Entropic Risk Measures: Convexity Without Homogeneity

Why Study the Entropic Risk Measure?

A risk measure converts a random loss into a single amount used to assess risk or determine a capital requirement. Different measures respond differently to uncertainty and to changes in exposure. The entropic risk measure is useful because it penalizes large losses through an exponential function and is convex, while generally not being positively homogeneous. This makes it a clear example of why convexity and coherence are not the same property.

Throughout, \(X\) denotes a loss: larger values are worse. A risk measure \(\rho(X)\) is therefore interpreted as the risk assigned to that loss. The risk-aversion parameter \(\theta\) is positive and has units reciprocal to the units of loss. For example, if losses are measured in dollars, \(\theta\) is measured per dollar.

Definition: For \(\theta>0\), the entropic risk measure of a loss \(X\) is \[ \rho_\theta(X)=\frac{1}{\theta}\ln E\!\left[e^{\theta X}\right], \] provided \(E[e^{\theta X}]<\infty\). The parameter \(\theta\) is held fixed when comparing losses or checking the risk-measure properties. If the exponential moment is infinite, the risk measure is infinite and the finite-valued properties below do not apply as stated.

For a finite-state loss with outcomes \(x_j\) and probabilities \(p_j\), the expectation in the definition is the probability-weighted sum \(\sum_j p_j e^{\theta x_j}\). The exponential makes large losses influential, and the logarithm and division by \(\theta\) return the result to the units of loss.

Three Core Properties

The relevant comparisons use the same positive parameter \(\theta\). Monotonicity compares losses ordered state by state; translation invariance describes adding a fixed amount of loss; convexity compares a mixture of two losses with the corresponding mixture of their risks.

Key properties: If the exponential moments involved are finite, then:

Monotonicity: if \(X\leq Y\) almost surely, then \(\rho_\theta(X)\leq\rho_\theta(Y)\).
Translation invariance: for any constant \(c\), \(\rho_\theta(X+c)=\rho_\theta(X)+c\).
Convexity: for \(0\leq\lambda\leq1\), \(\rho_\theta(\lambda X+(1-\lambda)Y)\leq\lambda\rho_\theta(X)+(1-\lambda)\rho_\theta(Y)\).

The translation formula uses the loss convention: adding \(c>0\) means adding a sure loss, so measured risk rises by \(c\). This is not a statement that adding cash to an insurer’s assets increases its loss.

Why Convexity Holds

The convexity inequality follows from Hölder’s inequality. It does not require the two losses to be independent. The losses must be defined on the same probability space so that their mixture is meaningful, and their exponential moments must be finite.

Exponentiate the mixture.
For \(0<\lambda<1\), \(e^{\theta(\lambda X+(1-\lambda)Y)}=(e^{\theta X})^\lambda(e^{\theta Y})^{1-\lambda}\).

Apply Hölder’s inequality.
For nonnegative random variables \(U,V\), \(E[U^\lambda V^{1-\lambda}]\leq(E[U])^\lambda(E[V])^{1-\lambda}\). Take \(U=e^{\theta X}\) and \(V=e^{\theta Y}\).

Take logarithms and divide.
The natural logarithm is increasing, and \(1/\theta>0\). These preserve the inequality and give the convexity result.

\begin{aligned} E\!\left[e^{\theta(\lambda X+(1-\lambda)Y)}\right] &=E\!\left[(e^{\theta X})^\lambda(e^{\theta Y})^{1-\lambda}\right]\\ &\leq \left(E[e^{\theta X}]\right)^\lambda \left(E[e^{\theta Y}]\right)^{1-\lambda},\\ \rho_\theta(\lambda X+(1-\lambda)Y) &\leq \lambda\rho_\theta(X)+(1-\lambda)\rho_\theta(Y). \end{aligned}

At \(\lambda=0\) or \(\lambda=1\), the inequality is equality. The proof also shows why dependence does not invalidate convexity: Hölder’s inequality applies to the joint random variables without an independence assumption.

Worked Applications

Worked Example: Monotonicity and a Sure Loss Addition

A hypothetical two-state portfolio has equal probabilities in the two states. Its loss \(X\) is 0 dollars in the first state and 1 dollar in the second. A second portfolio has loss \(Y=X+1\) dollars in each state. Use \(\theta=\ln 2\) per dollar to calculate both risks and check monotonicity and translation invariance.

Method. For each loss, calculate the probability-weighted exponential moment, then apply the definition. Since \(Y=X+1\), \(X\leq Y\) in every state and translation invariance predicts an increase of exactly 1 dollar.

For \(X\), the exponential moment is \[ E[e^{\theta X}]=\frac12e^0+\frac12e^{\ln 2} =\frac12+\frac12(2)=1.5. \] Thus \[ \rho_\theta(X)=\frac{\ln(1.5)}{\ln 2} \approx 0.5850\text{ dollars}. \] For \(Y\), the losses are 1 and 2 dollars, so \[ E[e^{\theta Y}]=\frac12(2)+\frac12(4)=3, \qquad \rho_\theta(Y)=\frac{\ln 3}{\ln 2}\approx1.5850\text{ dollars}. \]

The check is \(\rho_\theta(Y)-\rho_\theta(X)=1.0000\) dollar, as required by translation invariance, and \(\rho_\theta(X)<\rho_\theta(Y)\), as required by monotonicity. Probabilities and risks are rounded to four decimal places.

Common mistake avoided: The added dollar is a sure loss, so it must be added to the risk. Reversing the sign would confuse a loss addition with a cash addition.

Worked Example: Convexity for Losses in Opposite States

Two hypothetical portfolios are evaluated in two equally likely states. Portfolio \(X\) has losses 0 and 6 dollars, while portfolio \(Y\) has losses 6 and 0 dollars, respectively. Let \(\theta=0.2\) per dollar. Calculate the risk of each portfolio and of the equally weighted mixture \(Z=(X+Y)/2\).

Method. First compute each exponential moment. Then calculate the mixture state by state. Convexity predicts \(\rho_\theta(Z)\leq(\rho_\theta(X)+\rho_\theta(Y))/2\).

The portfolios have the same two outcomes with the same probabilities, so \[ E[e^{\theta X}]=E[e^{\theta Y}] =\frac{1+e^{0.2(6)}}{2} =\frac{1+e^{1.2}}{2} \approx 2.16005846. \] Since \(1/\theta=5\) dollars, \[ \rho_\theta(X)=\rho_\theta(Y) =5\ln(2.16005846) \approx 5(0.7701353) =3.8507\text{ dollars}. \]

In the first state, \(Z=(0+6)/2=3\) dollars. In the second state, \(Z=(6+0)/2=3\) dollars. Thus \(Z\) is a certain loss of 3 dollars, and \[ \rho_\theta(Z)=\frac{1}{0.2}\ln(e^{0.2(3)}) =5\ln(e^{0.6})=3\text{ dollars}. \] Therefore \(3.0000\leq(3.8507+3.8507)/2=3.8507\), which checks convexity. The risk values are rounded to four decimal places.

Common mistake avoided: Do not average the two standalone risk values and call that the mixture’s risk. The mixture must first be formed state by state; here it produces a certain loss of 3 dollars.

Worked Example: Positive Homogeneity Fails

A hypothetical loss \(X\) is 0 or 2 dollars with equal probability. Take \(\theta=0.5\) per dollar. Compare \(\rho_\theta(2X)\) with \(2\rho_\theta(X)\).

Method. Apply the entropic formula separately to \(X\) and \(2X\), keeping \(\theta\) fixed. Positive homogeneity would require the two results to be equal.

For \(X\), the outcomes in the exponential are 0 and \(0.5(2)=1\), so \[ E[e^{\theta X}]=\frac{1+e}{2} \approx\frac{1+2.7182818}{2}=1.8591409. \] Hence \[ \rho_\theta(X)=\frac{1}{0.5}\ln(1.8591409) =2\ln(1.8591409)\approx1.2402\text{ dollars}, \] and \(2\rho_\theta(X)\approx2.4805\) dollars.

The loss \(2X\) takes values 0 and 4 dollars. Therefore \[ E[e^{\theta(2X)}]=\frac{1+e^2}{2} \approx\frac{1+7.3890561}{2}=4.1945281, \] and \[ \rho_\theta(2X)=2\ln(4.1945281)\approx2.8676\text{ dollars}. \] Thus \(\rho_\theta(2X)\neq2\rho_\theta(X)\). Values are rounded to four decimal places.

Common mistake avoided: Scaling the loss by 2 does not scale its risk by 2 at fixed \(\theta\). The exponential changes from \(e^{\theta X}\) to \(e^{2\theta X}\), so the expectation inside the logarithm changes nonlinearly.

Why Convexity Does Not Mean Homogeneity

Positive homogeneity requires \(\rho(cX)=c\rho(X)\) for every \(c\geq0\). Convexity instead requires that the risk of a weighted mixture not exceed the same weighted mixture of the risks. The worked scaling example violates positive homogeneity while leaving the convexity result intact.

For a fixed positive parameter, the entropic measure has the scaling identity \[ \rho_\theta(cX)=c\,\rho_{c\theta}(X)\qquad(c>0), \] when the relevant exponential moments are finite. Indeed, both sides equal \(\frac{1}{\theta}\ln E[e^{c\theta X}]\). The parameter on the right is \(c\theta\), not \(\theta\). Replacing \(\rho_{c\theta}(X)\) by \(\rho_\theta(X)\) is generally unjustified. Positive homogeneity does hold for certain special losses, such as a constant loss, but not for all random losses.

Key takeaway: For fixed \(\theta>0\), the entropic risk measure is monotone, translation invariant, and convex on losses with finite exponential moments. It is generally not positively homogeneous, so it is not a coherent risk measure under the four-axiom definition.

Common Exam Traps

Changing the sign convention: With losses, \(X+c\) adds a sure loss and increases risk by \(c\). Do not apply an asset-value sign convention without changing the definition.
Ignoring units: The product \(\theta X\) must be dimensionless. Changing the loss unit from dollars to thousands of dollars requires a corresponding change in the numerical value of \(\theta\).
Assuming independence is needed: The convexity proof uses Hölder’s inequality and does not assume independent losses.
Confusing convexity with positive homogeneity: The convexity inequality is not the equality \(\rho(cX)=c\rho(X)\). Test each property using its own definition.
Using an infinite exponential moment as if it were finite: The finite-valued formula and its stated properties require \(E[e^{\theta X}]<\infty\) for the losses being evaluated.

Check Your Understanding

For each problem, use the loss convention and keep the stated risk-aversion parameter fixed unless the problem says otherwise.

A certain hypothetical loss is 7 dollars. For any \(\theta=0.3\) per dollar, calculate its entropic risk. (Answer: 7 dollars.)
A hypothetical loss takes values 0 and 1 dollar with equal probability, and \(\theta=\ln 2\) per dollar. Calculate its risk to four decimal places. (Answer: \(\ln(1.5)/\ln 2\approx0.5850\) dollars.)
A hypothetical loss \(X\) is 1 or 4 dollars with equal probability, and \(\theta=0.5\) per dollar. A sure loss of 2 dollars is then added. Give the risk before and after the addition in exact form, and verify the difference. (Answer: Before, \(2\ln((e^{0.5}+e^2)/2)\) dollars; after, \(2+2\ln((e^{0.5}+e^2)/2)\) dollars; the increase is 2 dollars.)
A hypothetical loss \(X\) is 0 or 2 dollars with equal probability, with \(\theta=0.5\) per dollar. Compare \(\rho_\theta(2X)\) with \(2\rho_\theta(X)\), rounding to four decimal places. (Answer: \(\rho_\theta(2X)\approx2.8676\) dollars, while \(2\rho_\theta(X)\approx2.4805\) dollars; they are unequal.)
In two equally likely states, hypothetical losses \(X\) and \(Y\) are respectively \((0,6)\) and \((6,0)\) dollars. With \(\theta=0.2\) per dollar, calculate the loss of \((X+Y)/2\) in each state and compare its risk with the average standalone risk. (Answer: The mixture loses 3 dollars in each state, so its risk is 3 dollars; each standalone risk is \(5\ln((1+e^{1.2})/2)\approx3.8507\) dollars, and \(3\leq3.8507\).)

What You'll Learn