From a Supporting Line to an Integral Inequality
For a finite weighted average, convexity says that applying a convex function after averaging cannot give a larger value than averaging the transformed values. The integral version has the same structure, but there are two points to handle carefully: the mean might be at the boundary of the interval on which the function is defined, and the integral of the transformed function might be infinite. A supporting line at the mean resolves both issues.
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, so \(\mathbb{P}(\Omega)=1\). A real-valued measurable random variable \(X\) is integrable when \(\mathbb{E}|X|=\int_\Omega |X|\,d\mathbb{P}<\infty\). Equalities and inequalities between random variables in this tutorial are understood \(\mathbb{P}\)-almost everywhere unless stated otherwise. An interval \(I\subseteq\mathbb{R}\) may be open, closed, half-open, unbounded, or a singleton. A finite convex function \(\varphi:I\to\mathbb{R}\) satisfies \(\varphi(\lambda s+(1-\lambda)t)\le\lambda\varphi(s)+(1-\lambda)\varphi(t)\) for all \(s,t\in I\) and \(\lambda\in[0,1]\).
We first establish the supporting-line fact used in the proof. For an interior point \(m\) of \(I\), a supporting line means an affine function \(t\mapsto\varphi(m)+a(t-m)\) lying below \(\varphi(t)\) for every \(t\in I\).
Proof. For \(x\in I\) with \(x<m\), define the left secant slope \(s(x)=(\varphi(m)-\varphi(x))/(m-x)\). For \(y\in I\) with \(y>m\), define the right secant slope \(r(y)=(\varphi(y)-\varphi(m))/(y-m)\). Convexity, applied to \(m\) as a convex combination of \(x\) and \(y\), gives \(s(x)\le r(y)\). Since \(m\) is interior, choose fixed \(x_0<m<y_0\) in \(I\). Every left slope is bounded above by \(r(y_0)\), and the set of left slopes is nonempty and contains the finite value \(s(x_0)\). Thus \(a=\sup_{x<m}s(x)\) is finite. For every \(x<m\), \(s(x)\le a\), which rearranges to \(\varphi(x)\ge\varphi(m)+a(x-m)\). For every \(y>m\), \(r(y)\ge s(x)\) for all \(x<m\), and hence \(r(y)\ge a\); this rearranges to \(\varphi(y)\ge\varphi(m)+a(y-m)\). At \(t=m\) equality holds. These inequalities prove the assertion. \(\square\)
Proof of the Integral Inequality
Proof. We first check that the mean belongs to \(I\), including the possibility of excluded endpoints. If \(I\) has a finite lower endpoint \(\alpha\), then \(X\ge\alpha\) almost everywhere, so \(m\ge\alpha\). If \(\alpha\notin I\), then \(X-\alpha>0\) almost everywhere. The measurable sets \(A_n=\{X-\alpha>1/n\}\) have union of probability one. At least one has positive probability, because a countable union of null sets would be null. For such an \(n\), \(\mathbb{E}(X-\alpha)\ge\mathbb{P}(A_n)/n>0\), so \(m>\alpha\). Thus an excluded finite lower endpoint cannot be the mean. If \(I\) has a finite upper endpoint \(\beta\), the same reasoning applied to \(\beta-X\) shows \(m\le\beta\), with strict inequality when \(\beta\notin I\). These endpoint bounds, together with the fact that \(I\) is an interval, give \(m\in I\).
If \(m\) is an included lower endpoint \(\alpha\), then \(X-\alpha\ge0\) and \(\mathbb{E}(X-\alpha)=m-\alpha=0\). For each \(n\ge1\), the set \(\{X-\alpha>1/n\}\) must have probability zero, since otherwise the expectation would be positive. Their union is \(\{X>\alpha\}\), so \(X=\alpha\) almost everywhere. Therefore \(\varphi(X)=\varphi(\alpha)=\varphi(m)\) almost everywhere, and Jensen holds with equality. If \(m\) is an included upper endpoint, apply the same argument to \(\beta-X\). If \(I\) is a singleton, the lower-endpoint argument applies directly.
It remains to consider \(m\) in the interior of \(I\). By the lemma, choose a finite \(a\in\mathbb{R}\) such that \(\varphi(X)\ge\varphi(m)+a(X-m)\) almost everywhere. The function \(\varphi\) is Borel measurable on \(I\) (finite convex functions on intervals are Borel measurable), so \(\varphi(X)\) is measurable. The affine function \(\varphi(m)+a(X-m)\) is integrable because \(X\) is integrable. The supporting-line bound implies that the negative part of \(\varphi(X)\) is bounded above by the absolute value of this integrable affine function. Thus \(\varphi(X)\) has a well-defined extended integral, possibly \(+\infty\), and we may integrate the pointwise inequality. Linearity for integrable functions and \(\mathbb{E}(X-m)=0\) give $$ \mathbb{E}[\varphi(X)]\ge \mathbb{E}[\varphi(m)+a(X-m)] =\varphi(m)+a(\mathbb{E}X-m)=\varphi(m). $$ This proves the theorem in the interior case as well as the endpoint cases. \(\square\)
Worked Applications
Worked Example: A Finite Probability Space
Let \(\Omega=\{1,2,3\}\), with \(\mathbb{P}(\{1\})=1/2\), \(\mathbb{P}(\{2\})=1/3\), and \(\mathbb{P}(\{3\})=1/6\). These probabilities sum to \(1/2+1/3+1/6=1\). Set \(X(1)=-1\), \(X(2)=0\), \(X(3)=2\), and take the convex function \(\varphi(t)=|t|\) on \(I=\mathbb{R}\).
First compute the mean: $$ \mathbb{E}X=\frac12(-1)+\frac13(0)+\frac16(2)=-\frac12+\frac13=-\frac16. $$ Thus \(\varphi(\mathbb{E}X)=|-1/6|=1/6\). The transformed expectation is $$ \mathbb{E}[\varphi(X)]=\frac12|-1|+\frac13|0|+\frac16|2|=\frac12+\frac13=\frac56. $$ Consequently \(1/6\le5/6\), as Jensen asserts. For an independent supporting-line check, at \(m=-1/6\) the line with slope \(-1\) is \(t\mapsto -t\), and \(|t|\ge -t\) for every real \(t\). Its expected value is \(-\mathbb{E}X=1/6\), no larger than the computed \(5/6\). Check: the probability-weighted mean of the three values is \(-1/6\), and applying absolute value gives exactly the left side \(1/6\).
Worked Example: A Supporting Line at a Corner
On \(\Omega=\{-1,1\}\), give each point probability \(1/2\), let \(X(\omega)=\omega\), and set \(\varphi(t)=\max\{0,t\}\) on \(\mathbb{R}\). This function is convex: it is the maximum of the two affine functions \(0\) and \(t\). We have \(\mathbb{E}X=(-1+1)/2=0\), and \(\varphi(\mathbb{E}X)=0\). At the two points, \(\varphi(-1)=0\) and \(\varphi(1)=1\), so \(\mathbb{E}[\varphi(X)]=(0+1)/2=1/2\). Hence \(0\le1/2\).
The mean is a corner of \(\varphi\), so a supporting slope need not be unique. The slope \(a=1/2\) works: for \(t\le0\), \(\varphi(t)=0\ge t/2\), and for \(t\ge0\), \(\varphi(t)=t\ge t/2\). Thus \(\varphi(t)\ge\varphi(0)+(1/2)t\) on all of \(\mathbb{R}\). Integrating gives \(1/2\ge0+(1/2)\mathbb{E}X=0\). Check: the direct computation and the supporting-line bound both give the asserted inequality.
Worked Example: The Transformed Integral Can Be Infinite
Take \(\Omega=\{1,2,\ldots\}\) with all subsets measurable and \(\mathbb{P}(\{n\})=2^{-n}\). The probabilities sum to \(1\), since \(\sum_{n=1}^{\infty}2^{-n}=1\). Set \(X(n)=2^{n/2}\) and \(\varphi(t)=t^2\) on \(\mathbb{R}\), a finite convex function. The variable \(X\) is integrable: with \(r=1/\sqrt2\), its expectation is \(\sum_{n\ge1}2^{-n/2}=\sum_{n\ge1}r^n=r/(1-r)=1/(\sqrt2-1)=\sqrt2+1\).
But \(\mathbb{E}[\varphi(X)]=\sum_{n\ge1}2^{-n}(2^{n/2})^2=\sum_{n\ge1}1=+\infty\). Therefore Jensen reads \((\sqrt2+1)^2\le+\infty\), with \((\sqrt2+1)^2=3+2\sqrt2\). Check: the partial sum of the transformed expectation through \(n=N\) is \(N\), which tends to \(+\infty\); the partial sums of the original expectation tend to the finite geometric-series value \(\sqrt2+1\). Thus integrability of \(X\) does not imply integrability of \(\varphi(X)\).
A Common Supporting-Line Error
A supporting slope must be checked on the whole domain, not inferred just from the value of the function at the mean. For \(\varphi(t)=\max\{0,t\}\) at \(m=0\), the proposed slope \(a=-1\) would require \(\varphi(t)\ge -t\) for every \(t\). At \(t=-1\), this says \(0\ge1\), which is false. The correction is to choose a valid slope, such as \(a=1/2\): for \(t\le0\), \(0\ge t/2\), and for \(t\ge0\), \(t\ge t/2\). The resulting line is genuinely supporting, and only then can it be integrated.
The proof’s essential step is the pointwise affine lower bound at the mean. It converts convexity into an inequality involving an integrable function of \(X\), whose expectation is determined by \(\mathbb{E}X=m\). The endpoint argument is not optional: an excluded finite endpoint cannot be the mean, whereas an included endpoint mean forces \(X\) to equal that endpoint almost everywhere. Keeping these cases separate prevents an unjustified use of an interior supporting line.
Check Your Understanding
For each question, specify which assumptions make the integral inequality meaningful.
- Why can the negative part of \(\varphi(X)\) not have infinite integral under the theorem’s hypotheses?
- If \(I=(0,1)\) and \(X\in I\) almost everywhere, explain why \(\mathbb{E}X\) cannot equal \(0\) or \(1\).
- For \(\varphi(t)=|t|\) at an interior mean \(m>0\), give a supporting slope and verify the supporting inequality on each side of \(m\).
- In the corner example, test whether slope \(a=-1\) is supporting by evaluating the required inequality at \(t=-1\). Then verify slope \(a=1/2\).
- Give an example or explanation showing why \(\mathbb{E}|X|<\infty\) does not by itself ensure \(\mathbb{E}[\varphi(X)]<\infty\).