Conditional Jensen Inequality

Convexity Under Conditional Expectation

Conditional expectation averages a random variable while retaining information in a sub-sigma-algebra. Convex functions respond to averaging in a definite direction: the convex function of an average is no greater than the average of the convex function. Conditional Jensen’s inequality makes this principle precise, including the integrability conditions needed for both sides to be meaningful.

Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, \(\mathcal{G}\) is a sub-sigma-algebra of \(\mathcal{F}\), and random variables are real-valued unless stated otherwise. Conditional expectations are defined up to almost-sure equality. We use the standard order and linearity properties of conditional expectation and its property of pulling out an integrable \(\mathcal{G}\)-measurable factor when the resulting product is integrable.

Conditional Jensen’s Inequality

Proof. We first establish the supporting-line representation needed for the argument. For each rational \(r\), convexity implies that the secant slopes to the left of \(r\) are no greater than the secant slopes to its right. In particular, the quantities \[ a_r=\sup_{x<r}\frac{\phi(r)-\phi(x)}{r-x}, \qquad b_r=\inf_{x>r}\frac{\phi(x)-\phi(r)}{x-r} \] satisfy \(a_r\le b_r\). They are finite: secant slopes near \(r\) are bounded by slopes to fixed points on either side, and slopes farther away are bounded using those same fixed points and finiteness of \(\phi\). Choose \(s_r\in[a_r,b_r]\). If \(x<r\), then \((\phi(r)-\phi(x))/(r-x)\le s_r\); if \(x>r\), then \((\phi(x)-\phi(r))/(x-r)\ge s_r\). Rearranging either inequality gives \[ \ell_r(x):=\phi(r)+s_r(x-r)\le\phi(x), \] with equality at \(x=r\). Thus \(\ell_r\) is a supporting affine function.

These countably many lines recover \(\phi\): for each \(y\in\mathbb{R}\), take rationals \(r_n\to y\). A finite convex function on \(\mathbb{R}\) is continuous and its supporting slopes are bounded on every compact subinterval of \(\mathbb{R}\). The latter follows from the secant-slope bounds using fixed points on either side of that subinterval. Hence \(s_{r_n}\) is bounded for all sufficiently large \(n\), so \(s_{r_n}(y-r_n)\to0\). Continuity gives \(\ell_{r_n}(y)\to\phi(y)\). Since every \(\ell_r\le\phi\), this proves \(\phi(y)=\sup_{r\in\mathbb{Q}}\ell_r(y)\).

Set \(Y=\mathbb{E}[X\mid\mathcal{G}]\) and \(Z=\mathbb{E}[\phi(X)\mid\mathcal{G}]\); both are integrable and \(Y\) has a \(\mathcal{G}\)-measurable version. For each rational \(r\), \(\ell_r(X)\le\phi(X)\) almost surely. The order and linearity properties of conditional expectation therefore give \[ \ell_r(Y)=\mathbb{E}[\ell_r(X)\mid\mathcal{G}]\le Z \quad\text{almost surely}. \] There are only countably many rationals, so these inequalities hold simultaneously outside one null set. Taking their supremum and using the representation just proved yields \(\phi(Y)\le Z\) almost surely. Finally, choose any one rational \(r_0\). Since \(\ell_{r_0}(Y)\le\phi(Y)\le Z\) and both bounds are integrable, \(\phi(Y)\) is integrable. This proves the claim and the inequality. \(\square\)

The proof needs \(\phi(X)\) to be integrable so that its conditional expectation is an ordinary integrable random variable. It does not require \(\mathbb{E}[X\mid\mathcal{G}]\) to have a particular distribution, nor does it require \(\mathbb{P}\) to be completed.

Worked Applications

Proof. The assumption \(X\in L^p(\mathbb{P})\) implies \(X\in L^1(\mathbb{P})\) because \(\mathbb{P}\) is a probability measure, and \(|X|^p\) is integrable. Thus conditional Jensen applies. Integrating its almost-sure inequality and using the defining integral property of conditional expectation gives \[ \mathbb{E}\big[|\mathbb{E}[X\mid\mathcal{G}]|^p\big] \le \mathbb{E}\big[\mathbb{E}[|X|^p\mid\mathcal{G}]\big] =\mathbb{E}[|X|^p]. \] Taking \(p\)-th roots proves the norm inequality. For \(p=1\), the same argument applies directly with \(\phi(x)=|x|\). \(\square\)

Conditional Variance Bounds

The preceding calculation holds generally. If \(X\) is square-integrable, then conditional Jensen with \(\phi(x)=x^2\) gives \(Y^2\le\mathbb{E}[X^2\mid\mathcal{G}]\) almost surely. In particular, \(Y\) is square-integrable, by integrating this inequality.

Proof. Since \(X,Y\in L^2(\mathbb{P})\), the products \(XY\) and \(Y^2\) are integrable. Expanding the square and using linearity gives \[ \mathbb{E}[(X-Y)^2\mid\mathcal{G}] =\mathbb{E}[X^2\mid\mathcal{G}] -2\mathbb{E}[XY\mid\mathcal{G}] +\mathbb{E}[Y^2\mid\mathcal{G}]. \] The pull-out property gives \(\mathbb{E}[XY\mid\mathcal{G}]=Y\mathbb{E}[X\mid\mathcal{G}]=Y^2\), and \(\mathbb{E}[Y^2\mid\mathcal{G}]=Y^2\) because \(Y^2\) is \(\mathcal{G}\)-measurable. Substitution yields the identity. The variance is nonnegative because it is the conditional expectation of a nonnegative random variable. The upper bound follows from the identity and \(Y^2\ge0\). \(\square\)

A Common Mistake

Conditional Jensen does not usually give equality. On a two-point space with probability \(1/2\) at each point, let \(\mathcal{G}=\{\varnothing,\Omega\}\) and let \(X\) take the values \(-1\) and \(1\). Then \(\mathbb{E}[X\mid\mathcal{G}]=0\), so \(|\mathbb{E}[X\mid\mathcal{G}]|=0\), whereas \(\mathbb{E}[|X|\mid\mathcal{G}]=(1+1)/2=1\).

The incorrect calculation would replace \(|\mathbb{E}[X\mid\mathcal{G}]|\) by \(\mathbb{E}[|X|\mid\mathcal{G}]\) as though they were equal. The correction is the inequality \(|\mathbb{E}[X\mid\mathcal{G}]|\le\mathbb{E}[|X|\mid\mathcal{G}]\); in this example it reads \(0\le1\), strictly. The assumptions matter as well: for the general theorem above, one must check that \(X\) and \(\phi(X)\) are integrable before writing the ordinary conditional expectations in the inequality.