Almost Sure

16 June 20

Pathwise Martingale Inequalities

Recall Doob’s inequalities, covered earlier in these notes, which bound expectations of functions of the maximum of a martingale in terms of its terminal distribution. Although these are often applied to martingales, they hold true more generally for cadlag submartingales. Here, I use {\bar X_t\equiv\sup_{s\le t}X_s} to denote the running maximum of a process.

Theorem 1 Let X be a nonnegative cadlag submartingale. Then,

  • {{\mathbb P}\left(\bar X_t \ge K\right)\le K^{-1}{\mathbb E}[X_t]} for all {K > 0}.
  • {\lVert\bar X_t\rVert_p\le (p/(p-1))\lVert X_t\rVert_p} for all {p > 1}.
  • {{\mathbb E}[\bar X_t]\le(e/(e-1)){\mathbb E}[X_t\log X_t+1]}.

In particular, for a cadlag martingale X, then {\lvert X\rvert} is a submartingale, so theorem 1 applies with {\lvert X\rvert} in place of X.

We also saw the following much stronger (sub)martingale inequality in the post on the maximum maximum of martingales with known terminal distribution.

Theorem 2 Let X be a cadlag submartingale. Then, for any real K and nonnegative real t,

\displaystyle  {\mathbb P}(\bar X_t\ge K)\le\inf_{x < K}\frac{{\mathbb E}[(X_t-x)_+]}{K-x}. (1)

This is particularly sharp, in the sense that for any distribution for {X_t}, there exists a martingale with this terminal distribution for which (1) becomes an equality simultaneously for all values of K. Furthermore, all of the inequalities stated in theorem 1 follow from (1). For example, the first one is obtained by taking {x=0} in (1). The remaining two can also be proved from (1) by integrating over K.

Note that all of the submartingale inequalities above are of the form

\displaystyle  {\mathbb E}[F(\bar X_t)]\le{\mathbb E}[G(X_t)] (2)

for certain choices of functions {F,G\colon{\mathbb R}\rightarrow{\mathbb R}^+}. The aim of this post is to show how they have a more general `pathwise’ form,

\displaystyle  F(\bar X_t)\le G(X_t) - \int_0^t\xi\,dX (3)

for some nonnegative predictable process {\xi}. It is relatively straightforward to show that (2) follows from (3) by noting that the integral is a submartingale and, hence, has nonnegative expectation. To be rigorous, there are some integrability considerations to deal with, so a proof will be included later in this post.

Inequality (3) is required to hold almost everywhere, and not just in expectation, so is a considerably stronger statement than the standard martingale inequalities. Furthermore, it is not necessary for X to be a submartingale for (3) to make sense, as it holds for all semimartingales. We can go further, and even drop the requirement that X is a semimartingale. As we will see, in the examples covered in this post, {\xi_t} will be of the form {h(\bar X_{t-})} for an increasing right-continuous function {h\colon{\mathbb R}\rightarrow{\mathbb R}}, so integration by parts can be used,

\displaystyle  \int h(\bar X_-)\,dX = h(\bar X)X-h(\bar X_0)X_0 - \int X\,dh(\bar X). (4)

The right hand side of (4) is well-defined for any cadlag real-valued process, by using the pathwise Lebesgue–Stieltjes integral with respect to the increasing process {h(\bar X)}, so can be used as the definition of {\int h(\bar X_-)dX}. In the case where X is a semimartingale, integration by parts ensures that this agrees with the stochastic integral {\int\xi\,dX}. Since we now have an interpretation of (3) in a pathwise sense for all cadlag processes X, it is no longer required to suppose that X is a submartingale, a semimartingale, or even require the existence of an underlying probability space. All that is necessary is for {t\mapsto X_t} to be a cadlag real-valued function. Hence, we reduce the martingale inequalities to straightforward results of real-analysis not requiring any probability theory and, consequently, are much more general. I state the precise pathwise generalizations of Doob’s inequalities now, leaving the proof until later in the post. As the first of inequality of theorem 1 is just the special case of (1) with {x=0}, we do not need to explicitly include this here.

Theorem 3 Let X be a cadlag process and t be a nonnegative time.

  1. For real {K > x},
    \displaystyle  1_{\{\bar X_t\ge K\}}\le\frac{(X_t-x)_+}{K-x}-\int_0^t\xi\,dX (5)

    where {\xi=(K-x)^{-1}1_{\{\bar X_-\ge K\}}}.

  2. If X is nonnegative and p,q are positive reals with {p^{-1}+q^{-1}=1} then,
    \displaystyle  \bar X_t^p\le q^p X^p_t-\int_0^t\xi dX (6)

    where {\xi=pq\bar X_-^{p-1}}.

  3. If X is nonnegative then,
    \displaystyle  \bar X_t\le\frac{e}{e-1}\left( X_t \log X_t +1\right)-\int_0^t\xi\,dX (7)

    where {\xi=\frac{e}{e-1}\log(\bar X_-\vee1)}.

The pathwise inequalities can be stated quite succinctly in discrete time, in which case they become simple inequalities for finite sequences of real numbers. Given any such sequence {x_0,x_1,\ldots,x_n}, set {\bar x_k=\max(x_0,\ldots,x_k)}. By considering the cadlag process {X_t=x_n} over {n\le t < n+1}, the inequalities stated in theorem 3 are of the form,

\displaystyle  F(\bar x_n)\le G(x_n) - \sum_{k=1}^nh(\bar x_{k-1})(x_k-x_{k-1}). (8)

The more general continuous-time inequalities can also be derived as a limit of the discrete-time case, so this is still quite general. Furthermore, if {x_k} is a discrete-time submartingale, then taking expectations of (8) gives the discrete version of Doob’s inequalities, from which the continuous-time versions can be obtained by taking limits. Alternatively, we can show directly that Doob’s inequalities follow from the pathwise versions by taking expectations in continuous time.

Lemma 4 Let X be a cadlag submartingale taking values in an interval {I\subseteq{\mathbb R}} and satisfying (3) for some {F,G\colon I\rightarrow{\mathbb R}^+}, where G is convex and increasing, and for some nonnegative and locally bounded predictable process {\xi}. Then,

\displaystyle  {\mathbb E}[F(\bar X_\tau)]\le{\mathbb E}[G(X_\tau)]

for all bounded stopping times {\tau}.

Proof: As {\xi} is locally bounded and nonnegative, {M\equiv\int\xi\,dX} is locally a submartingale and, hence, there exists a sequence of stopping times {\tau_n} increasing to infinity such that {M^{\tau_n}} are submartingales. Therefore, {{\mathbb E}[M_{\tau_n\wedge\tau}]\ge0}. Next, as G is convex and increasing, {G(X)} is a submartingale. So, taking expectations of (3) at time {\tau\wedge\tau_n},

\displaystyle  {\mathbb E}[1_{\{\tau_n\ge\tau\}}F(\bar X_\tau)]\le{\mathbb E}[F(\bar X_{\tau\wedge\tau_n})]\le{\mathbb E}[G(X_{\tau\wedge\tau_n})]\le{\mathbb E}[G(X_\tau)].

Letting n increase to infinity and applying monotone convergence on the left hand side gives the result. ⬜

See Gushchin, On pathwise counterparts of Doob’s maximal inequalities, and the references contained therein, for more information on the pathwise approach.

To prove the pathwise inequalities, I start with the following simple, but very neat, identity. This actually extends to all measurable and locally bounded functions h in the case where X is a semimartingale (applying the monotone class theorem). However, I state it only for the required situation where {h} is right-continuous and increasing, in which case the integral makes sense for all cadlag processes X.

Lemma 5 Let X be a cadlag process and {h\colon{\mathbb R}\rightarrow{\mathbb R}} be increasing and right-continuous. Then,

\displaystyle  \int h(\bar X_-)dX= \int h(\bar X_-)d\bar X - h(\bar X)(\bar X-X). (9)

Proof: Applying (4),

\displaystyle  h(\bar X)(\bar X-X)=\int h(\bar X_-)\,d(\bar X-X)+\int(\bar X-X)dh(\bar X).

So, to prove (9) it only needs to be shown that

\displaystyle  \int(\bar X-X)dh(\bar X)=0

and, as {h(\bar X)} is an increasing process, this integral is understood in the pathwise Lebesgue-Stieltjes sense.

For any {t\ge0} let {\tau_t} be the last time before t at which {X=\bar X_t},

\displaystyle  \tau_t=\sup\left\{s\le t\colon X_s\ge\bar X_t\right\}.

Then, {\bar X_s=\bar X_t} is constant on {[\tau_t,t)} and, hence,

\displaystyle  \int_0^\infty 1_{[\tau_t,t)}(\bar X-X)\,dh(\bar X)=(\bar X_{\tau_t}-X_{\tau_t})\Delta h(\bar X_{\tau_t}).

As {\bar X_{\tau_t}=\bar X_{\tau_t-}\vee X_{\tau_t}}, we see that {\Delta h(\bar X_{\tau_t})=0} whenever {X_{\tau_t} < \bar X_{\tau_t}} and, hence, the above equality is identically zero. Next, if s is any time at which {X_s < \bar X_s} then, by right-continuity, {X_t < \bar X_s=\bar X_t} for all {t > s} sufficiently close to s, giving {s\in[\tau_t,t)}. Hence, using the fact that the positive rationals {{\mathbb Q}_+} are dense in {{\mathbb R}_+}, by countable additivity,

\displaystyle  \int(\bar X-X)dh(\bar X)\le\sum_{t\in{\mathbb Q}_+}\int_0^\infty 1_{[\tau_t,t)}(\bar X-X)\,dh(\bar X)=0

as required. ⬜

We obtain the following consequence of lemma 9. I am using {H^\prime} to denote the right-hand derivative which, as {H} is convex, is right-continuous and increasing.

Corollary 6 Let X be a cadlag process and {H\colon{\mathbb R}\rightarrow{\mathbb R}^+} be convex with {h=H^\prime}. Then,

\displaystyle  \begin{aligned} \int h(\bar X_-)\,dX & \le H(\bar X)-h(\bar X)(\bar X-X)\\ &= h(\bar X)X -(h(\bar X)\bar X-H(\bar X)). \end{aligned} (10)

Proof: By (9) it is sufficient to show that {H(\bar X)\ge\int h(\bar X_-)\,d\bar X}. See, for example, lemma 1 of the post on local times. Alternatively, by approximating {h} by continuous functions, we can suppose that it is continuous (and increasing). So, by convexity,

\displaystyle  H(\bar X_t)-H(\bar X_s)\ge h(\bar X_s)(\bar X_t-\bar X_s).

Hence, for any sequence of times {0=t_0\le t_1\le\cdots\le t_n=t},

\displaystyle  H(\bar X_t)\ge\sum_{k=1}^n h(\bar X_{t_{k-1}})(\bar X_{t_k}-\bar X_{t_{k-1}}).

Letting the mesh {\max_k\lvert t_k-t_{k-1}\rvert} go to zero, bounded convergence shows that the right hand side of the above inequality tends to {\int_0^t h(\bar X_-)d\bar X}, giving the result. ⬜

I did not include terms involving the starting value {X_0} in the pathwise inequalities, just in order to have very simple statements. However, they can be included if desired, in which case we should subtract {H(X_0)} from the right hand side of (10) and, the proof given still holds. In fact, if we do this, then (10) is very sharp, and is an equality whenever {\bar X} is continuous. I finally give proofs of the pathwise inequalities, as an application of corollary 6.

Proof of Theorem 3: Applying (10) with {h(x)=1_{\{x\ge K\}}} and {H(x)=(x-K)_+} gives,

\displaystyle  \begin{aligned} \int_0^t h(\bar X_-)\,dX &\le 1_{\{\bar X\ge K\}}X-1_{\{\bar X\ge K\}}(\bar X - (\bar X-K))\\ &=1_{\{\bar X\ge K\}}(X-x)-1_{\{\bar X\ge K\}}(K-x)\\ &\le(X-x)_+-1_{\{\bar X\ge K\}}(K-x). \end{aligned}

Dividing through by {K-x} gives (5).

Next, apply (10) with {h(x)=pqx^{p-1}_+} and {H(x)=qx^p_+},

\displaystyle  \begin{aligned} \int h(\bar X_-)\,dX &\le pq\bar X^{p-1}X - (pq\bar X^p - q\bar X^p)\\ &\le (p-1)\bar X^p+(qX)^p-(p-1)q\bar X^p\\ &=q^p X^p-\bar X^p, \end{aligned}

using the AM-GM inequality. This gives (6) as required.

Finally, apply (10) with {h(x) = \log_+x\equiv\log(x\vee1)} and {H(x)=x\log_+x-(1-x)_+},

\displaystyle  \begin{aligned} \int h(\bar X_-)dX &\le X\log_+\bar X - (\bar X\log_+ X-\bar X\log_+\bar X+(1-\bar X)_+)\\ &=X\log_+\bar X-(1-\bar X)_+\\ &\le X\log X +1-(1-e^{-1})\bar X. \end{aligned}

The last inequality is using (2) of the original post on martingale inequalities, and we obtain (7). ⬜

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