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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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    @mpiktas: I wish it was correct )2011-06-23

1 Answers 1

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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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    @mpiktas: I think that's it. There is also written that if $X_n$ is a sign-semidefinite supermartingale, then $C = 1$ - so inequality which I stated in topic http://math.stackexchange.com/questions/45886/bounds-for-submartingale is still correct. Anyway, either in the book of 2006 the proved that $C=1$ for all cases - but then there is a counterexample by George.2011-06-23