# Almost Sure

## 18 June 12

### Rao’s Quasimartingale Decomposition

In this post I’ll give a proof of Rao’s decomposition for quasimartingales. That is, every quasimartingale decomposes as the sum of a submartingale and a supermartingale. Equivalently, every quasimartingale is a difference of two submartingales, or alternatively, of two supermartingales. This was originally proven by Rao (Quasi-martingales, 1969), and is an important result in the general theory of continuous-time stochastic processes.

As always, we work with respect to a filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge0},{\mathbb P})}$. It is not required that the filtration satisfies either of the usual conditions — the filtration need not be complete or right-continuous. The methods used in this post are elementary, requiring only basic measure theory along with the definitions and first properties of martingales, submartingales and supermartingales. Other than referring to the definitions of quasimartingales and mean variation given in the previous post, there is no dependency on any of the general theory of semimartingales, nor on stochastic integration other than for elementary integrands.

Recall that, for an adapted integrable process X, the mean variation on an interval ${[0,t]}$ is

$\displaystyle {\rm Var}_t(X)=\sup{\mathbb E}\left[\int_0^t\xi\,dX\right],$

where the supremum is taken over all elementary processes ${\xi}$ with ${\vert\xi\vert\le1}$. Then, X is a quasimartingale if and only if ${{\rm Var}_t(X)}$ is finite for all positive reals t. It was shown that all supermartingales are quasimartingales with mean variation given by

 $\displaystyle {\rm Var}_t(X)={\mathbb E}\left[X_0-X_t\right].$ (1)

Rao’s decomposition can be stated in several different ways, depending on what conditions are required to be satisfied by the quasimartingale X. As the definition of quasimartingales does differ between texts, there are different versions of Rao’s theorem around although, up to martingale terms, they are equivalent. In this post, I’ll give three different statements with increasingly stronger conditions for X. First, the following statement applies to all quasimartingales as defined in these notes. Theorem 1 can be compared to the Jordan decomposition, which says that any function ${f\colon{\mathbb R}_+\rightarrow{\mathbb R}}$ with finite variation on bounded intervals can be decomposed as the difference of increasing functions or, equivalently, of decreasing functions. Replacing finite variation functions by quasimartingales and decreasing functions by supermartingales gives the following.

Theorem 1 (Rao) A process X is a quasimartingale if and only if it decomposes as

 $\displaystyle X=Y-Z$ (2)

for supermartingales Y and Z. Furthermore,

• this decomposition can be done in a minimal sense, so that if ${X=Y^\prime-Z^\prime}$ is any other such decomposition then ${Y^\prime-Y=Z^\prime-Z}$ is a supermartingale.
• the inequality
 $\displaystyle {\rm Var}_t(X)\le{\mathbb E}[Y_0-Y_t]+{\mathbb E}[Z_0-Z_t],$ (3)

holds, with equality for all ${t\ge0}$ if and only if the decomposition is minimal.

• the minimal decomposition is unique up to a martingale. That is, if ${X=Y-Z=Y^\prime-Z^\prime}$ are two such minimal decompositions, then ${Y^\prime-Y=Z^\prime-Z}$ is a martingale.

By replacing Y and Z by –Z and –Y, this theorem can alternatively be written in terms of decompositions into the difference of submartingales, which is how it is sometimes stated. I stick to using supermartingales in this post, in order to be consistent with the alternative statements of Rao’s theorem given below. Also, in the statement above, the processes Y and Z occurring in the decomposition are not required to be cadlag. This is because we are not requiring X to satisfy any pathwise properties or putting any restrictions on the underlying filtration. However, as we will show in Lemma 5 below, if X is cadlag and the filtration satisfies the usual conditions then Y and Z do have cadlag modifications.

In the statement of Theorem 1, the fact that inequality (3) holds for all such decompositions follows immediately from identity (1) for the mean variation of a supermartingale together with the fact that ${{\rm Var}_t(\cdot)}$ is subadditive on the space of integrable processes. However, the fact that equality holds for the minimal decomposition is more advanced, and this will be left until the proof of the theorem given further below. Also, the statement that the minimal decomposition is unique up to a martingale follows easily from the definitions. If ${X=Y-Z=Y^\prime-Z^\prime}$ were two minimal decompositions then, by definition, both ${Y-Y^\prime}$ and ${-(Y-Y^\prime)}$ are supermartingales, so ${Y-Y^\prime}$ is a martingale.

One possible problem with the decomposition in Theorem 1 is that it is not unique. We only obtain uniqueness up to a martingale term. However, by placing slightly stronger conditions on X, it is possible to obtain a unique minimal decomposition. The mean variation of an integrable process X on ${[0,\infty)}$ was defined as

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}(X)&\displaystyle=\lim_{t\rightarrow\infty}{\rm Var}_t(X)\smallskip\\ &\displaystyle=\sup{\mathbb E}\left[\int_0^\infty\xi\,dX\right]. \end{array}$

Here, the supremum is taken over all elementary processes ${\xi}$ with ${\vert\xi\vert\le1}$. In particular, for a nonnegative supermartingale X, applying (1) to ${-X}$ gives

 $\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}(X)&\displaystyle=\lim_{t\rightarrow\infty}{\mathbb E}[X_0-X_t]\smallskip\\ &\displaystyle\le{\mathbb E}[X_0] \end{array}$ (4)

with equality if and only if ${X_t\rightarrow0}$ in ${L^1}$ as t goes to infinity.

The following version of Rao’s decomposition can be compared to the statement that any function ${f\colon{\mathbb R}_+\rightarrow{\mathbb R}}$ with bounded total variation decomposes as ${f=c+g-h}$, where c is constant and ${g,h}$ are nonnegative functions decreasing to zero. Similarly as above, the statement of Theorem 2 is obtained by replacing functions of bounded total variation by processes with ${{\rm Var}(X) < \infty}$, replacing the constant term by a martingale, and replacing decreasing functions by supermartingales. One technical point should be made before stating this theorem. As I am not assuming that any of the processes have nice pathwise properties, such as being cadlag, it does not make sense to consider two processes X and Y to be equal only if they are equal up to evanescence (i.e., that ${X_t=Y_t}$ simultaneously for all t, with probability 1). This is not determined by the joint distributions of X and Y at finite sets of times, and need not even be a measurable property. Instead, I consider X and Y to be equal if the weaker condition ${{\mathbb P}(X_t=Y_t)=1}$ for each time t is satisified. Similarly, a process X will be called nonnegative if ${{\mathbb P}(X_t\ge0)=1}$ for each time t.

Theorem 2 (Rao) An integrable adapted process X satisfies ${{\rm Var}(X) < \infty}$ if and only if it decomposes as

 $\displaystyle X=M+Y-Z$ (5)

for a martingale M and nonnegative supermartingales Y, Z. Furthermore,

• this decomposition can be done in a minimal sense, so that if ${X=M^\prime+Y^\prime-Z^\prime}$ is any other such decomposition then ${Y^\prime-Y}$ and ${Z^\prime-Z}$ are nonnegative supermartingales.
• the inequality
 $\displaystyle {\rm Var}(X)\le{\mathbb E}[Y_0+Z_0].$ (6)

holds, with equality if and only if the decomposition is minimal.

• the minimal decomposition is unique in the sense that, if ${X=M^\prime+Y^\prime-Z^\prime}$ is any other such minimal decomposition then, ${M_t=M_t}$, ${Y_t=Y^\prime_t}$ and ${Z_t=Z^\prime_t}$ (almost surely) for each ${t\in{\mathbb R}_+}$.
• in the minimal decomposition, the limits ${\lim_{t\rightarrow\infty}Y_t=\lim_{t\rightarrow\infty}Z_t=0}$ hold almost surely and in ${L^1}$, and the martingale M is almost surely zero at each time if and only if ${\lim_{t\rightarrow\infty}X_t=0}$ in ${L^1}$.

As we will see in Lemma 4 below, the supermartingales Y,Z in minimal decomposition (5) do agree with those in (2) up to martingale terms. The proof of Theorem 2 will be left until later on in this post. Note, however that inequality (6) is an immediate consequence of (4) for the mean variation of a supermartingale together with the fact that ${{\rm Var}(\cdot)}$ is a subadditive function on the integrable processes and that the martingale M has zero mean variation. Uniqueness of the minimal decomposition is also immediate. If ${X=M^\prime+Y^\prime-Z^\prime}$ was any other minimal decomposition then, by definition, both ${Y-Y^\prime}$ and ${Y^\prime-Y}$ are nonnegative, so ${Y=Y^\prime}$. Similarly, ${Z=Z^\prime}$ and, hence, ${M=M^\prime}$.

Let us now move on to the third version of Rao’s decomposition, which gives a unique minimal decomposition, but does not involve the additional martingale term as in (5). For any integrable adapted process X, we defined the following increasing function of t,

$\displaystyle {\rm Var}^*_t(X)={\rm Var}_t(X)+{\mathbb E}[\vert X_t\vert].$

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}^*(X)&\displaystyle=\lim_{t\rightarrow\infty}{\rm Var}^*_t(X)\smallskip\\ &\displaystyle={\mathbb E}\left[\int_0^\infty\xi\,dX\right], \end{array}$

where the supremum is taken over all elementary process ${\xi}$ with index set ${[0,\infty]}$ and satisfying ${\vert\xi\vert\le1}$. In the case where X is not integrable, then ${{\rm Var}^*(X)}$ is defined to be infinite. For nonnegative supermartingales,

$\displaystyle {\rm Var}^*_t(X)={\mathbb E}[X_0-X_t]+{\mathbb E}[\vert X_t\vert]={\mathbb E}[X_0]$

independently of t. So, every nonnegative supermartingale satisfies

 $\displaystyle {\rm Var}^*(X)={\mathbb E}[X_0].$ (7)

We now state the third version of Rao’s decomposition, which applies to all integrable adapted processes with ${{\rm Var}^*(X)}$ finite.

Theorem 3 (Rao) An adapted process X satisfies ${{\rm Var}^*(X) < \infty}$ if and only if it decomposes as

 $\displaystyle X=Y-Z$ (8)

for nonnegative supermartingales Y and Z. Furthermore,

• this decomposition can be done in a minimal sense, so that if ${X=Y^\prime-Z^\prime}$ is any other such decomposition then ${Y^\prime-Y=Z^\prime-Z}$ is a nonnegative supermartingale.
• the inequality
 $\displaystyle {\rm Var}^*(X)\le{\mathbb E}[Y_0+Z_0].$ (9)

holds, with equality if and only if the decomposition is minimal.

• the minimal decomposition is unique in the sense that, if ${X=Y-Z=Y^\prime-Z^\prime}$ are two such minimal decompositions, then ${Y_t=Y^\prime_t}$ and ${Z_t=Z^\prime_t}$ (almost surely) for each ${t\ge0}$.

Note that inequality (9) is a straightforward consequence of identity (7) for the mean variation of a nonnegative supermartingale together with the fact that ${{\rm Var}^*(\cdot)}$ is subadditive. Also, as in the statement of Theorem 2, uniqueness of the minimal decomposition follows immediately from the definition of minimality.

We can show that the supermartingales Y and Z in each of the three alternative statements of Rao’s decomposition do agree up to a martingale term.

Lemma 4 Let X be a quasimartingale and ${X=Y^\prime-Z^\prime}$ be a decomposition as the difference of supermartingales which is minimal in the sense of Theorem 1.

If the minimal decomposition (5) exists then ${Y-Y^\prime}$ and ${Z-Z^\prime}$ are martingales. Similarly, if the minimal decomposition (8) exists then ${Y-Y^\prime}$ and ${Z-Z^\prime}$ are martingales.

Proof: If ${X=Y-Z}$ is the minimal decomposition (8) then, it needs to be shown that this is also minimal in the sense of Theorem 1, so that uniqueness of minimal decompositions will imply that ${Y-Y^\prime=Z-Z^\prime}$ is a martingale.

By definition of minimality in (8), if ${X=\tilde Y-\tilde Z}$ for nonnegative supermartingales ${\tilde Y,\tilde Z}$, then ${\tilde Y-Y=\tilde Z-Z}$ is a supermartingale. It just needs to be shown that this still holds when the processes are not assumed to be nonnegative. However, in that case, we can stop ${\tilde Y,\tilde Z}$ at any fixed time T to define new supermartingales

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y^{\prime\prime}_t=\tilde Y_{T\wedge t}+(Y_t-Y_{T\wedge t})+{\mathbb E}[Y_T-\tilde Y_T\;\vert\mathcal{F}_t],\smallskip\\ &\displaystyle Z^{\prime\prime}_t=\tilde Z_{T\wedge t}+(Z_t-Z_{T\wedge t})+{\mathbb E}[Z_T-\tilde Z_T\;\vert\mathcal{F}_t]. \end{array}$

Then, ${Y^{\prime\prime}_t=Y_t}$ for all ${t\ge T}$, from which we have

$\displaystyle Y^{\prime\prime}_t\ge{\mathbb E}[Y_T\;\vert\mathcal{F}_t]\ge0$

for all ${t\le T}$. So ${Y^{\prime\prime}}$ and, similarly, ${Z^{\prime\prime}}$ are nonnegative supermartingales. Furthermore, the equality ${X=Y-Z=\tilde Y-\tilde Z}$ implies that ${X=Y^{\prime\prime}-Z^{\prime\prime}}$. By minimality in decomposition (8),

$\displaystyle Y^{\prime\prime}_t-Y_t=\tilde Y_t-Y_t+{\mathbb E}[Y_T-\tilde Y_T\;\vert\mathcal{F}_t]$

is a (nonnegative) supermartingale over ${t\le T}$. As the final term on the right hand side is a martingale and T was arbitrary, it follows that ${\tilde Y-Y}$ is a supermartingale and, similarly, so is ${\tilde Z-Z}$, as was required.

Next, suppose that the minimal decomposition (5) exists. Then, setting ${X^\prime=X-M}$, minimality implies that if ${X^\prime=\tilde Y-\tilde Z}$ is any other decomposition as the difference of nonnegative supermartingales then, ${\tilde Y-Y=\tilde Z-Z}$ are nonnegative supermartingales. So, the decomposition ${X^\prime=Y-Z}$ is also minimal in the sense of Theorem 3 and, by what we have already shown above, is also minimal in the sense of Theorem 1. So, ${X=Y-(Z+M)}$ is a minimal decomposition of X as the difference of supermartingales in the sense of Theorem 1 and, hence, ${Y-Y^\prime}$ is a martingale. Similarly, ${Z-Z^\prime}$ is a martingale. ⬜

Finally, before moving on to the proof of Rao’s decomposition, we show that the supermartingales Y and Z do indeed have a cadlag modification under suitable conditions.

Lemma 5 Suppose that X is a cadlag process and that the filtration satisfies the usual conditions.

Whenever either of the minimal decompositions (2), (5) and (8) exist then, the supermartingales Y and Z have a cadlag modification.

Proof: First, as the underlying filtration is assumed to satisfy the usual conditions, every martingale has a cadlag modification. As Lemma 4 says that the minimal decompositions (5) and (8) reduce to (2) up to a martingale term, it is enough to just consider decomposition (2).

So, suppose that ${X=Y-Z}$ is a minimal decomposition of X as the difference of supermartingales, and let T be any fixed time. Choose a sequence of times ${T_n}$ strictly decreasing to T. Then, as in the proof of existence of cadlag modifications of submartingales and supermartingales, the limits

$\displaystyle Y_{T+}=\lim_{n\rightarrow\infty}Y_{T_n},\ Z_{T+}=\lim_{n\rightarrow\infty}Z_{T_n}$

exist in ${L^1}$. Furthermore, as the filtration satisfies the usual conditions, ${Y_{T+}}$ and ${Z_{T+}}$ are ${\mathcal{F}_T}$-measurable. The supermartingale property implies that

$\displaystyle Y_T\ge{\mathbb E}[Y_{T+}\;\vert\mathcal{F}_T]=Y_{T+}$

and, similarly, ${Z_T\ge Z_{T+}}$. We now define new supermartingales by subtracting out the jump at time T,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y^\prime_t = Y_t+1_{\{t > T\}}(Y_T-Y_{T+}),\smallskip\\ &\displaystyle Z^\prime_t = Z_t+1_{\{t > T\}}(Z_T-Z_{T+}). \end{array}$

These are indeed supermartingales. If ${s\le T < t}$ then, for any ${T_n < t}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}[Y^\prime_t\;\vert\mathcal{F}_s]&\displaystyle={\mathbb E}[Y_t+Y_T-Y_{T+}\;\vert\mathcal{F}_s]\smallskip\\ &\displaystyle\le{\mathbb E}[Y_{T_n}-Y_{T+}+Y_T\;\vert\mathcal{F}_s]. \end{array}$

Taking the limit as n goes to infinity gives,

$\displaystyle {\mathbb E}[Y^\prime_t\;\vert\mathcal{F}_s]\le{\mathbb E}[Y_T\;\vert\mathcal{F}_s]\le Y_s=Y^\prime_s.$

This inequality also follows immediately from the supermartingale property for Y in the case that T is not contained in the interval ${[s,t)}$. So ${Y^\prime}$ and, similarly, ${Z^\prime}$ are supermartingales. As X is right-continuous,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle Y^\prime_t-Z^\prime_t&\displaystyle=Y_t-Z_t+1_{\{t > T\}}(Y_T-Z_T-Y_{T+}+Z_{T+})\smallskip\\ &\displaystyle=X_t+1_{\{t > T\}}(X_T-X_{T+})=X_t. \end{array}$

By minimality of the decomposition, this implies that ${X^\prime-X}$ is a supermartingale and, hence,

$\displaystyle {\mathbb E}[Y_T-Y_{T+}]={\mathbb E}[Y^\prime_T-Y_T]\le{\mathbb E}[Y^\prime_0-Y_0]=0.$

As ${Y_T-Y_{T+}}$ is nonnegative, this shows that ${Y_{T+}=Y_T}$ and, hence, Y is right-continuous in probability. Therefore, it has a cadlag modification. Similarly, Z has a cadlag modification. ⬜

#### Proof of Rao’s Decomposition

In discrete time, Rao’s decomposition is particularly easy to prove. So, I’ll start by describing the decomposition in this setting and, then, generalize to continuous time by taking limits along a sequence of partitions.

Consider any integrable adapted process ${X_t}$ with time index t running through a finite set ${\{t_0 < t_1 < \cdots < t_n\}}$, and define the nonnegative adapted processes

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y_{t_j}={\mathbb E}\left[\sum_{k=j+1}^n{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]_-\;\bigg\vert\mathcal{F}_{t_j}\right],\smallskip\\ &\displaystyle Z_{t_j}={\mathbb E}\left[\sum_{k=j+1}^n{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]_+\;\bigg\vert\mathcal{F}_{t_j}\right]. \end{array}$

Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle Y_{t_j}-Z_{t_j}&\displaystyle={\mathbb E}\left[-\sum_{k=j+1}^n{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\big\vert\mathcal{F}_{t_{k-1}}\right]\;\bigg\vert\mathcal{F}_{t_j}\right]\smallskip\\ &\displaystyle=\sum_{k=j+1}^n{\mathbb E}\left[X_{t_{k-1}}-X_{t_k}\;\vert\mathcal{F}_{t_j}\right]\smallskip\\ &\displaystyle=X_{t_j}-{\mathbb E}\left[X_{t_n}\;\vert\mathcal{F}_{t_j}\right]. \end{array}$

So, let us suppose now that ${X_{t_n}=0}$, in which case we obtain the decomposition ${X=Y-Z}$. Next, the inequalities

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y_{t_{k-1}}-{\mathbb E}\left[Y_{t_k}\;\vert\mathcal{F}_{t_{k-1}}\right]={\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]_-\ge0,\smallskip\\ &\displaystyle Z_{t_{k-1}}-{\mathbb E}\left[Z_{t_k}\;\vert\mathcal{F}_{t_{k-1}}\right]={\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]_+\ge0, \end{array}$

show that Y and Z are supermartingales. This decomposition is also minimal in the sense that if ${X=Y^\prime-Z^\prime}$ is any other such decomposition for supermartingales ${Y^\prime}$ and ${Z^\prime}$ then, ${Y^\prime-Y}$ and ${Z^\prime-Z}$ are supermartingales.

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[Y_{t_{k-1}}-Y_{t_k}\;\vert\mathcal{F}_{t_{k-1}}\right] &\displaystyle={\mathbb E}\left[Y^\prime_{t_k}-Y^\prime_{t_{k-1}}-(Z^\prime_{t_k}-Z^\prime_{t_{k-1}})\;\big\vert\mathcal{F}_{t_{k-1}}\right]_-\smallskip\\ &\displaystyle\le{\mathbb E}\left[Y^\prime_{t_k}-Y^\prime_{t_{k-1}}\;\big\vert\mathcal{F}_{t_{k-1}}\right]_-+{\mathbb E}\left[Z^\prime_{t_k}-Z^\prime_{t_{k-1}}\;\big\vert\mathcal{F}_{t_{k-1}}\right]_+\smallskip\\ &\displaystyle={\mathbb E}\left[Y^\prime_{t_{k-1}}-Y^\prime_{t_k}\;\big\vert\mathcal{F}_{t_{k-1}}\right]. \end{array}$

The final equality follows from the fact that ${Y^\prime}$ and ${Z^\prime}$ are supermartingales and, hence, the conditional expectations on the right hand side of the inequality are nonpositive. Rearranging this slightly as

$\displaystyle Y^\prime_{t_{k-1}}-Y_{t_{k-1}}\ge{\mathbb E}\left[Y^\prime_{t_k}-Y_{t_k}\;\big\vert\mathcal{F}_{t_{k-1}}\right],$

shows that ${Y^\prime-Y}$ is a supermartingale and, similarly for ${Z^\prime-Z}$. If is furthermore assumed that ${Y^\prime,Z^\prime}$ are nonnegative then,

$\displaystyle Y^\prime_{t_k}-Y_{t_k}\ge{\mathbb E}\left[Y^\prime_{t_n}-Y_{t_n}\;\big\vert\mathcal{F}_{t_k}\right]=0.$

So, ${Y^\prime-Y}$ and, similarly, ${Z^\prime-Z}$ are nonnegative. The mean variation of X can be written as

$\displaystyle Y_{t_0}+Z_{t_0}={\mathbb E}\left[\sum_k\left\lvert{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]\right\rvert\right].$

Now, let us turn our attention to continuous-time processes by approximation with the discrete-time case described above. Suppose that X is an integrable adapted process with time index ${[0,\infty)}$. We extend this to a process on ${[0,\infty]}$ by setting ${X_\infty=0}$. A partition ${P}$ of the interval ${[0,\infty]}$ is defined to be an increasing sequence of times ${0=t_0 < t_1 < \cdots < t_n=\infty}$.

Now, using the partition ${P}$, define the processes ${Y^P,Z^P}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle Y^P_t&\displaystyle={\mathbb E}\left[\sum_{k=1}^n1_{\{t_k > t\}}{\mathbb E}\left[X_{t_k}-X_{t_{k-1}\vee t}\;\vert\mathcal{F}_{t_{k-1}\vee t}\right]_-\;\bigg\vert\mathcal{F}_t\right],\smallskip\\ \displaystyle Z^P_t&\displaystyle={\mathbb E}\left[\sum_{k=1}^n1_{\{t_k > t\}}{\mathbb E}\left[X_{t_k}-X_{t_{k-1}\vee t}\;\vert\mathcal{F}_{t_{k-1}\vee t}\right]_+\;\bigg\vert\mathcal{F}_t\right]. \end{array}$

By comparing with the discrete-time case above, we see that, restricted to the times in ${P}$, ${Y^P}$ and ${Z^P}$ are supermartingales satisfying ${X^P=Y^P-Z^P}$. Also, for any time t not in P, the definition of ${Y^P_t}$ and ${Z^P_t}$ given above is unchanged if we refine P by adding t. So, the identity ${X=Y^P-Z^P}$ holds (almost surely) at each time t.

Define the mean variation on P as

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}^P(X)&\displaystyle= {\mathbb E}\left[\sum_{k=1}^n\left\lvert{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]\right\rvert\right]\smallskip\\ &\displaystyle={\mathbb E}\left[Y^P_0+Z^P_0\right] \end{array}$

and recall that ${{\rm Var}^*(X)}$ is the supremum of ${{\rm Var}^P(X)}$ taken over all partitions P of ${[0,\infty]}$. Taking the limit of this decomposition along a suitable sequence of decompositions gives the continuous-time version of Rao’s decomposition stated in Theorem 3.

Lemma 6 Suppose that ${{\rm Var}^*(X) < \infty}$, and let ${P_m}$ be a sequence of partitions of ${[0,\infty]}$ such that

$\displaystyle {\rm Var}^{P_m}(X)\rightarrow{\rm Var}^*(X).$

as m goes to infinity. Then, the limits

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle Y_t&\displaystyle=\lim_{m\rightarrow\infty}Y^{P_m}_t,\smallskip\\ \displaystyle Z_t&\displaystyle=\lim_{m\rightarrow\infty}Z^{P_m}_t \end{array}$

exist in ${L^1}$ independently of the choice of partitions, and, ${Y}$ and ${Z}$ are nonnegative supermartingales satisfying

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X_t=Y_t-Z_t,\smallskip\\ &\displaystyle{\rm Var}^*(X)={\mathbb E}[Y_0+Z_0]. \end{array}$

Furthermore, if ${X=Y^\prime-Z^\prime}$ is any other decomposition of X as the difference of nonnegative supermartingales, then ${Y^\prime-Y}$ and ${Z^\prime-Z}$ are also nonnegative supermartingales.

Proof: First consider two partitions P and Q of ${[0,\infty]}$ such that Q is a refinement of Q. As described above, restricted to the times in Q, ${Y^Q}$ and ${Z^Q}$ are supermartingales satisfying the identity ${X=Y^Q-Z^Q}$. In particular, ${Y^Q}$ and ${Z^Q}$ are supermartingales when restricted to the times in P. So, by the minimality of the decomposition ${X=Y^P-Z^P}$, the processes ${Y^Q-Y^P}$ and ${Z^Q-Z^P}$ are nonnegative supermartingales at the times in P. For each time t in P this gives,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[\left\lvert Y^Q_t-Y^P_t\right\rvert+\left\lvert Z^Q_t-Z^P_t\right\rvert\right] &\displaystyle={\mathbb E}\left[Y^Q_t-Y^P_t+Z^Q_t-Z^P_t\right]\smallskip\\ &\displaystyle\le{\mathbb E}\left[Y^Q_0-Y^P_0+Z^Q_0-Z^P_0\right]\smallskip\\ &\displaystyle={\rm Var}^Q(X)-{\rm Var}^P(X)\smallskip\\ &\displaystyle\le{\rm Var}^*(X)-{\rm Var}^P(X). \end{array}$

Note, also, that adding any fixed time t to the partition P does not affect the definitions of ${Y^P_t,Z^P_t}$ given above. So, this inequality holds for all ${t\in[0,\infty]}$.

Now consider any two partitions P and Q of ${[0,\infty]}$. Then, letting R be a refinement of both P and Q.

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle{\mathbb E}\left[\left\lvert Y^Q_t-Y^P_t\right\rvert+\left\lvert Z^Q_t-Z^P_t\right\rvert\right] &\displaystyle\le &\displaystyle{\mathbb E}\left[\left\lvert Y^R_t-Y^P_t\right\rvert+\left\lvert Z^R_t-Z^P_t\right\rvert\right]\smallskip\\ &&\displaystyle+{\mathbb E}\left[\left\lvert Y^Q_t-Y^R_t\right\rvert+\left\lvert Z^Q_t-Z^R_t\right\rvert\right]\smallskip\\ &\displaystyle\le&\displaystyle2{\rm Var}^*(X)-{\rm Var}^P(X)-{\rm Var}^Q(X) \end{array}$

In particular if ${P_m}$ is a sequence of partitions with ${{\rm Var}^{P_m}(X)\rightarrow{\rm Var}^*(X)}$ then this shows that ${Y^{P_m}_t}$ and ${Z^{P_m}_t}$ is Cauchy in ${L^1}$. It also shows that if ${Q_m}$ is any other such sequence then ${Y^{Q_m}_t-Y^{P_m}_t}$ and ${Z^{Q_m}_t-Z^{P_m}_t}$ tends to zero in ${L^1}$. So, the limits defining ${Y_t}$ and ${Z_t}$ do indeed exist independently of the specific choice of partitions ${P_m}$ used. Also, taking limits in ${L^1}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y_t-Z_t=\lim_{m\rightarrow\infty}(Y^{P_m}_t-Y^{P_m}_t)=X_t,\smallskip\\ & \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[Y_0+Z_0\right]&\displaystyle=\lim_{m\rightarrow\infty}{\mathbb E}\left[Y^{P_m}_0+Z^{P_m}_0\right]\smallskip\\ &\displaystyle=\lim_{m\rightarrow\infty}{\rm Var}^{P_m}(X)={\rm Var}^*(X) \end{array} \end{array}$

as required.

It remains to show that Y and Z are supermartingales. Fixing times ${s < t}$, we can refine the partitions ${P_m}$ to obtain new partitions ${Q_m}$ containing these times. As shown above, ${Y^{Q_m}-Y^{P_m}}$ and ${Z^{Q_m}-Z^{P_m}}$ will be nonnegative supermartingales at the times in ${P_m}$, so that

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}^*(X)&\displaystyle\ge{\rm Var}^{Q_m}(X)={\mathbb E}\left[Y^{Q_m}_0+Z^{Q_m}_0\right]\smallskip\\ &\displaystyle\ge {\mathbb E}\left[Y^{P_m}_0+Z^{P_m}_0\right]={\rm Var}^{P_m}(X)\smallskip\\ &\displaystyle\rightarrow{\rm Var}^*(X). \end{array}$

By the first part of the lemma, proven above, this shows that ${Y^{Q_m}_u,Z^{Q_m}_u}$ converges to ${Y_u,Z_u}$ in ${L^1}$ for all times u. As ${Y^{Q_m}}$ is a supermartingale at the times in ${Q_m}$, taking limits in ${L^1}$ gives the inequality

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle Y_s=\lim_{m\rightarrow\infty}Y^{Q_m}_s&\displaystyle\ge\lim_{m\rightarrow\infty}{\mathbb E}\left[Y^{Q_m}_t\;\big\vert\mathcal{F}_s\right]\smallskip\\ &\displaystyle={\mathbb E}\left[Y_t\;\vert\mathcal{F}_s\right]. \end{array}$

So Y and, similarly, Z are supermartingales.

Finally, suppose that ${X=Y^\prime-Z^\prime}$ for nonnegative supermartingales ${Y^\prime}$, ${Z^\prime}$. By the minimality of the discrete-time decomposition, ${Y^\prime-Y^{Q_m}}$ and ${Z^\prime-Z^{Q_m}}$ are nonnegative supermartingales at the times in ${Q_m}$. Again, taking limits in ${L^1}$, this shows that ${Y^\prime-Y}$ and ${Z^\prime-Z}$ are nonnegative supermartingales. ⬜

It is now a simple matter of applying Lemma 6 to prove Rao’s decomposition in the form of Theorem 3.

Proof of Theorem 3: First, existence of the decomposition (8) gives

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}^*(X)&\displaystyle\le{\rm Var}^*(Y)+{\rm Var}^*(Z)\smallskip\\ &\displaystyle={\mathbb E}[Y_0]+{\mathbb E}[Z_0] < \infty. \end{array}$

So, finiteness of ${{\rm Var}^*(X)}$ is a necessary condition, and inequality (9) holds. Conversely, suppose that ${{\rm Var}^*(X)}$ is finite. Lemma 6 states the existence of a decomposition ${X=Y-Z}$ for nonnegative supermartingales ${Y,Z}$, which is minimal in the required sense and satisfies

$\displaystyle {\rm Var}^*(X)={\mathbb E}[Y_0+Z_0].$

So, equality does indeed hold in (9) for the minimal decomposition.

Finally, suppose that ${X=Y-Z}$ is any decomposition of X as the difference of supermartingales such that equality holds in (9) and that ${X=Y^\prime-Z^\prime}$ is the minimal decomposition. Then, ${Y-Y^\prime}$, ${Z-Z^\prime}$ are nonnegative supermartingales and,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}[(Y_t-Y^\prime_t)+(Z_t-Z^\prime_t)]&\displaystyle\le{\mathbb E}[(Y_0-Y^\prime_0)+(Z_0-Z^\prime_0)]\smallskip\\ &\displaystyle={\mathbb E}[Y_0+Z_0]-{\mathbb E}[Y^\prime_0+Z^\prime_0]\smallskip\\ &\displaystyle={\rm Var}^*(X)-{\rm Var}^*(X)=0. \end{array}$

So, ${Y=Y^\prime}$, ${Z=Z^\prime}$ and the decomposition is minimal as claimed. ⬜

We now move on to proving the alternative statement of Rao’s decomposition given in Theorem 2. This can be done in a similar way to above, by first explicitly writing out the decomposition in discrete time and them taking the limit along a sequence of partitions. However, as we have already given a proof of Theorem 3, we will instead take the approach of reducing decomposition (5) to (8) by subtracting out a martingale term. The statement that X can be decomposed as the sum of a martingale and a term tending to zero in ${L^1}$ is sometimes referred to as the Riesz decomposition of X.

Lemma 7 Let X be an integrable process such that ${{\rm Var}(X) < \infty}$. Then, for each ${t\in{\mathbb R}_+}$, the limit

$\displaystyle M_t=\lim_{T\rightarrow\infty}{\mathbb E}[X_T\;\vert\mathcal{F}_t]$

exists in ${L^1}$. Furthermore, M is a martingale and ${X_t-M_t\rightarrow0}$ in ${L^1}$ as ${t\rightarrow\infty}$.

Proof: Consider any times ${S\le T}$ and sequence ${0=t_0\le t_1\le\cdots\le t_n=S}$. Then, by the definition of ${{\rm Var}_T(X)}$ in terms of partitions,

$\displaystyle {\mathbb E}\left[\sum_{k=1}^n\left\lvert{\mathbb E}\left[X_{t_k}-X_{t_{k-1}}\;\vert\mathcal{F}_{t_{k-1}}\right]\right\rvert+\left\lvert{\mathbb E}\left[X_T-X_S\;\vert\mathcal{F}_S\right]\right\rvert\right]\le{\rm Var}_T(X).$

Taking the supremum over all such finite sequences for the ${t_k}$ gives

$\displaystyle {\mathbb E}\left[\left\lvert{\mathbb E}\left[X_T-X_S\;\vert\mathcal{F}_S\right]\right\rvert\right]\le{\rm Var}_T(X)-{\rm Var}_S(X).$

Using ${M^{(T)}_t}$ to denote ${{\mathbb E}[X_T\vert\mathcal{F}_t]}$, then ${M^{(T)}_t-M^{(S)}_t}$ is a martingale and, hence, ${\lvert M^{(T)}-M^{(S)}\rvert}$ is a submartingale. So, for ${t\le S}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[\left\lvert M^{(T)}_t-M^{(S)}_t\right\rvert\right] &\displaystyle\le{\mathbb E}\left[\left\lvert M^{(T)}_S-M^{(S)}_S\right\rvert\right]\smallskip\\ &\displaystyle={\mathbb E}\left[\left\lvert{\mathbb E}\left[X_T-X_S\;\vert\mathcal{F}_S\right]\right\rvert\right]\smallskip\\ &\displaystyle\le{\rm Var}_T(X)-{\rm Var}_S(X), \end{array}$

which tends to zero as S,T go to infinity. So, ${M^{(T)}_t}$ is Cauchy in ${L^1}$ as ${T\rightarrow\infty}$ and, hence, ${M_t=\lim_{T\rightarrow\infty}M^{(T)}_t}$ exists in ${L^1}$. Also, as it is an ${L^1}$ limit of martingales, M is itself a martingale.

Finally, for ${S\le T}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}\left[\left\lvert X_S-M^{(T)}_S\right\rvert\right]&\displaystyle={\mathbb E}\left[\left\lvert{\mathbb E}\left[X_S-X_T\;\vert\mathcal{F}_S\right]\right\rvert\right]\smallskip\\ &\displaystyle\le{\rm Var}_T(X)-{\rm Var}_S(X). \end{array}$

Letting T increase to infinity,

$\displaystyle {\mathbb E}\left[\left\lvert X_S-M_S\right\rvert\right]\le{\rm Var}(X)-{\rm Var}_S(X).$

This tends to zero as S goes to infinity, so ${X_S-M_S}$ tends to zero in ${L^1}$. ⬜

We can now prove Theorem 2 by subtracting out a martingale term as in Lemma 7 and applying the decomposition already proven above.

Proof of Theorem 2: If decomposition (5) exists then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}(X)&\displaystyle\le{\rm Var}(M)+{\rm Var}(Y)+{\rm Var}(Z)\smallskip\\ &\displaystyle\le0+{\mathbb E}[Y_0]+{\mathbb E}[Z_0] < \infty. \end{array}$

So, inequality (6) holds and ${{\rm Var}(X) < \infty}$ is a necessary condition. Conversely, suppose that ${{\rm Var}(X)}$ is finite. Then, Lemma 7 says that there exists a martingale M such that ${X_t-M_t\rightarrow0}$ in ${L^1}$ as t goes to infinity. Setting ${X^\prime=X-M}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}^*(X^\prime)&\displaystyle={\rm Var}(X^\prime)+\lim_{t\rightarrow\infty}{\mathbb E}[\lvert X^\prime_t\rvert]\smallskip\\ &\displaystyle={\rm Var}(X-M)\smallskip\\ &\displaystyle={\rm Var}(X) < \infty. \end{array}$

So, Theorem 3 applies, and there exists a minimal decomposition ${X^\prime=Y-Z}$ for supermartingales Y and Z. This gives ${X=M+Y-Z}$ as in decomposition (5) and, by (9),

$\displaystyle {\rm Var}(X)={\rm Var}^*(X^\prime)={\mathbb E}[Y_0+Z_0].$

In particular, this gives

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\mathbb E}[Y_0+Z_0]&\displaystyle={\rm Var}(X^\prime)\smallskip\\ &\displaystyle\le{\rm Var}(Y)+{\rm Var}(Z)\smallskip\\ &\displaystyle={\mathbb E}[X_0+Y_0]-\lim_{t\rightarrow\infty}{\mathbb E}[Y_t+Z_t]. \end{array}$

As they are nonnegative, this implies that ${Y_t}$ and ${Z_t}$ tend to zero in ${L^1}$ as ${t\rightarrow\infty}$. Now, suppose that ${X=M^\prime+Y^\prime-Z^\prime}$ is any other decomposition for martingale ${M^\prime}$ and nonnegative supermartingales ${Y^\prime,Z^\prime}$. Then,

$\displaystyle X^\prime=(M^\prime+Y^\prime)-(M+Z^\prime).$

expresses ${X^\prime}$ as the difference of supermartingales. By Lemma 4, minimality of the decomposition ${X^\prime=Y-Z}$ implies that ${(M^\prime+Y^\prime)-Y}$ and ${(M+Z^\prime)-Z}$ are supermartingales. In particular, ${Y^\prime-Y}$ is a supermartingale so, for any ${T\ge t}$,

$\displaystyle Y^\prime_t-Y_t\ge{\mathbb E}[Y^\prime_T-Y_T\;\vert\mathcal{F}_t]\ge-{\mathbb E}[Y_T\;\vert\mathcal{F}_t].$

As ${Y_T\rightarrow0}$ in ${L^1}$, this shows that ${Y^\prime-Y}$ and, similarly, ${Z^\prime-Z}$ are nonnegative. So, the decomposition is minimal as claimed. We also see that ${Y_t\rightarrow0}$ and, similarly, ${Z_t\rightarrow0}$ in ${L^1}$ as ${t\rightarrow\infty}$. Also, ${M_t-X_t\rightarrow0}$ in ${L^1}$, so if M is identically zero then ${X_t\rightarrow0}$ and, conversely, if ${X_t\rightarrow0}$ in ${L^1}$ then,

$\displaystyle M_t=\lim_{T\rightarrow\infty}{\mathbb E}[X_T\;\vert\mathcal{F}_t]=0.$

Finally, suppose that ${X=M+Y-Z}$ is any decomposition for martingale M and nonnegative supermartingales Y,Z such that equality holds in (6), and that ${X=M^\prime+Y^\prime-Z^\prime}$ is the minimal decomposition. Then, ${Y-Y^\prime}$,${Z-Z^\prime}$ are nonnegative supermartingales and,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle E\left[(Y_t-Y^\prime_t)+(Z_t-Z^\prime_t)\right]&\displaystyle\le{\mathbb E}\left[(Y_0-Y^\prime_0)+(Z_0-Z^\prime_0)\right]\smallskip\\ &\displaystyle={\mathbb E}[Y_0+Z_0]-{\mathbb E}[Y^\prime_0+Z^\prime_0]\smallskip\\ &\displaystyle={\rm Var}(X)-{\rm Var}(X)=0. \end{array}$

So, ${Y=Y^\prime}$, ${Z=Z^\prime}$, and the decomposition is minimal. ⬜

Finally, we move on to proving Rao’s decomposition in the form of Theorem 1. This can be done by applying the decomposition (8) over an increasing sequence of bounded intervals, and patching these together to obtain (2) for arbitrary quasimartingales.

Proof of Theorem 1: First, as supermartingales are quasimartingales and linear combinations of quasimartingales are quasimartingales, if X is the difference of supermartingales then it is a quasimartingale.

Conversely, let X be a quasimartingale. Let us start by assuming that the minimal decomposition (2) does indeed exist, and that ${T\ge0}$ is any fixed time. The stopped process ${X^T=Y^T-Z^T}$ is a difference of supermartingales. If ${X^T=Y^\prime-Z^\prime}$ is any other such decomposition then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle Y^{\prime\prime}_t=Y^\prime_t+(Y_t-Y_{T\wedge t}),\smallskip\\ &\displaystyle Z^{\prime\prime}_t=Z^\prime_t+(Z_t-Z_{T\wedge t}), \end{array}$

are again supermartingales satisfying ${X=Y^{\prime\prime}-Z^{\prime\prime}}$. Therefore, ${Y^{\prime\prime}-Y}$ and ${Z^{\prime\prime}-Z}$ are supermartingales and, stopping at time T, ${Y^\prime-Y^T}$ and ${Z^\prime-Z^T}$ are also supermartingales. This shows that the decomposition ${X^T=Y^T-Z^T}$ is also minimal, so stopping the processes at deterministic times preserves minimal decompositions.

Now, suppose that X is any quasimartingale and, again, that ${T\ge0}$ is any fixed time. Then, the stopped process ${X^T}$ satisfies

$\displaystyle {\rm Var}^*(X^T)={\rm Var}_T(X)+{\mathbb E}[\lvert X_T\rvert] < \infty.$

So, Theorem 3 applies to ${X^T}$, and we can let ${X^T=Y-Z}$ be a decomposition which is minimal in the sense of (8). By Lemma 4, it is also minimal in the sense of Theorem 1. So, the minimal decomposition does always exist if we stop the process at a finite time. Furthermore, as equality holds in (9),

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle{\rm Var}_T(X)&\displaystyle={\rm Var}^*(X^T)-{\mathbb E}[\lvert X_T\rvert]\smallskip\\ &\displaystyle={\mathbb E}[Y_0+Z_0]-{\mathbb E}[\lvert Y_T-Z_T\rvert]\smallskip\\ &\displaystyle\ge{\mathbb E}[Y_0+Z_0-Y_T-Z_T]. \end{array}$

So, equality holds in (3) and, as the minimal decomposition is unique up to a martingale term, it holds for any minimal decomposition.

Next, let ${T_n}$ be a sequence of deterministic times increasing to infinity and, for each n, let

$\displaystyle X^{T_n}=Y^n-Z^n$

be a minimal decomposition as the difference of supermartingales. As minimal decompositions are preserved by stopping at deterministic times, ${X^{T_n}=(Y^{n+1}))^{T_n}-(Z^{n+1})^{T_n}}$ is also a minimal decomposition. Then, as minimal decompositions are unique up to a martingale, this implies that ${Y^n-(Y^{n+1})^{T_n}=Z^n-(Z^{n+1})^{T_n}}$ is a martingale. We can define new minimal decompositions by setting ${\tilde Y^1=Y^1}$, ${\tilde Z^1=Z^1}$ and, for each ${n\ge1}$,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\tilde Y^{n+1}_t=Y^{n+1}_t+{\mathbb E}\left[\tilde Y^n_{T_n}-Y^{n+1}_{T_n}\;\big\vert\mathcal{F}_t\right],\smallskip\\ &\displaystyle\tilde Z^{n+1}_t=Z^{n+1}_t+{\mathbb E}\left[\tilde Z^n_{T_n}-Z^{n+1}_{T_n}\;\big\vert\mathcal{F}_t\right]. \end{array}$

As ${Y^n_t-Y^{n+1}_t}$ is a martingale on ${t\le T_n}$, we also have ${\tilde Y^{n+1}_t=\tilde Y^n_t}$ on this range. Also, ${X^{T_n}=\tilde Y^n-\tilde Z^n}$ and, as ${\tilde Y^n-Y^n}$, ${\tilde Z^n-Z^n}$ are martingales, these are also minimal decompositions. As ${\tilde Y^{n+1}_t=\tilde Y^n_t}$ and ${\tilde Z^{n+1}_t=\tilde Z^n_t}$ for all ${t\le T_n}$, we can set

$\displaystyle Y_t=\lim_{n\rightarrow\infty}\tilde Y^n_t,\ Z_t=\lim_{n\rightarrow\infty}\tilde Z^n_t.$

For each t, this is just the limit of a constant sequence once ${T_n\ge t}$. Then, ${Y,Z}$ are supermartingales with ${X=Y-Z}$. Suppose that ${X=Y^\prime-Z^\prime}$ is any other such decomposition. By stopping at time ${T_n}$ and using the fact that ${X^{T_n}=Y^{T_n}-Z^{T_n}}$ is a minimal decomposition then, it follows that ${(Y^\prime)^{T_n}-Y^{T_n}=(Z^\prime)^{T_n}-Z^{T_n}}$ are supermartingales. As this holds for each n, ${Y^\prime-Y=Z^\prime-Z}$ is a supermartingale and the decomposition is minimal.

Finally, suppose that ${X=Y-Z}$ is any decomposition as a difference of supermartingales such that equality holds in (3) and that ${X=Y^\prime-Z^\prime}$ is a minimal decomposition. Then,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle{\mathbb E}[(Y_t-Y^\prime_t)+(Z_t-Z^\prime_t)]&\displaystyle=&\displaystyle{\mathbb E}[Y^\prime_0+Z^\prime_0-Y^\prime_t-Z^\prime_t]\smallskip\\ &&\displaystyle{}-{\mathbb E}[Y_0+Z_0-Y_t-Z_t]\smallskip\\ &&\displaystyle{}+{\mathbb E}[Y_0+Z_0-Y^\prime_0+Z^\prime_0]\smallskip\\ &\displaystyle=&\displaystyle{\rm Var}_t(X)-{\rm Var}_t(X)\smallskip\\ &&\displaystyle{}+{\mathbb E}[Y_0+Z_0-Y^\prime_0+Z^\prime_0] \end{array}$

is independent of t. As ${Y-Y^\prime=Z-Z^\prime}$ are supermartingales their expectations are nonincreasing functions of t and, hence, are constant in t. As a supermartingale with constant expectation is a martingale, this shows that ${Y-Y^\prime=Z-Z^\prime}$ is a martingale, so the decomposition is indeed minimal. ⬜

1. Hi, I want to use some theorems in your blogs, but how I add the references of these theorems? Could you tell me a name of a book or something? Thank you very much

Comment by Wenfeng Dong — 6 July 14 @ 11:40 AM

2. Hi, I have a personal question. Do you ever find it hard to balance piracy and general marauding with your mathematics career? How do you strike a good pirate-math balance?

Thank you very much,
Kevin Stangl

Comment by kevinmstangl — 27 May 16 @ 12:02 AM

• Avast ye! The last landlubber to ask personal questions like ye had to walk the plank and pay a visit to Davy Jones’ Locker. Aaaarrrrgggghhhh! Yo ho ho, and all that.

Comment by George Lowther — 6 June 16 @ 1:14 AM

3. Hi George! First of all thank you for your blog, it is very complete and precise.
I have a doubt about one step of the proof of Rao’s decomposition. It refers to the discrete time, when you are proving the “minimality” of the decomposition, and in particular when you want to prove that $Y^\prime-Y$ is nonnegative, are you assuming that $Y_{t_n}=0$ (and similarly $Z_{t_n}=0$)? Above you said “let us assume $X_{t_n}=0$” which is weaker than saying $Y_{t_n}=0$ and $Z_{t_n}=0$ but otherwise I don’t see how you could prove that $\mathbb{E}\left[Y^\prime_{t_n}-Y_{t_n}|\mathcal{F}_{t_k}\right]=0$. If you assume $Y_{t_n}=0$ (let’s say by construction) then you can say $\mathbb{E}\left[Y^\prime_{t_n}-Y_{t_n}|\mathcal{F}_{t_k}\right]\geq 0$, since $Y^\prime$ is nonnegative, and this is sufficient to prove that $Y^\prime-Y$ is nonnegative.
Am I right or am I missing something?

Cheers,
Alessandro Milazzo

Comment by Alessandro Milazzo — 7 February 17 @ 11:38 AM

• Hi Alessandro. I assume you are talking about the argument immediately following the heading “Proof of Rao’s decomposition”. I don’t need to assume anything about $Y$ because I explicitly define it. The fact that $Y_{t_n}=0$ follows from the definition (note that the summation on the rhs of the definition for $Y_{t_n}$ has no terms). You are correct to bring up the equality $\mathbb{E}[Y^\prime_{t_n}-Y_{t_n}\,\vert\mathcal{F}_{t_k}]=0$ though. This is a mistake and should be an inequality, as you point out, which is all that is required by the argument anyway.
I’ll update to fix this point and to clarify when I have some time.

Many thanks!
George

Comment by George Lowther — 28 February 17 @ 12:19 PM

• You are right, George!

Thanks again,
Alessandro

Comment by Anonymous — 22 March 17 @ 11:05 AM

4. Hi George, I have a question about the construction of Rao’s decomposition in discrete time. In the proof you assume that $X_{t_n}=0$. What if the process you’re considering does not satisfy this property? How would you construct the Rao’s decomposition in that case? My intuitive guess is to construct a process $\tilde{X}$ which is the same as $X$ but extended at a further time $\tilde{X}_{t_{n+1}}=0$. Then, define $Y$ and $Z$ as you did where now the sums go up to $n+1$ instead of $n$. This seems to work to me. Is that the right procedure?

Thanks,
Alessandro

Comment by Alessandro Milazzo — 1 December 17 @ 4:18 PM

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