game once? (Recitation 5's accompanying handout simulates this game over 20 and 200 plays; the running average behaves very differently from the usual law-of-large-numbers picture, which is worth keeping in mind while answering the second question.)

Sources¶

Lecture 5 slides (lectures/05-slides.md): the definition of a random variable, the PMF and its two defining properties, the geometric-PMF example, the tetrahedral-die-minimum table, the binomial PMF, the definition of expectation, the properties of expectation, and the definition and properties of variance. This file is a model reconstruction of a PDF with no text layer, so every displayed equation should be treated as unverified against the original slide.
Lecture 5 transcript (recordings/lectures/05.md), by caption timestamp: random variables as functions, the height/weight example and the height-to-centimeters example ($00{:}00$–$11{:}04$); the PMF and the geometric-PMF derivation ($11{:}04$–$17{:}51$); the tetrahedral-die-minimum example ($17{:}51$–$20{:}00$); the binomial-PMF recap and the aside on the bell-shaped limiting curve ($21{:}02$–$24{:}20$); expectation, the frequency and center-of-gravity interpretations, the $\$1/\$2/\$4$ payoff example, and the uniform-symmetry argument ($24{:}20$–$30{:}59$); the expected value rule and the law of the unconscious statistician ($30{:}59$–$37{:}44$); linearity of expectation and the height-in-centimeters derivation ($37{:}44$–$42{:}24$); and variance, its shortcut formula, and the shift/scale argument for its properties ($42{:}24$–$50{:}15$).
Exercises: Recitation 5 slides (recitations/05-slides.md, dated September 23, 2010), Problems 1–4, and the accompanying extra handout's simulation note for Problem 4.
Not used: Problem Set 5 (psets/05-questions/, parts 01–03) and Tutorial 5 (tutorials/05-slides.md) both work with continuous random variables and joint PDFs — material this lecture does not reach — so none of their problems are reproduced above.
Referred to but not contained here: the lecturer points forward to the emergence of a bell-shaped (normal) curve for the binomial PMF at large $n$, promising a proof "in a couple of months"; that result is not covered in this lecture and is not reproduced beyond the observation itself.

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