Checking Convexity and Concavity of Risk Payoffs

Why Convexity Appears in Statistical Risk

Minimax arguments compare two kinds of choices: a statistician selects a decision rule, while nature may select a parameter value or a distribution over parameter values. The shape of the payoff as a function of these choices matters. In particular, Sion’s minimax theorem asks for convexity in one strategy and concavity in the other, along with additional hypotheses. In statistical decision problems, these shape conditions often come directly from how randomization and averaging work.

The key is to keep the two ways of mixing distinct. Mixing randomized rules means randomizing between procedures, and it averages their risks. Mixing distributions over parameters means averaging the risk across those distributions. Both operations make the payoff affine in the strategy being mixed, provided the risks are well-defined and finite. They do not, in general, make the payoff jointly convex or jointly concave in both choices.

Payoff and Strategy Definitions

Let $\theta\in\Theta$ be the parameter, let $X$ have distribution $P_\theta$, and let $a\in\mathcal A$ be an action. A randomized decision rule $Q$ assigns a probability distribution $Q(\cdot\mid x)$ on actions to each observation $x$. A deterministic rule is the special case that selects one action with probability one at each $x$. Let $L(\theta,a)$ be the loss.

Definition: The risk of a randomized rule $Q$ at parameter $\theta$ is its expected loss, $$ R(Q,\theta)=\int_{\mathcal X}\int_{\mathcal A}L(\theta,a)\,Q(da\mid x)\,P_\theta(dx). $$ For a probability distribution $\pi$ on $\Theta$, the expected-loss payoff is $$ \Phi(Q,\pi)=\int_\Theta R(Q,\theta)\,\pi(d\theta). $$

For the calculations below, assume the relevant integrands are measurable and the displayed risks and payoffs are finite real numbers. These conditions hold, for example, in finite problems with finite losses. When mixing two rules, assume their risks are finite at the parameters under consideration; when mixing two parameter distributions, assume the risk is integrable under each distribution. These hypotheses ensure that the averages below are defined. No topological or compactness conclusion is implied by the algebraic convexity calculations alone.

The relevant strategy sets are convex when they contain mixtures of their elements. For rules, the mixture is defined by $Q_t(\cdot\mid x)=(1-t)Q_0(\cdot\mid x)+tQ_1(\cdot\mid x)$, for $0\leq t\leq1$. For parameter distributions, it is $\pi_t=(1-t)\pi_0+t\pi_1$. A real-valued function $g$ on a convex set is convex if $g((1-t)u+tv)\leq(1-t)g(u)+tg(v)$; it is concave if the inequality is reversed. A function is affine if equality holds for every such pair and every $t$.

Mixing Rules Averages Risk

Theorem: For fixed $\theta$, risk is affine in the randomized rule. If $Q_0,Q_1$ have finite risks at $\theta$, then for $Q_t=(1-t)Q_0+tQ_1$, $$ R(Q_t,\theta)=(1-t)R(Q_0,\theta)+tR(Q_1,\theta). $$ Consequently, for any fixed $\pi$ for which the risks are integrable, $\Phi(Q,\pi)$ is affine, and therefore convex, as a function of $Q$.

Proof. For each observation $x$, the mixture $Q_t$ assigns to an action set the corresponding mixture of its probabilities under $Q_0$ and $Q_1$. Integrating the loss against that mixture gives $$ \int_{\mathcal A}L(\theta,a)\,Q_t(da\mid x) =(1-t)\int_{\mathcal A}L(\theta,a)\,Q_0(da\mid x) +t\int_{\mathcal A}L(\theta,a)\,Q_1(da\mid x). $$ Integrate both sides with respect to $P_\theta(dx)$. Finiteness of the two risks permits linearity of the integral, yielding the asserted identity. Integrating that identity with respect to $\pi(d\theta)$, under the stated integrability conditions, gives $\Phi(Q_t,\pi)=(1-t)\Phi(Q_0,\pi)+t\Phi(Q_1,\pi)$. Equality implies convexity. $\square$

The argument does not require the loss to be convex as a function of the action. It uses linearity of integration with respect to a mixture of probability distributions on actions. This distinction matters: a randomized rule chooses an action at random, whereas replacing its possible actions with their numerical average is a different operation.

Worked Example: Mixing Two Rules in a Finite Decision Problem

Suppose there are two parameter values, $\theta_0,\theta_1$, and two actions, $a_0,a_1$. The loss is $0$ for the matching action and $2$ for the other action. There is no informative observation. Rule $Q_0$ always chooses $a_0$, and rule $Q_1$ always chooses $a_1$. Their risks are $R(Q_0,\theta_0)=0,\ R(Q_0,\theta_1)=2,\ R(Q_1,\theta_0)=2,\ R(Q_1,\theta_1)=0$.

Mix the rules with weight $t=1/4$ on $Q_1$. The resulting rule chooses $a_1$ with probability $1/4$ and $a_0$ with probability $3/4$. At $\theta_0$, its risk is $ (3/4)0+(1/4)2=1/2$. At $\theta_1$, its risk is $ (3/4)2+(1/4)0=3/2$. These are exactly the mixtures of the original risks: $ (3/4)R(Q_0,\theta_0)+(1/4)R(Q_1,\theta_0)=1/2$, and $ (3/4)R(Q_0,\theta_1)+(1/4)R(Q_1,\theta_1)=3/2$.

For a distribution assigning probability $w$ to $\theta_0$ and $1-w$ to $\theta_1$, the mixture rule’s payoff is $w(1/2)+(1-w)(3/2)=3/2-w$. The calculation checks the risk identity at each parameter and then averages those risks under the chosen distribution.

Mixing Parameter Distributions Averages the Payoff

Theorem: For fixed $Q$, expected-loss payoff is affine in the parameter distribution. If $\pi_0,\pi_1$ give finite payoffs, then for $\pi_t=(1-t)\pi_0+t\pi_1$, $$ \Phi(Q,\pi_t)=(1-t)\Phi(Q,\pi_0)+t\Phi(Q,\pi_1). $$ Thus $\Phi(Q,\pi)$ is both concave and convex in $\pi$.

Proof. By definition, $\pi_t$ assigns to each measurable parameter set the mixture of its probabilities under $\pi_0$ and $\pi_1$. Linearity of integration, with the assumed finite integrals, gives $$ \Phi(Q,\pi_t)=\int_\Theta R(Q,\theta)\big((1-t)\pi_0+t\pi_1\big)(d\theta) =(1-t)\int_\Theta R(Q,\theta)\,\pi_0(d\theta) +t\int_\Theta R(Q,\theta)\,\pi_1(d\theta). $$ The right-hand side is $(1-t)\Phi(Q,\pi_0)+t\Phi(Q,\pi_1)$, as required. Equality in the defining inequalities gives both convexity and concavity. $\square$

Worked Example: A Mixture of Parameter Distributions

Consider the estimator that always reports $0$, with squared-error loss $L(\theta,a)=(a-\theta)^2$. Its risk is $R(Q,\theta)=\theta^2$. Let $\pi_0$ put all its mass at $0$, let $\pi_1$ put all its mass at $2$, and define $\pi_t=(1-t)\pi_0+t\pi_1$.

Under $\pi_0$, the payoff is $R(Q,0)=0^2=0$. Under $\pi_1$, it is $R(Q,2)=2^2=4$. Directly under the mixture, $\Phi(Q,\pi_t)=(1-t)0+t4=4t$. For $t=1/4$, the mixture puts probabilities $3/4$ at $0$ and $1/4$ at $2$, so its expected loss is $(3/4)0^2+(1/4)2^2=1$, agreeing with $4(1/4)=1$.

This is an average of risks at parameter values. It is not the risk evaluated at the mean parameter: the mean is $2t$, and the risk there is $(2t)^2=4t^2$. At $t=1/4$, that plug-in risk is $1/4$, not the mixture payoff $1$. A distribution over parameters and a single parameter value are different inputs to the payoff.

Separate Convexity Is Not Joint Convexity

For a minimax interchange, the relevant shape conditions are applied one strategy at a time: with the parameter distribution fixed, the payoff should be convex in the rule; with the rule fixed, it should be concave in the distribution. The affine identities above supply both conditions. They do not assert joint convexity in the pair $(Q,\pi)$. Indeed, the payoff often contains products between the probabilities describing the two strategies.

Worked Example: A Bilinear Payoff Need Not Be Jointly Convex

Use the two-parameter, two-action loss from the earlier calculation. Let $p$ be the probability of choosing $a_1$, and let $w$ be the probability assigned to $\theta_0$. The risks are $2p$ at $\theta_0$ and $2(1-p)$ at $\theta_1$. Thus $$ \Phi(p,w)=2wp+2(1-w)(1-p)=2-2p-2w+4pw. $$

For each fixed $w$, this is affine in $p$; for each fixed $p$, it is affine in $w$. To test joint convexity, compare the two strategy pairs $(p,w)=(0,0)$ and $(1,1)$. The payoff at each endpoint is $2$. Their midpoint is $(1/2,1/2)$, where the payoff is $2-1-1+4(1/2)(1/2)=1$. Convexity would require the midpoint payoff to be at most the average endpoint payoff, which is $2$; this test does not disprove convexity. Instead compare $(0,1)$ and $(1,0)$: the payoff at both endpoints is $0$, while at their midpoint it is $1$. Convexity would require $1\leq(0+0)/2=0$, which is false. For concavity, the first pair has midpoint payoff $1$, less than its endpoint average $2$, so compare $(0,0)$ and $(1,1)$ under concavity: it requires midpoint payoff at least $2$, but the midpoint payoff is $1$. Thus the payoff is neither jointly convex nor jointly concave.

The failure of joint convexity or concavity does not contradict the separate affine calculations. Each strategy is tested while the other is held fixed; changing both at once is a different test.

Randomizing Actions Is Not Averaging Actions

A common source of incorrect convexity claims is to replace a probability distribution over actions by its mean action. Those operations need not have the same risk. Randomization averages the losses of the actions actually selected. Averaging the actions first evaluates the loss at a potentially different action.

Worked Example: Nonlinear Loss and the Average Action

There is one parameter value, $\theta=1$, no informative observation, and squared-error loss $L(\theta,a)=(a-\theta)^2$. Suppose the rule selects action $0$ with probability $1/2$ and action $2$ with probability $1/2$. Its expected loss is $$ \frac12(0-1)^2+\frac12(2-1)^2=\frac12(1)+\frac12(1)=1. $$

The average of the two actions is $1$. If action $1$ is available and the procedure instead always selects it, the loss is $(1-1)^2=0$. Therefore the randomized rule and the deterministic rule at the average action do not have the same risk. The affine risk result concerns mixtures of distributions over actions, not substitution of an average action into the loss function.

A Practical Convexity Check

Before using a minimax theorem for an expected-loss game, check the strategy and payoff in a fixed order. This separates the algebraic shape conditions from other requirements of the theorem, such as its hypotheses about the sets and the payoff’s regularity.

Identify what is being mixed.
Specify whether the statistician mixes randomized rules, and whether nature mixes parameter values through probability distributions. Do not substitute averaged actions for randomized actions.

Write the payoff as an integral of risk.
Use $\Phi(Q,\pi)=\int R(Q,\theta)\,\pi(d\theta)$. Verify measurability and finiteness, or state conditions that make the integrals well-defined.

Mix one argument at a time.
Hold $\pi$ fixed and check the risk under a mixture of rules. Then hold $Q$ fixed and check the payoff under a mixture of distributions.

State the conclusion narrowly.
The identities establish separate affinity, hence convexity in $Q$ and concavity in $\pi$. They do not establish joint convexity, compactness, semicontinuity, or the other hypotheses needed for minimax interchange.

The main benefit of this check is diagnostic. If a calculation appears to violate convexity in randomized rules, determine whether the object being mixed is really a probability kernel or whether the calculation has instead averaged actions, changed the model, or left the class of admissible rules. If the parameter-side calculation is nonlinear, check whether it evaluates risk under a mixture distribution or substitutes a summary such as the mean parameter.

Key takeaway: Under finite, well-defined expected loss, randomizing between rules averages their risks, and mixing parameter distributions averages their payoffs. These facts give the separate convexity and concavity conditions used in minimax arguments; they do not replace the other hypotheses of a minimax theorem.

Check Your Understanding

For each question, distinguish the strategy being varied from the one held fixed.

State the risk identity for a mixture of two randomized rules and explain why it implies convexity.
Why is the payoff affine in a parameter distribution when the rule is fixed?
In the squared-error example with risk $R(Q,\theta)=\theta^2$, why does a mixture of point masses at $0$ and $2$ generally not have the same payoff as risk evaluated at the mean parameter?
Does separate affinity of a payoff imply joint convexity? Give the calculation from the two-action example that answers this.
What is the difference between randomizing equally between actions $0$ and $2$, and always choosing their average under squared-error loss at $\theta=1$?