# Almost Sure

## 21 December 09

### Martingale Inequalities

Martingale inequalities are an important subject in the study of stochastic processes. The subject of this post is Doob’s inequalities which bound the distribution of the maximum value of a martingale in terms of its terminal distribution, and is a consequence of the optional sampling theorem. We work with respect to a filtered probability space ${(\Omega,\mathcal{F},\{\mathcal{F}_t\}_{t\ge 0},{\mathbb P})}$. The absolute maximum process of a martingale is denoted by ${X^*_t\equiv\sup_{s\le t}\vert X_s\vert}$. For any real number ${p\ge 1}$, the ${L^p}$-norm of a random variable ${Z}$ is

$\displaystyle \Vert Z\Vert_p\equiv{\mathbb E}[|Z|^p]^{1/p}.$

Then, Doob’s inequalities bound the distribution of the maximum of a martingale by the ${L^1}$-norm of its terminal value, and bound the ${L^p}$-norm of its maximum by the ${L^p}$-norm of its terminal value for all ${p>1}$.

Theorem 1 Let ${X}$ be a cadlag martingale and ${t>0}$. Then

1. for every ${K>0}$,

$\displaystyle {\mathbb P}(X^*_t\ge K)\le\frac{\Vert X_t\Vert_1}{K}.$

2. for every ${p>1}$,

$\displaystyle \Vert X^*_t\Vert_p\le \frac{p}{p-1}\Vert X_t\Vert_p$

We can define a topology on the space of cadlag martingales so that a sequence ${X^n}$ of martingales converges to a limit ${X}$ if ${\Vert X^n_t-X_t\Vert_p\rightarrow 0}$ as ${n\rightarrow\infty}$. It is clear that ${\Vert X_t\Vert_p\le\Vert X^*_t\Vert_p}$ for any process and, consequently Doob’s second inequality above shows that this topology is equivalent to the seemingly much stronger condition that ${\Vert(X^n-X)^*_t\Vert_p\rightarrow 0}$, for any ${p>1}$.

The second statement above does not generalize to ${p=1}$ and, in fact, there exist martingales ${X}$ such that the maximum ${X^*_t}$ has infinite expectation at all positive times.

Doob’s martingale inequalities are a consequence of the following inequalities applied to the submartingale ${|X_t|}$.

Theorem 2 Let ${X}$ be a nonnegative cadlag submartingale. Then,

1. ${{\mathbb P}(X^*_t\ge K)\le {\mathbb E}[X_t]/K}$ for each ${K>0}$.
2. ${\Vert X^*_t\Vert_p\le(p/(p-1))\Vert X_t\Vert_p}$ for each ${p>1}$.

Proof: By completing the filtration if necessary, without loss of generality we assume that it is complete. Consider the first time at which the process reaches the positive value ${K}$,

$\displaystyle \tau=\inf\left\{t\in{\mathbb R}_+\colon X_t\ge K\right\}$

which, by the debut theorem, is a stopping time. Then ${X_\tau=X^*_\tau\ge K}$ whenever ${\tau\le t}$ and, in particular, ${\{\tau\le t\}=\{X^*_t\ge K\}}$. Optional sampling gives

$\displaystyle K1_{\{X^*_t \ge K\}}\le 1_{\{\tau\le t\}}X_\tau \le {\mathbb E}[1_{\{X^*_t\ge K\}}X_t\vert\mathcal{F}_\tau].$

Take expectations

$\displaystyle K{\mathbb P}(X^*_t\ge K)\le {\mathbb E}[1_{\{X^*_t\ge K\}}X_t].$

Bounding the right hand side above by ${{\mathbb E}[X_t]}$ gives the first inequality. Multiply by ${K^{p-2}}$ and integrate up to some limit ${L>0}$ to get,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle{\mathbb E}\left[(L\wedge X^*_t)^p\right] &\displaystyle=&\displaystyle p{\mathbb E}\left[\int_0^L K^{p-1} 1_{\{X^*_t\ge K\}}\,dK\right]\smallskip\\ &\displaystyle=&\displaystyle p\int_0^L K^{p-1}{\mathbb P}(X^*_t\ge K)\,dK\smallskip\\ &\displaystyle\le &\displaystyle p\int_0^L K^{p-2}{\mathbb E}\left[1_{\{X^*_t\ge K\}}X_t\right]\,dK\smallskip\\ &\displaystyle=&\displaystyle p{\mathbb E}\left[X_t\int_0^{L\wedge X^*_t}K^{p-2}\,dK\right]\smallskip\\ &\displaystyle=&\displaystyle\frac{p}{p-1}{\mathbb E}\left[X_t(L\wedge X^*_t)^{p-1}\right]. \end{array}$

Setting ${q=p/(p-1)}$ so that ${1/p+1/q=1}$, Hölder’s inequality states that

$\displaystyle E[X_t(L\wedge X^*_t)^{p-1}]\le\Vert X_t\Vert_p\Vert(L\wedge X^*_t)^{p-1}\Vert_q = \Vert X_t\Vert_p\Vert L\wedge X^*_t\Vert_p^{p-1}.$

Substituting into the previous inequality,

$\displaystyle \Vert L\wedge X^*_t\Vert_p^{p}\le\frac{p}{p-1}\Vert X_t\Vert_p\Vert L\wedge X^*_t\Vert_p^{p-1}.$

Finally, cancel ${\Vert L\wedge X^*_t\Vert_p^{p-1}}$ from both sides and take the limit ${L\uparrow\infty}$ to get the second inequality in the statement of the theorem. ⬜

PS: in the last line the constant p/(p-1) is missing…

Comment by Mike — 15 February 12 @ 12:12 PM

2. Hi George,

Just a small comment:
I think one needs to first consider a finite time grid to be able to assume that {X_t^* >= K} = {\tau <= t} (second line in proof of theorem 2) and the use MCT to get the result for countable time and continuous time in the cadlag case, like you mentioned in the planetmath proof.

Best,
Tigran

Comment by Anonymous — 1 March 12 @ 11:06 AM

• Hi. You can do it that way, but it is not necessary to start by restricting to the finite case. I already proved the Debut theorem and optional sampling for continuous-time cadlag processes. As we assume the process is cadlag, these can be applied directly to the continuous time case.

Comment by George Lowther — 1 March 12 @ 11:50 PM

Blog at WordPress.com.