Asymmetric payoffs and exposure

A forecast of an event and the payoff attached to that event answer different questions. Let $X$ be a real random variable with CDF $F$ and let $g(X)$ be its payoff. We use $g$ for the transformation, reserving $F$ for the CDF as in the shared notation. An ordinary finite expected payoff requires $\mathbb E|g(X)|<\infty$.

The frequency of gains does not determine their mean¶

Consider this finite payoff table, in arbitrary payoff units:

Direct summation gives $\mathbb E[A]=-0.62$ and $\mathbb E[B]=0.62$. The higher probability of a gain belongs to the lower expected payoff. Both payoffs are bounded; fat tails are unnecessary for this distinction. More generally, a payoff equal to $a>0$ with probability $1-q$ and to $-b<0$ with probability $q$ has mean $(1-q)a-qb$.

This is a one-period expectation calculation. A repeated strategy with a finite resource constraint needs a model of ruin as well.

Curvature and Jensen’s inequality¶

Suppose $X$ takes values in an open interval $I$, its mean $\mu=\mathbb E X$ belongs to $I$, and $\mathbb E|X|<\infty$. If $g:I\to\mathbb R$ is differentiable and convex, and $\mathbb E|g(X)|<\infty$, then

We prove this differentiable case. Convexity gives the supporting-line inequality $g(x)\ge g(\mu)+g'(\mu)(x-\mu)$ for every $x\in I$: the secant slopes on either side of $\mu$ bracket $g'(\mu)$. Taking expectations of this integrable inequality proves the claim. Applying it to $-g$ reverses the inequality for concave $g$.

For a symmetric two-point perturbation $X=\mu\pm s$, each with probability $1/2$, the mean remains $\mu$. With $\mu=4$ and $s=2$,

For the quadratic payoff $g(x)=x^2$, the gain in expectation is exactly $s^2$; for $g(x)=-x^2$, it is $-s^2$. These are changes relative to the constant input $\mu$. Comparing two arbitrary distributions requires a specified ordering of their spreads; variance alone does not order expectations of every convex function.

For example, compare the symmetric law at $\pm1$ with the law assigning probability $3/4$ to zero and $1/8$ to each of $\pm2$. Both have mean zero and variance one. Their expectations of $|X|$ are 1 and $1/2$, whereas their expectations of $X^4$ are 1 and 4. Both functions are convex, so neither law gives the larger expectation for every convex payoff.

A transformation changes the tail and its moment requirements¶

For exact Pareto $X$ with $x_m>0$, $\alpha>0$, and a power payoff $g(x)=x^r$ with $r>0$, direct inversion of the event gives

The payoff exponent is $\alpha/r$. Its mean is finite exactly when $r<\alpha$, by the Pareto moment proof. For example, $\alpha=3/2$ gives a finite mean for $X$ and an infinite mean for $X^2$. The finite-expectation version of Jensen above cannot be applied to that quadratic payoff with both sides finite.

If instead $g(x)=\min(x,c)$ on $x\ge0$ with $c>0$, the payoff lies in $[0,c]$, so $\mathbb E e^{t g(X)}\le e^{|t|c}$ for every real $t$. For $X$ Pareto with $x_m=1$, $\alpha=3/2$, and $c=4$, integration of the survival gives

This cap bounds the realized payoff and changes its mean. It does not estimate the uncapped mean. The inference note shows why bounded payoffs have a different contamination sensitivity.

Convexity alone does not bound downside: $g(x)=e^x+x$ has $g''(x)=e^x>0$ but tends to $-\infty$ as $x\to-\infty$. An explicit bound such as $g(x)=\max(x-u,0)-d\ge-d$, for fixed $u$ and $d\ge0$, is a separate property. Repetition, costs and the distribution of $X$ still matter.

Worked calculation and sources¶

Use finite_payoff_mean.

from xmath.incerto.lecture_risk import finite_payoff_mean

probabilities = [0.98, 0.02]
print(f"A: {finite_payoff_mean(probabilities, [1, -80]):.4g}")
print(f"B: {finite_payoff_mean(probabilities, [-1, 80]):.4g}")
print(f"Convex gap: {finite_payoff_mean([0.5, 0.5], [2**2, 6**2]) - 4**2:.4g}")

Output

A: -0.62
B: 0.62
Convex gap: 4

A’s higher gain frequency does not imply a higher expected payoff.

The lower panel of the shared figure plots $\mathbb E[(4\pm s)^2]-16=s^2$ and its negative for $0\le s\le3$. Reproduce the figure and finite payoff table with uv run python demos/incerto/lecture_risk.py. These are exact finite sums on 121 spread values, without simulation or external data.

The source context is Taleb (2025), ch. 3, §3.4.1, pp. 41–43, and §3.10, pp. 60–63. The chapter motivates distinguishing state from exposure. The finite-table calculation, differentiable Jensen argument, power-tail transformation and capped-mean calculation are proved here.