Three Conditional-Expectation Inequalities
Martingales encode a balance of conditional expectations: given the present information, the future value has the present value as its mean. Submartingales and supermartingales relax that equality in opposite directions. The distinction matters when transforming a process. Convexity alone often gives a useful inequality, but whether the transformed process remains a submartingale can also depend on monotonicity.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), carrying a Brownian motion \(B_t\). The definitions below apply to adapted processes, not only to Brownian motion. No stochastic integral is needed here. A random variable is integrable when its absolute value has finite expectation; conditional expectations and inequalities between random variables are understood up to almost-sure equality or inequality.
The inequalities compare conditional means, not individual sample paths. Taking expectations in the submartingale inequality gives \(\mathbb{E}[X_t]\ge\mathbb{E}[X_s]\); for a supermartingale the inequality is reversed. These unconditional comparisons are necessary but do not by themselves establish the defining conditional inequalities. For \(s=t\), all three definitions reduce to equality, since \(X_t\) is \(\mathcal{F}_t\)-measurable.
Worked Example: Deterministic Drift and Classification
Let \(X_t=t\) and \(Y_t=-t\), with the processes adapted to the given filtration. For \(s\le t\), both variables are deterministic, so \(\mathbb{E}[X_t\mid\mathcal{F}_s]=t\) and \(\mathbb{E}[Y_t\mid\mathcal{F}_s]=-t\). Since \(t\ge s\), \(X\) is a submartingale; since \(-t\le -s\), \(Y\) is a supermartingale. If \(s<t\), then \(t\ne s\) and \(-t\ne -s\), so neither process is a martingale. At \(s=t\), equality holds as required. Both are integrable because their values are finite constants.
Check: At \(s=1,t=3\), the conditional means are \(3\) and \(-3\), while the present values are \(1\) and \(-1\). The required inequalities are \(3\ge1\) and \(-3\le-1\), respectively.
Conditional Jensen’s Inequality
The key tool is the conditional form of Jensen’s inequality. If \(\mathcal{G}\subseteq\mathcal{F}\) is a sigma-field, \(U\) is integrable, \(\varphi:\mathbb{R}\to\mathbb{R}\) is finite and convex, and \(\varphi(U)\) is integrable, then \(\varphi(\mathbb{E}[U\mid\mathcal{G}])\le\mathbb{E}[\varphi(U)\mid\mathcal{G}]\) almost surely. The finiteness and integrability conditions ensure that both sides are defined and finite almost surely.
Proof. A finite convex function on \(\mathbb{R}\) has a supporting affine function at every point: for each \(q\in\mathbb{Q}\), choose a slope \(m_q\) such that \(\ell_q(x)=\varphi(q)+m_q(x-q)\le\varphi(x)\) for all \(x\). The supporting-line property follows from the ordering of the left and right slopes of a convex function. Moreover, \(\varphi(x)=\sup_{q\in\mathbb{Q}}\ell_q(x)\): rational \(q\) can approach \(x\), and the slopes of supporting lines are bounded when \(q\) stays in a compact interval, so \(\ell_q(x)\) approaches \(\varphi(x)\). For each \(q\), conditional expectation preserves inequalities and affine functions, giving \(\ell_q(\mathbb{E}[U\mid\mathcal{G}])=\mathbb{E}[\ell_q(U)\mid\mathcal{G}]\le\mathbb{E}[\varphi(U)\mid\mathcal{G}]\) almost surely. There are only countably many rational \(q\), so these inequalities hold simultaneously outside one null set. Taking their supremum proves the stated inequality. \(\square\)
One practical consequence is that applying a convex function to a martingale produces a submartingale, provided the transformed variables remain integrable. A submartingale needs an additional condition: the convex function must be nondecreasing. The next result records both facts, as well as the corresponding concave transformation for supermartingales.
- If \(X\) is a martingale and \(\varphi:\mathbb{R}\to\mathbb{R}\) is convex, then \(\varphi(X)\) is a submartingale whenever \(\varphi(X_t)\) is integrable for every \(t\).
- If \(X\) is a submartingale and \(\varphi\) is convex and nondecreasing, then \(\varphi(X)\) is a submartingale whenever \(\varphi(X_t)\) is integrable for every \(t\).
- If \(X\) is a supermartingale and \(\varphi\) is concave and nondecreasing, then \(\varphi(X)\) is a supermartingale whenever \(\varphi(X_t)\) is integrable for every \(t\).
Proof. Each transformed process is adapted because \(\varphi\) is a measurable real-valued function and \(X_t\) is \(\mathcal{F}_t\)-measurable. Integrability is assumed in each assertion. For the first assertion, fix deterministic \(s\le t\). Conditional Jensen and the martingale equality give \(\mathbb{E}[\varphi(X_t)\mid\mathcal{F}_s]\ge\varphi(\mathbb{E}[X_t\mid\mathcal{F}_s])=\varphi(X_s)\). For the second, conditional Jensen gives \(\mathbb{E}[\varphi(X_t)\mid\mathcal{F}_s]\ge\varphi(\mathbb{E}[X_t\mid\mathcal{F}_s])\). Since \(X\) is a submartingale, \(\mathbb{E}[X_t\mid\mathcal{F}_s]\ge X_s\); because \(\varphi\) is nondecreasing, applying it preserves this inequality. Therefore the conditional expectation is at least \(\varphi(X_s)\). For the third assertion, conditional Jensen for concave \(\varphi\) reverses direction: \(\mathbb{E}[\varphi(X_t)\mid\mathcal{F}_s]\le\varphi(\mathbb{E}[X_t\mid\mathcal{F}_s])\). The supermartingale inequality gives \(\mathbb{E}[X_t\mid\mathcal{F}_s]\le X_s\), and monotonicity gives \(\varphi(\mathbb{E}[X_t\mid\mathcal{F}_s])\le\varphi(X_s)\). This is the required supermartingale inequality. \(\square\)
Convex Transformations in Examples
Worked Example: The Square of Brownian Motion
Show that \(B_t^2\) is a submartingale. Brownian motion is a martingale, and \(x\mapsto x^2\) is convex. For each finite \(t\), \(B_t\) is normally distributed with mean zero and variance \(t\), so \(\mathbb{E}[B_t^2]=t<\infty\). The theorem therefore applies. A direct conditional calculation also gives the inequality: for \(s\le t\), write \(B_t=B_s+(B_t-B_s)\). The increment is independent of \(\mathcal{F}_s\), has mean zero, and has second moment \(t-s\). Expanding the square and conditioning, \(\mathbb{E}[B_t^2\mid\mathcal{F}_s]=B_s^2+2B_s\mathbb{E}[B_t-B_s\mid\mathcal{F}_s]+\mathbb{E}[(B_t-B_s)^2\mid\mathcal{F}_s]=B_s^2+(t-s)\ge B_s^2\). Thus \(B^2\) is a submartingale.
Check: Taking expectations in the conditional identity gives \(\mathbb{E}[B_t^2]=\mathbb{E}[B_s^2]+(t-s)=s+(t-s)=t\), agreeing with the Brownian variance. Equality in the submartingale inequality holds when \(s=t\); it is strict when \(s<t\).
Worked Example: Absolute Value of a Martingale
For Brownian motion, \(|B_t|\) is a submartingale. The function \(\varphi(x)=|x|\) is convex, and \(\mathbb{E}[|B_t|]<\infty\) because \(B_t\) is normal with finite first absolute moment. The convex transformation result for martingales gives \(\mathbb{E}[|B_t|\mid\mathcal{F}_s]\ge|\mathbb{E}[B_t\mid\mathcal{F}_s]|=|B_s|\). At \(s=0\), \(B_0=0\) and \(\mathbb{E}[|B_t|]=\sqrt{2t/\pi}\) for \(t>0\); at \(t=0\), the expectation is \(0\). These values are consistent with the submartingale inequality from the initial time.
Check: Since \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\), the conditional Jensen inequality gives exactly the needed comparison, including when \(B_s\) is negative. No assumption that the process stays nonnegative is required.
Worked Example: A Convex Increasing Transformation of a Submartingale
Let \(Z\) equal \(-1\) or \(3\), each with probability \(1/2\). Take \(\mathcal{F}_t\) to be trivial for \(t<1\) and to contain the information in \(Z\) for \(t\ge1\). Define \(X_t=0\) for \(t<1\) and \(X_t=Z\) for \(t\ge1\). This process is adapted and integrable. If \(s<1\le t\), then \(\mathbb{E}[X_t\mid\mathcal{F}_s]=\mathbb{E}[Z]=((-1)+3)/2=1\ge0=X_s\). If \(s\le t<1\), both values are zero; if \(1\le s\le t\), the process is already known and constant, so the conditional expectation equals \(X_s\). These cases prove that \(X\) is a submartingale.
Now take \(\varphi(x)=x^+=\max(x,0)\), which is convex and nondecreasing. Before time \(1\), \(\varphi(X_t)=0\). At and after time \(1\), it equals \(0\) when \(Z=-1\) and \(3\) when \(Z=3\). For \(s<1\le t\), its conditional expectation is \((0+3)/2=3/2\ge0=\varphi(X_s)\); after time \(1\) it is known, and before time \(1\) it is zero. Hence the transformed process is also a submartingale.
Check: The original conditional mean across the jump is \(1\), while the transformed conditional mean is \(3/2\). Both are at least their respective pre-jump values, which are zero.
Why Absolute Values Need Care
For a martingale \(M\), conditional Jensen applies directly to the convex function \(x\mapsto|x|\), so \(|M|\) is a submartingale when each \(|M_t|\) is integrable. This does not mean absolute value preserves the submartingale property for every submartingale. The transformation theorem for submartingales requires the function to be both convex and nondecreasing; absolute value decreases on the negative half-line.
For a concrete failure, let \(X_t=-1\) for \(t<1\) and \(X_t=0\) for \(t\ge1\), with a deterministic filtration. This is a submartingale: its value increases from \(-1\) to \(0\), and is otherwise constant. But \(|X_t|\) decreases from \(1\) to \(0\). For \(s<1\le t\), \(\mathbb{E}[|X_t|\mid\mathcal{F}_s]=0<1=|X_s|\), so \(|X|\) is not a submartingale. In contrast, \(x^+=\max(x,0)\) is convex and nondecreasing, and is therefore a valid transformation of any submartingale for which the positive parts are integrable.
A common mistake is to cite convexity alone when transforming a submartingale. For example, \(x^2\) is convex, but it is not nondecreasing on all of \(\mathbb{R}\); the submartingale transformation theorem therefore does not apply to \(X^2\) solely from the assumption that \(X\) is a submartingale. For a martingale, convexity alone does suffice, as the square of Brownian motion demonstrates. In every case, check integrability separately: a conditional inequality is not a submartingale assertion unless the transformed process has finite first absolute moment at every time.
| Starting process | Transformation condition | Conclusion |
|---|---|---|
| Martingale | Convex; transformed values integrable | Submartingale |
| Submartingale | Convex and nondecreasing; transformed values integrable | Submartingale |
| Supermartingale | Concave and nondecreasing; transformed values integrable | Supermartingale |
| Submartingale | Convex but not known to be nondecreasing | No general conclusion from this theorem |
Check Your Understanding
For each question, specify the conditional inequality and any integrability or monotonicity condition that is needed.
- If \(X\) is a martingale and \(X_t^2\) is integrable at each time, what property does \(X_t^2\) have, and which feature of the square function is used?
- Why does the conditional Jensen argument for \(|M_t|\) apply when \(M\) is a martingale but not automatically when \(M\) is only a submartingale?
- For a submartingale \(X\), explain why \(X_t^+=\max(X_t,0)\) is a submartingale whenever it is integrable at each time.
- For a supermartingale \(X\), what assumptions on \(\varphi\) allow \(\varphi(X)\) to remain a supermartingale?
- Give a deterministic submartingale whose absolute value is not a submartingale, and identify the pair of times where the defining inequality fails.