Why Martingale Increments Are Orthogonal
A martingale may have increments that depend strongly on the information already observed. Nevertheless, once an increment is compared with any square-integrable quantity known before that increment occurs, their expected product is zero. In particular, distinct increments have zero \(L^2\) inner product. This fact is useful when studying the geometry of martingales in \(L^2\), and it follows directly from the defining conditional-expectation property rather than from independence.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with an increasing filtration \((\mathcal{F}_n)_{n\ge0}\), meaning \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\subseteq\mathcal{F}\). A real-valued process \((M_n)_{n\ge0}\) is a martingale with respect to this filtration if each \(M_n\) is \(\mathcal{F}_n\)-measurable and integrable, and \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n]=M_n\) almost surely. It is square-integrable at time \(n\) if \(\mathbb{E}[M_n^2]<\infty\). For each \(n\ge1\), define its increment by \(D_n=M_n-M_{n-1}\). The \(L^2\) inner product of square-integrable random variables \(U,V\) is \(\langle U,V\rangle_{L^2}=\mathbb{E}[UV]\); they are orthogonal when this inner product is zero.
Proof. Since \(M_j\) and \(M_{j-1}\) are square-integrable, their difference \(D_j\) is square-integrable: \(\mathbb{E}[D_j^2]\le 2\mathbb{E}[M_j^2]+2\mathbb{E}[M_{j-1}^2]<\infty\). By the martingale property and the measurability of \(M_{j-1}\) with respect to \(\mathcal{F}_{j-1}\), \(\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]=\mathbb{E}[M_j\mid\mathcal{F}_{j-1}]-M_{j-1}=0\) almost surely. For the second assertion, \(UD_j\) is integrable by Cauchy–Schwarz, since \(\mathbb{E}|UD_j|\le(\mathbb{E}[U^2])^{1/2}(\mathbb{E}[D_j^2])^{1/2}<\infty\). If \(U\) is bounded, the defining property of conditional expectation gives \(\mathbb{E}[UD_j]=\mathbb{E}[U\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]]=0\). For a general square-integrable \(U\), let \(U_r=\max(-r,\min(U,r))\). Each \(U_r\) is bounded and \(\mathcal{F}_{j-1}\)-measurable, and \(\mathbb{E}[(U-U_r)^2]\to0\) as \(r\to\infty\), because \((U-U_r)^2\le U^2\mathbf{1}_{\{|U|>r\}}\) and the expectations on the right tend to zero. Therefore \(\mathbb{E}[UD_j]=\lim_{r\to\infty}\mathbb{E}[U_rD_j]=0\), with convergence justified by \(|\mathbb{E}[(U-U_r)D_j]|\le(\mathbb{E}[(U-U_r)^2])^{1/2}(\mathbb{E}[D_j^2])^{1/2}\). \(\square\)
Proof. The increment \(D_i=M_i-M_{i-1}\) is \(\mathcal{F}_i\)-measurable and square-integrable. Since \(i<j\), the filtration property gives \(\mathcal{F}_i\subseteq\mathcal{F}_{j-1}\). Apply the proposition with \(U=D_i\) to obtain \(\mathbb{E}[D_iD_j]=0\). For indices in the reverse order, commutativity of multiplication gives \(\mathbb{E}[D_jD_i]=\mathbb{E}[D_iD_j]=0\). \(\square\)
The key is the timing of the information: \(D_i\) is already known by time \(j-1\), while \(D_j\) has conditional mean zero given that information. The conclusion is an expectation of a product, not a claim that the increments cannot be related. The next examples show how the result can be used both for direct computations and for increments with evident dependence.
Worked Example: Increments of a Fair-Coin Martingale
Let \(\Omega=\{-1,1\}^3\), with each of its eight outcomes assigned probability \(1/8\). Let \(\xi_1,\xi_2,\xi_3\) be the three coordinate values, and let \(\mathcal{F}_n\) be the sigma-algebra generated by the first \(n\) coordinates, with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Define \(M_0=0\) and \(M_n=\sum_{k=1}^n\xi_k\) for \(1\le n\le3\). Show that the distinct increments have zero inner products.
Each coordinate takes values \(1\) and \(-1\) with equal probability, independently of the preceding coordinates. Thus \(\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=0\), so \(M_{n+1}=M_n+\xi_{n+1}\) has conditional expectation \(M_n\). Each \(M_n\) is square-integrable because the sample space is finite, and \(D_n=M_n-M_{n-1}=\xi_n\).
For a direct calculation of the first inner product, the four equally likely pairs \((\xi_1,\xi_2)\) are \((1,1),(1,-1),(-1,1),(-1,-1)\). The products are respectively \(1,-1,-1,1\), so
For \(D_1,D_3\), the four equally likely pairs \((\xi_1,\xi_3)\) have the same possible values and products, so \(\mathbb{E}[D_1D_3]=(1-1-1+1)/4=0\). For \(D_2,D_3\), the four equally likely pairs \((\xi_2,\xi_3)\) also give \(\mathbb{E}[D_2D_3]=(1-1-1+1)/4=0\). These explicit calculations agree with the theorem. As a check, conditioning on \(\mathcal{F}_1\) gives \(\mathbb{E}[D_2\mid\mathcal{F}_1]=0\), and therefore \(\mathbb{E}[D_1D_2]=0\) because \(D_1\) is \(\mathcal{F}_1\)-measurable.
Worked Example: A Conditional-Expectation Martingale
Let \(\Omega=\{1,2,3,4\}\), assigning probability \(1/4\) to each outcome. Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) distinguish the sets \(\{1,2\}\) and \(\{3,4\}\), and let \(\mathcal{F}_2\) contain all subsets of \(\Omega\). Define \(Y\) by \(Y(1)=0,Y(2)=2,Y(3)=1,Y(4)=5\), and set \(M_n=\mathbb{E}[Y\mid\mathcal{F}_n]\) for \(n=0,1,2\). Compute \(\mathbb{E}[D_1D_2]\) directly.
First, \(\mathbb{E}[Y]=(0+2+1+5)/4=2\), so \(M_0=2\) on every outcome. On \(\{1,2\}\), the average of \(Y\) is \((0+2)/2=1\); on \(\{3,4\}\), it is \((1+5)/2=3\). Thus \(M_1\) is \(1,1,3,3\) on outcomes \(1,2,3,4\), respectively, and \(M_2=Y\). The increments \(D_1=M_1-M_0\) and \(D_2=M_2-M_1\) are therefore
Consequently, \(\mathbb{E}[D_1D_2]=(1-1-2+2)/4=0\). To check the conditional-expectation argument, on \(\{1,2\}\) the two values of \(D_2\) average to \((-1+1)/2=0\), and on \(\{3,4\}\) they average to \((-2+2)/2=0\). Hence \(\mathbb{E}[D_2\mid\mathcal{F}_1]=0\), while \(D_1\) is constant on each of the two sets in \(\mathcal{F}_1\). The proposition applies and gives the same inner product.
Worked Example: Orthogonal but Dependent Increments
Let \(\Omega=\{-1,1\}\times\{-1,1\}\), with each of the four pairs \((x,\varepsilon)\) having probability \(1/4\). Let \(\mathcal{F}_0\) be trivial, let \(\mathcal{F}_1\) reveal \(x\), and let \(\mathcal{F}_2\) reveal both coordinates. Set \(M_0=0\), \(M_1=x\), and \(M_2=x+D_2\), where \(D_2=2\varepsilon\) if \(x=1\), and \(D_2=\varepsilon\) if \(x=-1\). Verify the martingale property, compute the inner product, and check whether the increments are independent.
The first increment is \(D_1=x\), and \(\mathbb{E}[D_1]=0\) because the two signs of \(x\) occur equally often. Given \(x\), the sign \(\varepsilon\) is equally likely to be either value, so \(\mathbb{E}[D_2\mid\mathcal{F}_1]=0\): when \(x=1\), the two possible values are \(2\) and \(-2\), with mean \(0\); when \(x=-1\), they are \(1\) and \(-1\), also with mean \(0\). Thus \(\mathbb{E}[M_1\mid\mathcal{F}_0]=0=M_0\) and \(\mathbb{E}[M_2\mid\mathcal{F}_1]=M_1\). Both variables are square-integrable on this finite space.
The four values of \(D_1D_2=xD_2\), listed for \((x,\varepsilon)=(1,1),(1,-1),(-1,1),(-1,-1)\), are \(2,-2,-1,1\). Therefore
Yet the increments are not independent. The event \(\{|D_2|=2\}\) occurs exactly when \(D_1=1\), so \(\mathbb{P}(|D_2|=2,D_1=1)=1/2\). If these events were independent, this probability would instead be \(\mathbb{P}(|D_2|=2)\mathbb{P}(D_1=1)=(1/2)(1/2)=1/4\). The unequal probabilities verify dependence directly. Orthogonality controls the expected product; it does not require independence.
Using the Conditional-Mean-Zero Test
In applications, it is often easier to check a conditional expectation than to list all possible products. To prove that an earlier variable \(U\) and a later increment \(D_j\) have zero inner product, check three points: \(U\) is square-integrable, \(U\) is measurable with respect to \(\mathcal{F}_{j-1}\), and \(\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]=0\). The proposition then gives \(\mathbb{E}[UD_j]=0\). In particular, when \(i<j\), the increment \(D_i\) is measurable with respect to \(\mathcal{F}_{j-1}\), which is precisely why the orthogonality theorem applies.
Common mistake: treating orthogonality as independence. In the preceding example, \(\mathbb{E}[D_1D_2]=0\), but \(|D_2|\) is determined by \(D_1\). For a concrete incorrect calculation, orthogonality might be misread as independence and used to assert \(\mathbb{E}[D_2^2\mid D_1=1]=\mathbb{E}[D_2^2]\). The unconditional value is \(\mathbb{E}[D_2^2]=(4+4+1+1)/4=10/4=5/2\), whereas conditional on \(D_1=1\), the two possible values of \(D_2\) are \(2\) and \(-2\), giving \(\mathbb{E}[D_2^2\mid D_1=1]=(4+4)/2=4\). The correction is to use only what orthogonality states: the expected product of the distinct increments is zero. No independence conclusion follows without additional assumptions.
The square-integrability assumptions ensure the inner products exist as finite expectations, and the filtration specifies what information is available before each increment. If an increment is identically zero, the same proof still applies: its conditional mean and every inner product involving it are zero. Thus no nondegeneracy assumption is needed.
Check Your Understanding
Use conditional expectation and the information available at each time to answer the following.
- For \(i<j\), why is \(D_i\) measurable with respect to \(\mathcal{F}_{j-1}\)?
- Which martingale property gives \(\mathbb{E}[D_j\mid\mathcal{F}_{j-1}]=0\)?
- In the four-outcome conditional-expectation example, compute \(\mathbb{E}[D_1D_2]\) by conditioning on the two sets in \(\mathcal{F}_1\).
- In the dependent-increments example, compute \(\mathbb{E}[D_2^2\mid D_1=-1]\) and compare it with \(\mathbb{E}[D_2^2]\).
- What integrability condition ensures that \(\mathbb{E}[UD_j]\) is finite when both variables are used in the inner product?