Convex-Order Comparisons for Normal and Bernoulli Laws

Convex Order Measures Spread at a Fixed Mean

Two distributions can have the same mean but differ substantially in how their probability is spread around it. Convex order makes this comparison precise: it asks whether every convex loss or payoff has a larger expectation under one distribution than under the other. This is useful when a decision-maker is especially sensitive to extreme outcomes, since convex functions assign relatively greater weight to them.

The comparisons here have a simple structure. Among normal distributions with a common mean, the one with the larger variance is larger in convex order. A Bernoulli variable is likewise larger in convex order than the constant equal to its mean. In both cases, the larger variable represents a mean-preserving increase in variability.

Definition: Let \(X\) and \(Y\) be integrable real-valued random variables. Write \(X\leq_{\mathrm{cx}}Y\) if \(E[X]=E[Y]\) and \(E[\phi(X)]\leq E[\phi(Y)]\) for every convex function \(\phi:\mathbb{R}\to\mathbb{R}\) for which both expectations exist and are finite. The notation compares distributions, so \(X\) and \(Y\) need not be defined on the same probability space.

The equal-mean requirement is essential: convex order compares spread while holding location fixed. For example, the function \(\phi(x)=x^2\) is convex. Thus, if \(X\leq_{\mathrm{cx}}Y\) and both second moments are finite, then \(E[X^2]\leq E[Y^2]\). With equal means, this also gives \(\operatorname{Var}(X)\leq\operatorname{Var}(Y)\). The converse is not true for arbitrary distributions; the special families considered below have stronger structure.

A useful diagnostic is the stop-loss function \(t\mapsto E[(X-t)_+]\), where \(u_+=\max(u,0)\). The stop-loss characterization says that for integrable \(X,Y\), \(X\leq_{\mathrm{cx}}Y\) exactly when their means agree and \(E[(X-t)_+]\leq E[(Y-t)_+]\) for every real \(t\). A stop-loss calculation can verify or refute a proposed comparison.

Normal Laws with a Common Mean

Suppose \(X\sim N(\mu,\sigma_1^2)\) and \(Y\sim N(\mu,\sigma_2^2)\), with \(0\leq\sigma_1^2\leq\sigma_2^2\). The comparison follows by expressing the more variable normal variable as the sum of the less variable one and independent, mean-zero normal noise.

Theorem: If \(X\sim N(\mu,\sigma_1^2)\) and \(Y\sim N(\mu,\sigma_2^2)\), where \(0\leq\sigma_1^2\leq\sigma_2^2\), then \(X\leq_{\mathrm{cx}}Y\). A variance of zero is interpreted as the constant random variable \(\mu\).

Proof. If \(\sigma_2^2=\sigma_1^2\), the two distributions are identical, so the claim holds. Otherwise, take independent random variables \(X_0\sim N(\mu,\sigma_1^2)\) and \(Z\sim N(0,\sigma_2^2-\sigma_1^2)\). The sum \(X_0+Z\) has distribution \(N(\mu,\sigma_2^2)\), so it can serve as \(Y\) for this comparison. For any convex \(\phi\) with finite expectations under these normal laws, conditional Jensen's inequality gives

E[\phi(X_0+Z)\mid X_0] \geq \phi\!\left(X_0+E[Z\mid X_0]\right) = \phi(X_0).

Here independence gives \(E[Z\mid X_0]=E[Z]=0\). Taking expectations yields \(E[\phi(Y)]\geq E[\phi(X_0)]\). Both means equal \(\mu\), proving the convex-order comparison. \(\square\)

The proof explains why the result is specific to this setting: the larger-variance normal law can be obtained by adding independent noise with mean zero. For a general pair of distributions, a larger variance does not guarantee this relationship or convex order.

Worked Example: Comparing Two Normal Laws

Let \(X\sim N(2,1)\) and \(Y\sim N(2,4)\). Their common mean is \(2\), and their variances satisfy \(1\leq4\). The theorem therefore gives \(X\leq_{\mathrm{cx}}Y\).

For a numerical check, use the convex stop-loss payoff \((x-3)_+\). Write \(\phi\) and \(\Phi\) for the standard normal density and distribution function. For \(W\sim N(\mu,\sigma^2)\), with \(\sigma>0\), standardizing the integral gives

E[(W-t)_+] =\int_{(t-\mu)/\sigma}^{\infty}(\mu+\sigma z-t)\phi(z)\,dz =(\mu-t)\Phi\!\left(\frac{\mu-t}{\sigma}\right) +\sigma\phi\!\left(\frac{\mu-t}{\sigma}\right).

In the last equality, use \(1-\Phi(c)=\Phi(-c)\) and \(\int_c^\infty z\phi(z)\,dz=\phi(c)\), the latter following from \(\phi'(z)=-z\phi(z)\). At \(t=3\), the standardized arguments are \(-1\) for \(X\) and \(-0.5\) for \(Y\). Thus

E[(X-3)_+]=-\Phi(-1)+\phi(-1) \approx-0.158655+0.241971=0.083316,

E[(Y-3)_+]=-\Phi(-0.5)+2\phi(-0.5) \approx-0.308538+0.704130=0.395592.

The larger stop-loss expectation for \(Y\) at this threshold agrees with the theorem. This single threshold is a check, not a proof of convex order by itself; the theorem establishes the comparison for every convex function.

A Constant and a Bernoulli Variable

Let \(B\) have a Bernoulli distribution with success probability \(p\), so \(P(B=1)=p\), \(P(B=0)=1-p\), and \(E[B]=p\). The deterministic variable equal to \(p\) has exactly the same mean. Convexity implies that randomizing between 0 and 1 cannot reduce the expectation of a convex function below its value at the average.

Theorem: For \(0\leq p\leq1\), the constant \(p\) is smaller than \(B\sim\operatorname{Bernoulli}(p)\) in convex order: \(p\leq_{\mathrm{cx}}B\).

Proof. The means agree, since \(E[B]=p\). For every convex \(\phi\), the defining convexity inequality applied to \(p=(1-p)0+p1\) gives

\phi(p)\leq(1-p)\phi(0)+p\phi(1)=E[\phi(B)].

This proves the claimed order. If \(p=0\) or \(p=1\), the Bernoulli variable is itself constant and the two distributions are identical. \(\square\)

Worked Example: A Bernoulli Variable with Mean 0.3

Let \(B\sim\operatorname{Bernoulli}(0.3)\). The comparison is \(0.3\leq_{\mathrm{cx}}B\). Check it first with the convex function \(\phi(x)=x^2\):

E[(0.3)^2]=0.09, \qquad E[B^2]=0.7(0^2)+0.3(1^2)=0.3.

Now use the stop-loss function. For every real threshold \(t\), both possible outcomes must be included:

$$ E[(B-t)_+] =0.7(0-t)_++0.3(1-t)_+ =0.7(-t)_++0.3(1-t)_+. $$

For \(t=-1\), this is \(0.7(1)+0.3(2)=1.3\), equal to the constant's stop-loss value \((0.3-(-1))_+=1.3\). For \(t=0.5\), it is \(0.7(0)+0.3(0.5)=0.15\), while the constant's value is \((0.3-0.5)_+=0\). The contribution from outcome 0 cannot be dropped for negative thresholds. In particular, the shortened expression \(0.3(1-t)_+\) is valid only for \(t\geq0\).

Worked Example: A Convex Loss of Deviation

Suppose a binary outcome \(B\) indicates whether an event occurs, with \(P(B=1)=0.4\). Compare the constant prediction \(0.4\) with the random outcome under the convex loss \(\phi(x)=(x-0.4)^2\).

For the constant, the loss is \((0.4-0.4)^2=0\). For the Bernoulli variable, its two outcomes give

$$ E[(B-0.4)^2] =0.6(0-0.4)^2+0.4(1-0.4)^2 =0.6(0.16)+0.4(0.36) =0.096+0.144 =0.24. $$

The expectation is larger for the Bernoulli variable, as convex order requires. This calculation is also its variance: \(\operatorname{Var}(B)=p(1-p)=0.4(0.6)=0.24\), whereas the constant has variance zero.

What the Comparisons Do—and Do Not—Say

These results give concrete examples of a mean-preserving spread. In the normal family, independent mean-zero noise increases the variance while preserving the mean. In the Bernoulli comparison, the constant mean is replaced by outcomes on either side of it. In both cases, every convex function has an expectation at least as large under the more variable distribution.

A common mistake is to infer convex order from equal means and ordered variances for any pair of distributions. Variance tests only one convex function, \(x^2\); convex order requires the inequality for all convex functions. For example, let \(X\) take values \(-1\) and \(1\), each with probability \(1/2\). Let \(Y\) take values \(-\sqrt{200}\), \(0\), and \(\sqrt{200}\) with probabilities \(0.005\), \(0.99\), and \(0.005\), respectively. Both means are zero, and

\operatorname{Var}(X)=1, \qquad \operatorname{Var}(Y)=0.01(200)=2.

Nevertheless, at threshold \(t=0.5\),

E[(X-0.5)_+]=\tfrac12(1-0.5)=0.25, \qquad E[(Y-0.5)_+]=0.005(\sqrt{200}-0.5)\approx0.0682.

The stop-loss inequality required for \(X\leq_{\mathrm{cx}}Y\) fails. The lesson is to use the normal theorem only for the stated normal family, and to check a proposed general comparison with convex functions or the stop-loss criterion rather than relying on variance alone.

Key takeaway: Equal means are part of convex order, not an optional detail. For normal laws with a common mean, increasing variance increases the law in convex order; for a Bernoulli variable, its equal-mean constant is below it in convex order. Outside these settings, variance ordering alone is insufficient.

Check Your Understanding

For each question, identify the relevant convex-order comparison and justify your answer.

If \(X\sim N(5,9)\) and \(Y\sim N(5,16)\), which variable is larger in convex order? State the condition that makes the theorem apply.
For \(B\sim\operatorname{Bernoulli}(0.6)\), calculate \(E[(B-t)_+]\) at \(t=-2\), including contributions from both outcomes.
Why does the convex function \(\phi(x)=x^2\) imply a variance inequality when two variables have equal means and finite second moments?
Does equal mean and larger variance, by itself, establish convex order for arbitrary distributions? Explain using the role of the stop-loss criterion.
For \(p=0\) or \(p=1\), what happens to the convex-order comparison between the constant \(p\) and a \(\operatorname{Bernoulli}(p)\) variable?

Convex-Order Comparisons for Normal and Bernoulli Laws

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