Why Convex Transformations Create Submartingales
A martingale has no predictable change in its conditional mean: given the information available at time \(n\), its next conditional expectation is its current value. A convex function, however, weights fluctuations in a way that tends to increase averages. This is why applying a convex function to a martingale generally produces a submartingale, not another martingale. Absolute values and squares are especially useful examples, but the conditions for using them are not identical.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \((\mathcal{F}_n)_{n\ge0}\) be a filtration, and let \((M_n)_{n\ge0}\) be an integrable martingale adapted to this filtration. Thus \(M_n\) is \(\mathcal{F}_n\)-measurable, \(\mathbb{E}|M_n|<\infty\), and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely for every \(n\). A submartingale is an adapted, integrable process \((Y_n)\) satisfying \(\mathbb{E}[Y_{n+1}\mid\mathcal{F}_n]\ge Y_n\) almost surely for every \(n\).
The integrability condition is substantive. If \(\varphi(M_n)\) is not integrable, it cannot be a submartingale under the definition above. For \(\varphi(x)=|x|\), integrability follows from integrability of \(M_n\). For \(\varphi(x)=x^2\), it requires the stronger condition \(\mathbb{E}[M_n^2]<\infty\).
Conditional Jensen’s Inequality
The key inequality is conditional Jensen’s inequality. In the form needed here, if \(X\) and \(\varphi(X)\) are integrable, \(\mathcal{G}\subseteq\mathcal{F}\) is a sigma-field, and \(\varphi:\mathbb{R}\to\mathbb{R}\) is finite and convex, then \(\varphi(\mathbb{E}[X\mid\mathcal{G}])\le\mathbb{E}[\varphi(X)\mid\mathcal{G}]\) almost surely. The conditional expectation is finite almost surely because \(X\) is integrable. Here is a proof for real-valued convex functions.
Proof. For each rational number \(q\), choose a subgradient \(a_q\) of \(\varphi\) at \(q\); that is, choose a real number such that \(\ell_q(x)=\varphi(q)+a_q(x-q)\le\varphi(x)\) for every real \(x\). Such a subgradient exists for a finite convex function on the real line. The supporting-line inequality follows from convexity: the slopes of secant lines on the left of \(q\) are no greater than the slopes on its right, so a slope between the left and right limiting slopes gives a line below the graph.
These countably many lines satisfy \(\varphi(x)=\sup_{q\in\mathbb{Q}}\ell_q(x)\) for every \(x\). To see the reverse inequality to \(\ell_q(x)\le\varphi(x)\), take rationals \(q\) tending to \(x\). A finite convex function is continuous, and its subgradients are bounded when their base points stay in a compact interval around \(x\); consequently \(\ell_q(x)\) tends to \(\varphi(x)\). Put \(Y=\mathbb{E}[X\mid\mathcal{G}]\). For each rational \(q\), \(\ell_q(X)\le\varphi(X)\), so monotonicity and the affine-function property of conditional expectation give \(\ell_q(Y)=\mathbb{E}[\ell_q(X)\mid\mathcal{G}]\le\mathbb{E}[\varphi(X)\mid\mathcal{G}]\) almost surely. There are only countably many rationals, so these inequalities hold together outside one null set. Taking their supremum yields \(\varphi(Y)\le\mathbb{E}[\varphi(X)\mid\mathcal{G}]\) almost surely. \(\square\)
Proof. Since \(M_n\) is \(\mathcal{F}_n\)-measurable and \(\varphi\) is continuous, \(\varphi(M_n)\) is \(\mathcal{F}_n\)-measurable. It is integrable by assumption. Apply conditional Jensen’s inequality with \(X=M_{n+1}\) and \(\mathcal{G}=\mathcal{F}_n\). The martingale property gives \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely, and hence \(\mathbb{E}[\varphi(M_{n+1})\mid\mathcal{F}_n]\ge\varphi(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n])=\varphi(M_n)\) almost surely. These are exactly the submartingale conditions. \(\square\)
Absolute Values and Squares
The function \(\varphi(x)=|x|\) is convex. Because \(|M_n|\) is integrable whenever \(M_n\) is, the theorem applies to every integrable martingale without an additional moment assumption. By contrast, \(x^2\) is convex but its transformation is integrable only if \(M_n\) has a finite second moment. If each \(M_n\) is square-integrable, the theorem says that \((M_n^2)\) is a submartingale.
For squares, an identity makes the source of the submartingale increase explicit. Define the increment \(\Delta M_{n+1}=M_{n+1}-M_n\). If \(M_n\) and \(M_{n+1}\) are square-integrable, then
Indeed, expand \(M_{n+1}^2=(M_n+\Delta M_{n+1})^2\). The term \(M_n\) is \(\mathcal{F}_n\)-measurable, and the martingale condition gives \(\mathbb{E}[\Delta M_{n+1}\mid\mathcal{F}_n]=0\). Thus the conditional expectation of the cross term \(2M_n\Delta M_{n+1}\) is zero, leaving the displayed identity. Its final term is nonnegative, which verifies the submartingale inequality directly.
Worked Example: Absolute Values on a Four-Outcome Space
Let \(\Omega=\{(++),(+-),(-+),(--)\}\), with probability \(1/4\) on each outcome. Let \(\mathcal{F}_0=\{\varnothing,\Omega\}\), let \(\mathcal{F}_1\) reveal the first sign, and let \(\mathcal{F}_2\) reveal both signs. Define \(M_0=0\), \(M_1=1\) for a first \(+\) and \(M_1=-1\) for a first \(-\). Given a first \(+\), let \(M_2=3\) for \(++\) and \(M_2=-1\) for \(+-\). Given a first \(-\), let \(M_2=0\) for \(-+\) and \(M_2=-2\) for \(--\).
First check the martingale conditions. At time zero, \(\mathbb{E}[M_1]=(1+(-1))/2=0=M_0\). Conditional on a first \(+\), the next conditional mean is \((3+(-1))/2=1=M_1\); conditional on a first \(-\), it is \((0+(-2))/2=-1=M_1\). Thus \((M_n)\) is a martingale. All values are finite, so each transformed variable is integrable.
For absolute values, \(\mathbb{E}|M_0|=0\) and \(\mathbb{E}|M_1|=(1+1)/2=1\). At time one, on the first-\(+\) event, \(\mathbb{E}[|M_2|\mid\mathcal{F}_1]=(3+1)/2=2\ge |M_1|=1\). On the first-\(-\) event, it is \((0+2)/2=1=|M_1|\). Averaging at time two gives \(\mathbb{E}|M_2|=(3+1+0+2)/4=3/2\ge1=\mathbb{E}|M_1|\).
Check: The conditional calculations hold on each of the two positive-probability events in \(\mathcal{F}_1\), including equality on the first-\(-\) event. The unconditional means also increase from \(0\) to \(1\) to \(3/2\), consistent with the submartingale property.
Worked Example: A Square Transformation and Its Conditional Variance
Use the same four equally likely sign pairs and filtration. Set \(M_0=1\); let \(M_1=0\) after a first \(+\) and \(M_1=2\) after a first \(-\). Given a first \(+\), set \(M_2=-1\) for \(++\) and \(M_2=1\) for \(+-\). Given a first \(-\), set \(M_2=1\) for \(-+\) and \(M_2=3\) for \(--\).
At time zero, \(\mathbb{E}[M_1]=(0+2)/2=1=M_0\). Conditional on a first \(+\), \(\mathbb{E}[M_2\mid\mathcal{F}_1]=((-1)+1)/2=0=M_1\); conditional on a first \(-\), it is \((1+3)/2=2=M_1\). Thus this is also a martingale. Its values are bounded, so squares are integrable.
The squared values at time one are \(0\) and \(4\), so \(\mathbb{E}[M_1^2]=(0+4)/2=2\), whereas \(M_0^2=1\). Conditional on a first \(+\), the next squared-value mean is \((1+1)/2=1\ge0=M_1^2\). Conditional on a first \(-\), it is \((1+9)/2=5\ge4=M_1^2\). Unconditionally, \(\mathbb{E}[M_2^2]=(1+1+1+9)/4=3\ge2=\mathbb{E}[M_1^2]\).
Check using the identity: In either branch, \(\Delta M_2\) takes the values \(-1\) and \(1\) with equal conditional probabilities, so \(\mathbb{E}[(\Delta M_2)^2\mid\mathcal{F}_1]=(1+1)/2=1\). The identity therefore gives conditional square means \(0^2+1=1\) and \(2^2+1=5\), matching the direct calculations.
Worked Example: Integrability Can Fail for the Square
Let \(r=1/(2\sqrt{2})\), and take \(\Omega=\{(k,s): k\ge1,\ s\in\{-1,1\}\}\). Assign probability \((1-r)r^{k-1}/2\) to each \((k,s)\). These probabilities sum to \((1-r)\sum_{k\ge1}r^{k-1}=1\), since \(0<r<1\). Set \(X(k,s)=s2^k\). With \(\mathcal{F}_0\) trivial and \(\mathcal{F}_1\) the full sigma-field, the process \(M_0=0,\ M_1=X\) is a martingale: the two signs have equal probability at each \(k\), and \(X\) is integrable.
In fact, \(\mathbb{E}|X|=(1-r)\sum_{k\ge1}r^{k-1}2^k=2(1-r)\sum_{k\ge1}(2r)^{k-1}=2(1-r)/(1-2r)<\infty\), because \(2r=1/\sqrt{2}<1\). But \(\mathbb{E}[X^2]=(1-r)\sum_{k\ge1}r^{k-1}4^k=4(1-r)\sum_{k\ge1}(4r)^{k-1}=\infty\), because \(4r=\sqrt{2}>1\), so the summands do not tend to zero.
Check: \(|M_1|\) is integrable, so the absolute-value result applies. The square \(M_1^2\) is not integrable, so the theorem does not assert that \((M_n^2)\) is a submartingale here. This distinction prevents using the square example under only a first-moment assumption.
Common Mistake: Expecting the Square to Remain a Martingale
A frequent error is to square the martingale identity and conclude that \(\mathbb{E}[M_{n+1}^2\mid\mathcal{F}_n]=M_n^2\). Conditional expectation does not commute with squaring. In the second worked example, \(M_0=1\) and \(M_1\) equals \(0\) or \(2\), each with probability \(1/2\). The incorrect calculation would give \(\mathbb{E}[M_1^2]=(\mathbb{E}[M_1])^2=1\); direct calculation gives \((0^2+2^2)/2=2\). The correction is the conditional square identity: the square of a martingale has an additional nonnegative conditional increment-square term. It is therefore a submartingale, and it is a martingale only when that term is zero almost surely at every step.
The same distinction applies to absolute values: convex transformations need not preserve the martingale property, even though they produce submartingales under the stated integrability condition. The conditional inequality can be strict when the next value fluctuates around its current conditional mean, as the first branch of the absolute-value example illustrates.
Check Your Understanding
For each question, state any integrability condition needed and distinguish an almost-sure conditional inequality from a comparison of unconditional expectations.
- Why does \((|M_n|)\) have the required integrability whenever \((M_n)\) is an integrable martingale?
- Suppose \((M_n)\) is square-integrable. Write the conditional square identity using \(\Delta M_{n+1}\), and explain why its final term is nonnegative.
- In the square example, calculate the conditional mean of \(M_2^2\) on the event where \(M_1=2\), and compare it with \(M_1^2\).
- Can the convex-transformation theorem be applied to \(M_n^2\) if only \(\mathbb{E}|M_n|<\infty\) is known? Explain.
- What condition on \(\mathbb{E}[(\Delta M_{n+1})^2\mid\mathcal{F}_n]\) would make the square process satisfy the martingale equality at that step?