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In the book "Optimal Stopping and Free-Boundary Problems" there is given Doob's inequality of the following form.

Let $X = (X_t,F_t)$ be a submartingale. Then for any $\varepsilon>0$ and each $T>0$ $$ \mathsf P\left\{\sup\limits_{t\leq T}|X_t|\geq \varepsilon\right\}\leq \frac 1\varepsilon \sup\limits_{t\leq T}\,\,\,\mathsf E\,|X_t|. $$

On the other hand, George Lowther presented the following example. Let $X_0 = 0$, $$ X_1 = \begin{cases} 1, &\quad p = 1/3; \\ 0, &\quad p = 1/3; \\ -1,&\quad p =1/3. \end{cases} $$ and $X_2 = 1$ if $X_1 = 1$, $X_2 = 0$ if $X_1 = -1$ and
$$ X_2 = \begin{cases} 1, &\quad p = 1/2; \\ -1,&\quad p =1/2. \end{cases} $$ if $X_1 = 0$. Finally, $X_n = X_2$ for all $n\geq 2$.

It is easy to check that this is a submartingale. On the other hand, for $\varepsilon = 1$ we have $$ \mathsf P\left\{\sup\limits_{n\leq 2}|X_n|\geq 1\right\} = 1 $$ but $\mathsf E[|X_0|] = 0$, $\mathsf E[|X_1|] = 2/3$ and $\mathsf E[|X_2|] = 2/3$ - so we have inequality $1\leq 2/3$.

Could you help with finding a mistake since I am confused?

The reference can be seen here: http://books.google.com/books?id=UinZbLqpUDEC&pg=PA60&source=gbs_toc_r&cad=4#v=onepage&q&f=false page 62.

This question raised from the discussion here: Bounds for submartingale

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I've deleted my incorrect answer. –  mpiktas Jun 23 '11 at 9:22
    
@mpiktas: I wish it was correct ) –  Ilya Jun 23 '11 at 9:28
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1 Answer

up vote 3 down vote accepted

Doob inequality is stated for nonnegative submartingales $X_t$:

$$E\sup_{t\le T}X_t\le EX_T$$

Note, no modulus. In your case you have a submartingale $X_t$, which is not nonnegative. Furthermore $|X_t|$ is not actualy a submartingale, so you cannot apply this inequality here.

The probability space here is $\Omega=\{\omega_1,...,\omega_4\}$, with $P(\{\omega_1\})=1/3$, $P(\{\omega_2\})=1/3$, $P(\{\omega_3\})=1/6$ and $P(\{\omega_4\})=1/6$.

We have $X_1(\omega_1,...,\omega_4)=(1,-1,0,0)$ and $X_2(\omega_1,...,\omega_4)=(1,0,1,-1)$. Now $Y_1=|X_1|=(1,1,0,0)$ and $Y_2=|X_2|=(1,0,1,1)$ and we have that

$$E(Y_2|Y_1)=(1/3,1/3,1,1)$$

Update: I've finaly found the correct answer in another book of Shyriaev (page 492, Probability, second edition, 1995). The actual inequality for any submartingales is

$$P(\sup_{t\le T}|X_t|>\varepsilon)\le \frac{3}{\varepsilon}\sup_{t\le T}E|X_t|$$

Note the missing constant 3. This means that there is an error in the book in the link.

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$|X_1(\omega_2)| = 1$, isn't it? –  Ilya Jun 23 '11 at 9:16
    
@Gortaur, ah yes :) Sorry –  mpiktas Jun 23 '11 at 9:21
    
@Gortaur, the error is somewhere else. I will try to find it and fix the answer. –  mpiktas Jun 23 '11 at 9:21
    
@Gortaur, here is the second try. Hopefully it is ok :) But with my recent luck of incorrect answers I do not hold my breath. –  mpiktas Jun 23 '11 at 9:33
    
@mpiktas: I am interested exactly in the correctness of the result from the book I quoted. Do you mean that there is a mistake? –  Ilya Jun 23 '11 at 9:37
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