proof of Doob’s inequalities

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

$$\mathbb{P}(\underset{s\le t}{sup}{X}_s\ge K)\le K^{-1}𝔼[{(X_t)}_{+}]$$(1)

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

$$\mathbb{P}(X_t^{*}\ge K)\le K^{-1}𝔼[|X_t|],$$(2)

$${\parallel X_t^{*}\parallel }_p\le \frac{p}{p-1}{\parallel X_t\parallel }_p.$$(3)

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

First, let us consider the case where $𝕋$ is finite. The first time at which $X_t\ge K$,

$$\tau =inf\{t\in 𝕋:X_t\ge K\}$$

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

$$K\mathbb{P}(\tau \le t)\le 𝔼[1_{\{\tau \le t\}}{X}_{\tau \wedge t}]\le 𝔼[1_{\{\tau \le t\}}{X}_t]$$

However, $\tau \le t$ if and only if ${sup}_{s\le t}{X}_s\ge K$ giving,

$$\mathbb{P}(\underset{s\le t}{sup}{X}_s\ge K)\le K^{-1}𝔼[1_{\{{sup}_{s\le t}{X}_s\ge K\}}{X}_t],$$(4)

where the supremum is understood to be over $s\in 𝕋$.Now suppose that $𝕋$ is countable. Then choose finite subsets ${𝕋}_n\subseteq 𝕋$ which increase to $𝕋$ as $n$ goes to infinity. Replacing $𝕋$ by ${𝕋}_n$ in inequality (4) and using the monotone convergence theorem to take the limit $n\to \mathrm{\infty }$ 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

$$𝔼[|X_t|\mid {ℱ}_s]\ge \left|𝔼[X_t\mid {ℱ}_s]\right|=|X_s|$$

for any , 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 $L>0$,

$${\int }_0^L{K}^{p-1}\mathbb{P}(X_t^{*}\ge K)dK\le {\int }_0^L{K}^{p-2}𝔼[1_{\{X_t^{*}\ge K\}}{X}_t]dK.$$(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),

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

Putting these back into (5),

$${\parallel L\wedge X_t^{*}\parallel }_p^p\le \frac{p}{p-1}𝔼[X_t{(L\wedge X_t^{*})}^{p-1}].$$(6)

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

$$𝔼[X_t{(L\wedge X_t^{*})}^{p-1}]\le {\parallel X_t\parallel }_p{\parallel {(L\wedge X_t^{*})}^{p-1}\parallel }_q={\parallel X_t\parallel }_p{\parallel L\wedge X_t^{*}\parallel }_p^{p-1}.$$

Substituting into (6), the finite term ${\parallel L\wedge X_t^{*}\parallel }_p^{p-1}$ cancels to get

$${\parallel L\wedge X_t^{*}\parallel }_p\le \frac{p}{p-1}{\parallel X_t\parallel }_p,$$

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