Case Study 2 — When Does an Average Exist? The St. Petersburg Paradox and Expected Value as a Series

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Case Study 2 — When Does an Average Exist? The St. Petersburg Paradox and Expected Value as a Series

Field: Probability and mathematical economics Calculus used: Divergence test (§22.1, Step 0 of §22.10), ratio test (§22.5), limit comparison (§22.4), the boundary $p$-series (§22.2.1)

A game that should be worth a fortune

A casino offers you a game. It flips a fair coin until the first heads appears. If heads comes on the first flip you win \$2; if it first appears on the second flip you win \$4; on the third, \$8; and in general, if the first heads is on flip $n$, you win $2^{n}$ dollars. The probability that the first heads lands exactly on flip $n$ is $(\tfrac12)^{n} = 2^{-n}$ (you need $n-1$ tails then a head). How much should you pay to play once?

The principled answer from decision theory is the expected value — the probability-weighted average payoff — and that average is an infinite series: $$E[\text{payoff}] = \sum_{n=1}^\infty (\text{payoff on flip }n)\cdot P(\text{first head on flip }n) = \sum_{n=1}^\infty 2^{n}\cdot 2^{-n}.$$ Here is where Chapter 22 walks onto the stage. Simplify the term: $a_n = 2^n \cdot 2^{-n} = 1$. So $$E[\text{payoff}] = \sum_{n=1}^\infty 1 = 1 + 1 + 1 + \cdots.$$ By the very first line of the decision framework — Step 0, the divergence test (§22.10) — since $a_n = 1 \not\to 0$, this series diverges. The expected payoff is infinite. A rational expected-value maximizer should be willing to pay any finite price — a thousand dollars, a million — to play a single round. This is the St. Petersburg paradox, posed by Nicolaus Bernoulli in 1713: no sane person would pay even \$20, yet the mathematics of expectation says the game is worth more than everything you own. The paradox is not a flaw in the arithmetic; the series genuinely diverges. It is a flaw in using a raw expected value when that value is the sum of a divergent series — and the divergence test is what exposes it in one glance.

Why each term is exactly $1$ — and why that is the whole story

It is worth lingering on the structure, because it is a perfect specimen for the chapter's central skill: read the shape, name the test. The payoff $2^n$ grows geometrically; the probability $2^{-n}$ shrinks geometrically at exactly the matching rate. Their product is constant. Whenever a reward grows at the same exponential rate that its probability decays, the expected-value terms do not shrink, and the divergence test kills the average instantly. Notice we needed none of the heavy machinery — no ratio test, no integral test. The framework's Step 0 exists precisely so you never reach for a sophisticated tool when the terms fail to approach zero. (Contrast the temptation: a student who jumped straight to the ratio test would find $a_{n+1}/a_n = 1 \to 1$, the inconclusive case, and learn nothing. Step 0 first, always.)

Fixing the game: when does the average come back to life?

The paradox dissolves the moment we change the payoff so the reward no longer keeps pace with the vanishing probability. Suppose the casino caps the growth: it pays $n^{2}$ dollars (instead of $2^n$) when the first head appears on flip $n$. Now $$E = \sum_{n=1}^\infty n^{2}\cdot 2^{-n} = \sum_{n=1}^\infty \frac{n^{2}}{2^{n}}.$$ The terms here do tend to zero (Step 0 passes — an exponential in the denominator beats a polynomial in the numerator), so we proceed. There is a factorial-or-exponential structure, which is Step 3 of the framework: reach for the ratio test (§22.5). $$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^{n}}{n^2} = \frac{1}{2}\left(1+\frac1n\right)^{2} \longrightarrow \frac12 = \rho < 1.$$ Since $\rho = \tfrac12 < 1$, the series converges absolutely, and the expected payoff is a finite number. (In fact $\sum_{n\ge1} n^2 x^n = x(1+x)/(1-x)^3$ for $|x|<1$; at $x=\tfrac12$ this gives $E = 6$, so a fair price for the capped game is exactly \$6 — a perfectly reasonable ante.) The lesson generalizes: an expected value exists precisely when its defining series converges, and the convergence test you choose is dictated by how the payoff competes with the probability tail.

The boundary case, where the test matters most

The interesting territory is the boundary, where convergence is delicate and the choice of test is everything. Consider a heavy-tailed reward model common in insurance and finance: the probability of a loss of size $n$ decays like a power law, $P(N=n) = C/n^{3}$ (the constant $C$ makes the probabilities sum to $1$; this is legitimate because $\sum 1/n^3$ converges by the $p$-series rule, §22.2.1). The expected loss is $$E[N] = \sum_{n=1}^\infty n\cdot\frac{C}{n^{3}} = C\sum_{n=1}^\infty \frac{1}{n^{2}}.$$ This is a $p$-series with $p = 2 > 1$ — it converges (in fact to $C\pi^2/6$), so the mean loss is finite. But now ask about the variance, which involves $E[N^2]$: $$E[N^{2}] = \sum_{n=1}^\infty n^{2}\cdot\frac{C}{n^{3}} = C\sum_{n=1}^\infty \frac{1}{n} = \infty.$$ That is the harmonic series — $p = 1$, the borderline case that diverges (§22.2.1). So this distribution has a finite mean but infinite variance. The consequence is not academic: standard statistical tools (confidence intervals, the central limit theorem in its usual form, ordinary least squares) silently assume finite variance. An analyst who never checked the convergence of $\sum n^2 P(N=n)$ would apply those tools to data where they are invalid, and would be badly surprised when sample variances refuse to stabilize no matter how much data arrives. The single exponent crossing from $p=2$ (mean) to $p=1$ (variance) is the line between "use the textbook" and "the textbook does not apply" — and the $p$-series rule draws that line exactly.

For payoff structures that are not clean powers — say $P(N=n)\sim C\,n/(n^{3}+5)$ — the right instrument is limit comparison (§22.4): the term behaves like $C/n^2$, so $E[N]=\sum nP(N=n)$ behaves like $\sum 1/n$ and the mean diverges, while a slightly lighter tail $\sim C/n^{4}$ would give $\sum 1/n^2$ and a finite mean. Limit comparison lets you read the verdict off the leading-order decay without fighting the exact algebra — exactly the workhorse role §22.4 advertised.

Putting it together

# Expected value = a SERIES. Whether the average "exists" is a convergence question.
# (do NOT need to run; outputs are hand-computed)

# St. Petersburg: payoff 2^n, prob 2^-n  ->  terms a_n = 1, DIVERGES (divergence test)
#   partial means 1, 2, 3, 4, ...  -> infinity   (no finite fair price)

# Capped game: payoff n^2, prob 2^-n  ->  sum n^2/2^n, ratio rho = 1/2 < 1, CONVERGES
#   exact value 6  -> fair price = $6

# Power-law loss P(N=n)=C/n^3:
#   mean   E[N]   = C * sum 1/n^2  (p=2 > 1)  -> FINITE
#   second E[N^2] = C * sum 1/n    (p=1)      -> INFINITE  => variance does not exist
print("St. Petersburg E:", "diverges (a_n = 1, divergence test)")
print("Capped game E    :", 6)
print("Power-law: mean finite (p=2), variance infinite (p=1)")

The code computes nothing the convergence tests have not already decided; it merely displays verdicts that §22.1–22.5 produced by hand. That is the point of the chapter in applied probability: before you can talk about an average, a variance, or a fair price, you must prove the series defining it converges — and the test you choose is dictated by how the payoff races against the probability tail.

Why this matters beyond a parlor game

The St. Petersburg paradox launched, in 1738, Daniel Bernoulli's resolution via expected utility: replace the raw payoff $2^n$ with its utility, say $\log(2^n) = n\log 2$, and the expected utility $\sum n\log 2\cdot 2^{-n}$ converges (ratio test, $\rho=\tfrac12$) to a finite, sensible number. Diminishing marginal utility — the idea that the millionth dollar is worth less than the first — is, at its mathematical core, the move that turns a divergent expectation into a convergent one. The same convergence question reappears wherever rewards are heavy-tailed: pricing catastrophe insurance, modeling financial crashes with power-law return distributions, sizing bets under the Kelly criterion. In every case the practitioner is, knowingly or not, asking Chapter 22's question — does the series defining this average converge? — and answering it with these tests.

Discussion questions

In the original game, each term of the expectation equals $1$. Explain why the divergence test (§22.10, Step 0) settles the paradox without any of the other six tests, and why jumping to the ratio test would have been a wasted move.
The capped game ($n^2$ payoff) converges with $\rho = \tfrac12$. Replace the payoff with $n!$ dollars. Does the expected value exist? Apply the ratio test and interpret the result as a statement about the game.
For the power-law loss $P(N=n)=C/n^3$, the mean exists but the variance does not. State the general rule: for $P(N=n)\sim C/n^{s}$, for which $s$ does the mean exist? The variance? Use the $p$-series rule.
Bernoulli's utility fix uses $\log(\text{payoff})$. Verify that $\sum n\log 2\cdot 2^{-n}$ converges and name the test. Why does taking a logarithm "tame" the divergence?

Your turn

A game pays $3^{n}$ with probability $2^{-n}$. Compute the term $a_n$, apply the divergence test, and decide whether any finite fair price exists.
A game pays $n$ dollars with probability $2^{-n}$. Show $\sum n\,2^{-n}$ converges (ratio test) and identify $\rho$. (Its sum is $2$, a fair price of \$2.)
For a payoff $P(N=n)\sim C/n^{2.5}$, decide by limit comparison whether the mean and the variance exist, and state the practical consequence for using the central limit theorem on samples from this distribution.

Case Study 2 — When Does an Average Exist? The St. Petersburg Paradox and Expected Value as a Series

A game that should be worth a fortune

Why each term is exactly $1$ — and why that is the whole story

Fixing the game: when does the average come back to life?

The boundary case, where the test matters most

Putting it together

Why this matters beyond a parlor game

Discussion questions

Your turn

Further reading