Equality In Jensen's Inequality

What Equality in Jensen’s Inequality Tells Us

Jensen’s inequality compares the value of a convex function at a mean with the mean of the function’s values. Its equality case is more informative than the inequality alone: if the convex function bends strictly, equality can occur only when the input has no genuine variation, apart from changes on a null set. When the function is convex but not strictly convex, equality can also occur for nonconstant inputs, provided their values lie along a portion where the function is affine.

We work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. A random variable $X:\Omega\to\mathbb{R}$ is real-valued and measurable, and is assumed integrable, meaning $\mathbb{E}[|X|]<\infty$. Write $a=\mathbb{E}[X]$. Let $I\subseteq\mathbb{R}$ be an open interval, and let $\phi:I\to\mathbb{R}$ be a finite convex function. Assume $X\in I$ $\mathbb{P}$-a.e. and $\phi(X)$ is integrable. These conditions ensure both sides of Jensen’s inequality are finite real numbers. In particular, equality means ordinary equality of those finite numbers, not an equality involving $+\infty$.

Theorem (Equality for Strictly Convex Functions): Under the assumptions above, suppose also that $a\in I$ and $\phi$ is strictly convex on $I$. Then $$ \mathbb{E}[\phi(X)]=\phi(\mathbb{E}[X]) \quad\Longleftrightarrow\quad X=\mathbb{E}[X]\ \mathbb{P}\text{-a.e.} $$

Here strictly convex means that for distinct $x,y\in I$ and every $t\in(0,1)$, $\phi((1-t)x+ty)<(1-t)\phi(x)+t\phi(y)$. The theorem concerns equality almost everywhere, not equality at every point: a random variable may differ from its mean on a null set without changing either expectation.

Supporting Lines and the Equality Criterion

The proof uses a supporting line at the mean. A line $x\mapsto\phi(a)+s(x-a)$ supports $\phi$ at $a\in I$ if it lies below the graph throughout $I$ and agrees with $\phi$ at $a$. For a finite convex function on an open interval, at least one such slope $s\in\mathbb{R}$ exists. To see why, consider the secant slopes to the left and right of $a$: convexity implies that every slope $(\phi(a)-\phi(x))/(a-x)$, for $x<a$, is at most every slope $(\phi(y)-\phi(a))/(y-a)$, for $y>a$. The supremum of the left slopes and the infimum of the right slopes are finite and ordered; choosing $s$ between them gives the supporting inequality on both sides.

Proposition (Equality via a Supporting Line): On the probability space and under the integrability and domain assumptions just stated, let $\phi$ be convex, let $a=\mathbb{E}[X]\in I$, and choose any supporting slope $s$ at $a$. Define $$ C_s=\{x\in I:\phi(x)=\phi(a)+s(x-a)\}. $$ Then $\mathbb{E}[\phi(X)]=\phi(a)$ if and only if $X\in C_s$ $\mathbb{P}$-a.e. The set $C_s$ is an interval, possibly empty except for the point $a$, and may be larger when $\phi$ has an affine portion.

Proof. Define $q(x)=\phi(x)-\phi(a)-s(x-a)$ for $x\in I$. The supporting-line property gives $q(x)\ge0$ and $q(a)=0$. The random variable $q(X)$ is measurable and integrable: it is a difference of the integrable random variables $\phi(X)$, the constant $\phi(a)$, and $s(X-a)$. Since $\mathbb{E}[X]=a$, linearity of expectation gives $\mathbb{E}[q(X)]=\mathbb{E}[\phi(X)]-\phi(a)$. A nonnegative integrable function has integral zero exactly when it is zero a.e.: indeed, if its integral is zero, each set $\{q(X)>1/n\}$ has probability zero because $\mathbb{E}[q(X)]\ge (1/n)\mathbb{P}(q(X)>1/n)$; their countable union is $\{q(X)>0\}$. Conversely, a function zero a.e. has integral zero. Thus equality of the Jensen quantities holds exactly when $q(X)=0$ a.e., which is precisely $X\in C_s$ a.e. Finally, if $x,y\in C_s$, convexity gives, for $z=(1-t)x+ty$ with $0\le t\le1$, $\phi(z)\le(1-t)\phi(x)+t\phi(y)=\phi(a)+s(z-a)$. The supporting-line inequality gives the reverse inequality, so $z\in C_s$. Hence $C_s$ is an interval. $\square$

Proof of the strict-convexity theorem. Choose a supporting slope $s$ at $a$. If $x\ne a$ and $q(x)=0$, then $\phi(x)$ equals the supporting line at both $a$ and $x$. For every point strictly between $a$ and $x$, convexity bounds $\phi$ above by the line segment joining these endpoint values, while the supporting-line inequality bounds it below by that same line. Thus $\phi$ agrees with a line throughout the segment, contradicting strict convexity. Therefore $q(x)=0$ only at $x=a$. By the proposition, equality in Jensen’s inequality implies $X=a$ a.e. Conversely, if $X=a$ a.e., then $\phi(X)=\phi(a)$ a.e., so $\mathbb{E}[\phi(X)]=\phi(a)$. $\square$

Worked Examples

Worked Example: Strict Convexity Gives Strict Inequality

Let $\Omega=\{\omega_1,\omega_2\}$, with $\mathcal{F}$ the set of all subsets and $\mathbb{P}(\{\omega_1\})=\mathbb{P}(\{\omega_2\})=1/2$. Set $X(\omega_1)=0$, $X(\omega_2)=2$, and $\phi(x)=x^2$ on $I=\mathbb{R}$. The random variable and its image under $\phi$ are integrable, and $\phi$ is strictly convex. Compute both sides:

\mathbb{E}[X]=\frac12(0)+\frac12(2)=1, \qquad \phi(\mathbb{E}[X])=1^2=1,

\mathbb{E}[\phi(X)] =\frac12(0^2)+\frac12(2^2) =\frac12(0)+\frac12(4)=2.

Thus the Jensen gap is $2-1=1$, so equality does not hold. As an independent check, the mean squared deviation is $\frac12(0-1)^2+\frac12(2-1)^2=\frac12+\frac12=1$, which agrees with the gap for $x^2$. Since $X$ is not constant a.e. on this two-point probability space, the strict-convexity theorem predicts strict inequality.

Worked Example: Equality Along an Affine Portion

On the same two-point probability space, let $X(\omega_1)=0$, $X(\omega_2)=2$, and define $\phi(x)=\max(0,x)$ on $I=(-2,3)$. This function is convex but not strictly convex: it equals the affine function $x$ for $x\ge0$. Its values at both outcomes lie in $I$, and all variables involved are integrable. The mean is $\mathbb{E}[X]=\frac12(0)+\frac12(2)=1$, so $\phi(\mathbb{E}[X])=\max(0,1)=1$. Also, $\mathbb{E}[\phi(X)]=\frac12\max(0,0)+\frac12\max(0,2)=\frac12(0)+\frac12(2)=1$. Equality holds even though $X$ is not constant a.e.

At the mean $a=1$, slope $s=1$ gives the supporting line $x\mapsto x$. Its contact set in $I$ is $C_1=[0,3)$, which contains both possible values of $X$. This verifies the contact-set criterion directly. Check: both sides are exactly $1$, and the mean of the two values $0$ and $2$ lies in the same affine portion.

Worked Example: Equality Means Almost-Everywhere Constancy

Take $\Omega=[0,1]$, $\mathcal{F}$ the Lebesgue sigma-algebra, and $\mathbb{P}$ Lebesgue measure restricted to $[0,1]$, which has total mass $1$. Let $X(x)=3$ for $x\ne1/2$ and $X(1/2)=7$, and let $\phi(t)=e^t$ on $I=\mathbb{R}$. The exceptional singleton has probability zero, so $X=3$ a.e. and $X$ is integrable. Directly, $\mathbb{E}[X]=3\cdot1=3$, and $\mathbb{E}[e^X]=e^3\cdot m([0,1]\setminus\{1/2\})+e^7\cdot m(\{1/2\})=e^3\cdot1+e^7\cdot0=e^3$. Consequently, $\phi(\mathbb{E}[X])=e^3=\mathbb{E}[\phi(X)]$. The exponential function is strictly convex, and $X=3$ a.e., as the theorem requires.

Check: changing $X$ at the single point $1/2$ does not change its integral or the integral of $e^X$, because that set has measure zero. Equality here does not assert that $X(x)=3$ at every point.

Common Mistake: Treating a Convex Function as Linear

A frequent incorrect calculation is to replace $\mathbb{E}[\phi(X)]$ by $\phi(\mathbb{E}[X])$ on the grounds that expectation “passes through” a function. Expectation is linear, but applying a nonlinear function is not a linear operation. In the first example, that mistaken step would give $\mathbb{E}[X^2]=(\mathbb{E}[X])^2=1$. The actual computation is $\mathbb{E}[X^2]=\frac12(0^2)+\frac12(2^2)=2$, not $1$. The difference $1$ is exactly the Jensen gap.

Strict convexity is essential to the conclusion that equality forces constancy. For a general convex function, equality only forces the values of $X$ to lie a.e. in the contact set of a supporting line at the mean. That set can contain many points, as in the affine-portion example. Conversely, strict convexity does not say that every nonconstant input gives equality; it says that a nonconstant input gives a strict inequality under the theorem’s hypotheses. Finiteness and integrability matter as well: the equality statement above compares finite expectations, so it does not interpret an undefined difference such as $+\infty-(+\infty)$.

Check Your Understanding

Use the supporting-line criterion and check the integrability and domain assumptions in each question.

On a two-point probability space with equal probabilities, let $X$ take values $1$ and $3$. Compute both sides of Jensen’s inequality for $\phi(x)=x^2$, and determine whether equality holds.
For the function $\phi(x)=\max(0,x)$, identify a supporting line at a positive mean $a$ and describe its contact set.
Explain why the strict-convexity theorem concludes that $X$ equals its mean a.e., rather than at every point.
Suppose a convex function agrees with a line on an interval containing all values of $X$. Explain why Jensen equality can hold even if $X$ is not constant a.e.
State the role of the assumption that $\phi(X)$ is integrable in the equality criterion.

Equality In Jensen's Inequality

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