# proof of Doob’s inequalities

Let $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\in\mathbb{T}},\mathbb{P})$ be a filtered probability space with countable index set $\mathbb{T}$. If $(X_{t})_{t\in\mathbb{T}}$ is a submartingale, we show that

 $\mathbb{P}\left(\sup_{s\leq t}X_{s}\geq K\right)\leq K^{-1}\mathbb{E}[(X_{t})_% {+}]$ (1)

and if $X$ is a martingale or nonnegative submartingale then,

 $\displaystyle\mathbb{P}(X^{*}_{t}\geq K)\leq K^{-1}\mathbb{E}[|X_{t}|],$ (2) $\displaystyle\|X^{*}_{t}\|_{p}\leq\frac{p}{p-1}\|X_{t}\|_{p}.$ (3)

for every $K>0$ and $p>1$.

First, let us consider the case where $\mathbb{T}$ is finite. The first time at which $X_{t}\geq K$,

 $\tau=\inf\left\{t\in\mathbb{T}:X_{t}\geq K\right\}$

is a stopping time (as hitting times are stopping times). By Doob’s optional sampling theorem for submartingales $X_{\tau\wedge t}\leq\mathbb{E}[X_{t}\mid\mathcal{F}_{\tau\wedge t}]$ and therefore,

 $K\mathbb{P}(\tau\leq t)\leq\mathbb{E}[1_{\{\tau\leq t\}}X_{\tau\wedge t}]\leq% \mathbb{E}[1_{\{\tau\leq t\}}X_{t}]$

However, $\tau\leq t$ if and only if $\sup_{s\leq t}X_{s}\geq K$ giving,

 $\mathbb{P}\left(\sup_{s\leq t}X_{s}\geq K\right)\leq K^{-1}\mathbb{E}[1_{\{% \sup_{s\leq t}X_{s}\geq K\}}X_{t}],$ (4)

where the supremum is understood to be over $s\in\mathbb{T}$. Now suppose that $\mathbb{T}$ is countable. Then choose finite subsets $\mathbb{T}_{n}\subseteq\mathbb{T}$ which increase to $\mathbb{T}$ as $n$ goes to infinity. Replacing $\mathbb{T}$ by $\mathbb{T}_{n}$ in inequality (4) and using the monotone convergence theorem to take the limit $n\rightarrow\infty$ extends (4) to arbitrary uncountable index sets. Then, inequality (1) follows immediately from (4).

Now, suppose that $X$ is a martingale. Jensen’s inequality gives

 $\mathbb{E}[|X_{t}|\mid\mathcal{F}_{s}]\geq\left|\mathbb{E}[X_{t}\mid\mathcal{F% }_{s}]\right|=|X_{s}|$

for any $s, so $|X|$ is a nonnegative submartingale. Therefore, it is enough to prove inequalities (2) and (3) for $X$ a nonnegative submartingale, and the martingale case follows by replacing $X$ by $|X|$.

So, we take $X$ to be a nonnegative submartingale in the following. In this case, (2) just reduces to (1) and it only remains to prove inequality (3).

For $p>1$, multiply (4) by $K^{p-1}$ and integrate up to some limit $L>0$,

 $\int_{0}^{L}K^{p-1}\mathbb{P}(X^{*}_{t}\geq K)\,dK\leq\int_{0}^{L}K^{p-2}% \mathbb{E}[1_{\{X^{*}_{t}\geq K\}}X_{t}]\,dK.$ (5)

The left hand side of this inequality can be computed by commuting the order of integration with respect to $\mathbb{P}$ and $dK$ (Fubini’s theorem),

 $\begin{split}\displaystyle\int_{0}^{L}K^{p-1}\mathbb{P}(X^{*}_{t}\geq K)\,dK&% \displaystyle=\mathbb{E}\left[\int_{0}^{L}K^{p-1}1_{\{X^{*}_{t}\geq K\}}\,dK% \right]\\ &\displaystyle=\frac{1}{p}\mathbb{E}[(L\wedge X^{*})^{p}].\end{split}$

The right hand side of (5) can be computed similarly,

 $\begin{split}\displaystyle\int_{0}^{L}K^{p-2}\mathbb{E}[1_{\{X^{*}_{t}\geq K\}% }X_{t}]\,dK&\displaystyle=\mathbb{E}\left[X_{t}\int_{0}^{L}K^{p-2}1_{\{X^{*}_{% t}\geq K\}}\,dK\right]\\ &\displaystyle=\frac{1}{p-1}\mathbb{E}[X_{t}(L\wedge X^{*}_{t})^{p-1}].\end{split}$

Putting these back into (5),

 $\|L\wedge X^{*}_{t}\|_{p}^{p}\leq\frac{p}{p-1}\mathbb{E}[X_{t}(L\wedge X^{*}_{% t})^{p-1}].$ (6)

Now let $q=p/(p-1)$, so that $p,q$ are conjugate (http://planetmath.org/ConjugateIndex) and the Hölder inequality gives

 $\mathbb{E}[X_{t}(L\wedge X^{*}_{t})^{p-1}]\leq\|X_{t}\|_{p}\|(L\wedge X^{*}_{t% })^{p-1}\|_{q}=\|X_{t}\|_{p}\|L\wedge X^{*}_{t}\|_{p}^{p-1}.$

Substituting into (6), the finite term $\|L\wedge X^{*}_{t}\|_{p}^{p-1}$ cancels to get

 $\|L\wedge X^{*}_{t}\|_{p}\leq\frac{p}{p-1}\|X_{t}\|_{p},$

and the result follows by letting $L$ increase to infinity.

Title proof of Doob’s inequalities ProofOfDoobsInequalities 2013-03-22 18:39:55 2013-03-22 18:39:55 gel (22282) gel (22282) 5 gel (22282) Proof msc 60G46 msc 60G44 msc 60G42