Doob's Martingale Inequality

In mathematics, Doob's martingale inequality, also known as Kolmogorov's submartingale inequality is a result in the study of stochastic processes. It gives a bound on the probability that a submartingale exceeds any given value over a given interval of time. As the name suggests, the result is usually given in the case that the process is a martingale, but the result is also valid for submartingales.

The inequality is due to the American mathematician Joseph L. Doob.

Statement of the inequality

The setting of Doob's inequality is a submartingale relative to a filtration of the underlying probability space. The probability measure on the sample space of the martingale will be denoted by . The corresponding expected value of a random variable , as defined by Lebesgue integration, will be denoted by .

Informally, Doob's inequality states that the expected value of the process at some final time controls the probability that a sample path will reach above any particular value beforehand. As the proof uses very direct reasoning, it does not require any restrictive assumptions on the underlying filtration or on the process itself, unlike for many other theorems about stochastic processes. In the continuous-time setting, right-continuity (or left-continuity) of the sample paths is required, but only for the sake of knowing that the supremal value of a sample path equals the supremum over an arbitrary countable dense subset of times.

Discrete time

Let be a discrete-time submartingale relative to a filtration $\mathcal_1,\ldots,\mathcal_n$ of the underlying probability space, which is to say:
: X_i \leq \operatorname E _ \mid \mathcal_i
The submartingale inequality says that
: P\left \max_ X_i \geq C \right\leq \frac
for any positive number . The proof relies on the set-theoretic fact that the event defined by may be decomposed as the disjoint union of the events defined by and for all . Then
:CP(E_i) = \int_C\,dP \leq \int_X_i\,dP\leq\int_\text _n\mid\mathcal_i,dP=\int_X_n\,dP,
having made use of the submartingale property for the last inequality and the fact that $E_i\in\mathcal_i$ for the last equality. Summing this result as ranges from 1 to results in the conclusion
:$CP(E)\leq\int_X_n\,dP,$
which is sharper than the stated result. By using the elementary fact that , the given submartingale inequality follows.

In this proof, the submartingale property is used once, together with the definition of conditional expectation. The proof can also be phrased in the language of stochastic processes so as to become a corollary of the powerful theorem that a stopped submartingale is itself a submartingale. In this setup, the minimal index appearing in the above proof is interpreted as a stopping time.

Continuous time

Now let be a submartingale indexed by an interval of real numbers, relative to a filtration of the underlying probability space, which is to say:
: X_s \leq \operatorname E _t \mid \mathcal_s
for all The submartingale inequality says that if the sample paths of the martingale are almost-surely right-continuous, then
: P\left \sup_ X_t \geq C \right\leq \frac
for any positive number . This is a corollary of the above discrete-time result, obtained by writing
:$\sup_X_t=\sup\=\lim_\sup\$
in which is any sequence of finite sets whose union is the set of all rational numbers. The first equality is a consequence of the right-continuity assumption, while the second equality is purely set-theoretic. The discrete-time inequality applies to say that
:P\left \sup_ X_t \geq C \right\leq \frac
for each , and this passes to the limit to yield the submartingale inequality. This passage from discrete time to continuous time is very flexible, as it only required having a countable dense subset of , which can then automatically be built out of an increasing sequence of finite sets. As such, the submartingale inequality holds even for more general index sets, which are not required to be intervals or natural numbers.

Further inequalities

There are further submartingale inequalities also due to Doob. Now let be a martingale or a positive submartingale; if the index set is uncountable, then (as above) assume that the sample paths are right-continuous. In these scenarios, Jensen's inequality implies that is a submartingale for any number , provided that these new random variables all have finite integral. The submartingale inequality is then applicable to say that
:\text X_t, \geq Cleq \frac.
for any positive number . Here is the ''final time'', i.e. the largest value of the index set. Furthermore, one has
:\text sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

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{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

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External links

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{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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*
*
*
*
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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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*
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*
*
*
*
*
*

External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

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{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

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*
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External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory> X_T , ^p\leq \text\left sup_ , X_s, ^p\right\leq \left(\frac\right)^p\text X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

*
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*
*
*
*
*

External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

*
*
*
*
*
*
*
*
*

External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

*
*
*
*
*
*
*
*
*

External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory>X_T, ^p/math>
if is larger than one. This, sometimes known as ''Doob's maximal inequality'', is a direct result of combining the layer cake representation with the submartingale inequality and the Hölder inequality.

In addition to the above inequality, there holds
:\text\left, \sup_ X_ \ \leq \frac \left( 1 + \text max\\right)

Related inequalities

Doob's inequality for discrete-time martingales implies Kolmogorov's inequality: if ''X''₁, ''X''₂, ... is a sequence of real-valued independent random variables, each with mean zero, it is clear that

:\begin
\operatorname E\left X_1 + \cdots + X_n + X_ \mid X_1, \ldots, X_n \right&= X_1 + \cdots + X_n + \operatorname E\left X_ \mid X_1, \ldots, X_n \right\\
&= X_1 + \cdots + X_n,
\end

so S''_n'' = ''X''₁ + ... + ''X_n'' is a martingale. Note that Jensen's inequality implies that , S''_n'', is a nonnegative submartingale if S''_n'' is a martingale. Hence, taking ''p'' = 2 in Doob's martingale inequality,

: P\left S_i \ \geq \lambda \right\leq \frac,

which is precisely the statement of Kolmogorov's inequality.

Application: Brownian motion

Let ''B'' denote canonical one-dimensional Brownian motion. Then

: P\left \sup_ B_t \geq C \right\leq \exp \left( - \frac \right).

The proof is just as follows: since the exponential function is monotonically increasing, for any non-negative λ,

:$\left\ = \left\.$

By Doob's inequality, and since the exponential of Brownian motion is a positive submartingale,

:\begin
P\left \sup_ B_t \geq C \right& = P\left \sup_ \exp ( \lambda B_t ) \geq \exp ( \lambda C ) \right\\ pt& \leq \frac \\ pt& = \exp \left( \tfrac \lambda^2 T - \lambda C \right) && \operatorname E\left \exp (\lambda B_t) \right= \exp \left( \tfrac \lambda^2 t \right)
\end

Since the left-hand side does not depend on ''λ'', choose ''λ'' to minimize the right-hand side: ''λ'' = ''C''/''T'' gives the desired inequality.

References

Sources

*
*
*
*
*
*
*
*
*

External links

*

{{Stochastic processes

Probabilistic inequalities
Statistical inequalities
Martingale theory