Martingale inequalities

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Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]X=(X_n)_{n\geq 0}[/math] be a stochastic process, such that [math]X_n[/math] is [math]\F_n[/math]-measurable for all [math]n\geq 0[/math]. We denote

[[math]] X_n^*:=\sup_{j\leq n}\vert X_j\vert. [[/math]]

Note that [math](X_n^*)_{n\geq 0}[/math] is increasing and [math]\F_n[/math]-adapted. Therefore if [math]X_n\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] for all [math]n\geq 0[/math], then [math](X_n^*)_{n\geq 0}[/math] is a submartingale.

Maximal inequality and Doob's inequality

Recall Markov's inequality in terms of [math](X_n^*)_{n\geq0}[/math], which is given by

[[math]] \p[X_n^*\geq \alpha]\leq \frac{\E[X_n^*]}{\alpha}, [[/math]]

with the obvious bound

[[math]] \E[X_n^*]\leq \sum_{j=1}^n\E[\vert X_j\vert]. [[/math]]

We shall see for instance that when [math](X_n)_{n\geq 0}[/math] is a martingale, one can replace [math]\E[X_n^*][/math] by [math]\E[\vert X_n\vert][/math].

Proposition

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a submartingale and let [math]\lambda \gt 0,k\in\N[/math]. Define

[[math]] \begin{align*} A&:=\left\{\max_{0\leq n\leq k}X_n\geq \lambda\right\}\\ B&:=\left\{\min_{0\leq n\leq k}X_n\leq -\lambda\right\}. \end{align*} [[/math]]

Then the following hold.

• [[math]] \lambda\p[A]\leq \E[X_k\one_A], [[/math]]
• [[math]] \lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_0]. [[/math]]

If [math](X_n)_{n\geq 0}[/math] is a martingale, then [math](\vert X_n\vert)_{n\geq 0}[/math] is a submartingale. Moreover, from [math](i)[/math] we get

[[math]] \lambda\p[X^*_k\geq \alpha]\leq \E[\vert X_k\vert\one_A]\leq \E[\vert X_k\vert] [[/math]]

and hence

[[math]] \p[X_k^*\geq \alpha]\leq \frac{\E[\vert X_k\vert]}{\alpha}. [[/math]]

Show Proof

We need to show both points.

• Let us introduce
  [[math]] T=\inf\{n\in\N\mid X_n\leq \lambda\}\land k. [[/math]]
  Then [math]T[/math] is a stopping time, which is bounded by [math]k[/math]. We thus have
  [[math]] \E[X_T]\leq \E[X_k]. [[/math]]
  We note that [math]X_T=X_k[/math] if [math]T=k[/math], which happens for [math]\omega\in A^C[/math]. Hence we get
  [[math]] \E[X_T]=\E[X_T\one_A+X_T\one_{A^C}]=\E[X_T\one_A]+\E[X_k\one_{A^C}]\leq \underbrace{\E[X_k]}_{\E[X_k(\one_A+\one_{A^C})]}. [[/math]]
  Now we note that
  [[math]] \E[X_T\one_A]\geq \lambda \E[\one_A]=\lambda \p[A]. [[/math]]
  Therefore we get
  [[math]] \lambda\p[A]\leq \E[X_k\one_A]. [[/math]]
• Let us define
  [[math]] S=\inf\{n\in\N\mid X_n\leq -\lambda\}\land k. [[/math]]
  Again [math]S[/math] is a stopping time, which is bounded by [math]k[/math]. We hence have
  [[math]] \E[X_S]\geq \E[X_0]. [[/math]]
  Thus
  [[math]] \E[X_0]\leq \E[X_S\one_B]+\E[X_S\one_{B^C}]\leq -\lambda \p[B]+\E[X_k\one_{B^C}]. [[/math]]
  Therefore we get
  [[math]] \lambda\p[B]\leq \E[X_k\one_{B^C}]-\E[X_0]. [[/math]]

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Proposition (Kolmogorov's inequality)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a martingale, such that for all [math]n\geq 0[/math] we have [math]\E[X_n^2] \lt \infty[/math]. Then

[[math]] \p\left[\max_{0\leq k\leq n}\vert X_k\vert\geq \lambda\right]\leq \frac{\E[X_k^2]}{\lambda^2}. [[/math]]

Show Proof

We use the fact that [math](X_n^2)_{n\geq 0}[/math] is a positive submartingale. Therefore we get

[[math]] \lambda^2\underbrace{\p\left[\max_{0\leq k\leq n}\vert X_k\vert^2\geq \lambda^2\right]}_{\p\left[\max_{0\leq k\leq k}\vert X_k\vert\geq \lambda\right]}\leq \E\left[X_k^2\one_{\left\{\max_{0\leq k\leq n}\vert X_k\vert^2\geq \lambda^2\right\}}\right]\leq \E[X_k^2] [[/math]]

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Theorem (Maximal inequality)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a submartingale. Then for all [math]\lambda\geq 0[/math] and [math]n\in\N[/math], we get

[[math]] \lambda\p\left[\max_{0\leq k\leq n}\vert X_k\vert\geq \lambda\right]\leq \E[X_0]+2\E[\vert X_n\vert]. [[/math]]

Show Proof

Let [math]A[/math] and [math]B[/math] be defined as in Proposition 9.1. Then

[[math]] \begin{align*} \lambda \p\left[\max_{0\leq k\leq n}\vert X_k\vert \geq \lambda\right]&=\lambda\p[A\cup B]\leq \E[X_k\one_A]-\E[X_0]+\E[X_k\one_{B^C}]\\ &\leq \E[\vert X_0\vert]+\E[\vert X_n\vert]+\E[\vert X_n\vert]=\E[\vert X_0\vert]+2\E[\vert X_n\vert]. \end{align*} [[/math]]

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Theorem (Doob's inequality)

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math]p \gt 1[/math] and [math]q \gt 1[/math], such that [math]\frac{1}{p}+\frac{1}{q}=1[/math].

• If [math](X_n)_{n\geq 0}[/math] is a submartingale, then for all [math]n\geq 0[/math] we have
  [[math]] \left\|\max_{0\leq k\leq n}X_k^+\right\|_p\leq q\left\|X_n^+\right\|_p. [[/math]]
• If [math](X_n)_{n\geq 0}[/math] is a martingale, then for all [math]n\geq 0[/math] we have
  [[math]] \left\|\max_{0\leq k\leq n}\vert X_k\vert\right\|_p\leq q\| X_n\|_p. [[/math]]

Recall that if [math]X\in L^1(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], then

[[math]] \|X\|_p=\E[\vert X\vert^p]^{1/p}. [[/math]]

Moreover, if [math]p=q=2[/math] and [math](X_n)_{n\geq 0}[/math] is a martingale, then for all [math]n\geq 0[/math] we have

[[math]] \E\left[\max_{0\leq k\leq n} X_k^2\right]\leq 4\E[X_n^2]. [[/math]]

In general, we have

[[math]] \vert X_n\vert^2\leq \max_{0\leq k\leq n}\vert X_k\vert^p. [[/math]]

Therefore we get

[[math]] \E[\vert X_n\vert^p]\leq \E\left[\max_{0\leq k\leq n}\vert X_k\vert^p\right]\leq ^{Doob}q^p\E[\vert X_n\vert^p]. [[/math]]

We shall also recall that for [math]X\in L^p(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], we can write

[[math]] \E[\vert X\vert^p]=\E\left[\int_0^{\vert X\vert}p\lambda^{p-1}d\lambda\right]=\E\left[\int_0^\infty \one_{\{\vert X\vert\geq \lambda\}}p\lambda^{p-1}d\lambda\right]=\int_0^\infty p\lambda^{p-1}\p[\vert X\vert \geq \lambda]d\lambda [[/math]]

by using Fubini's theorem.

Show Proof

It is enough to prove [math](ii)[/math]. Since [math](X_n)_{n\geq 0}[/math] is a submartingale, we know that [math](X_n^+)_{n\geq 0}[/math] is a submartingale. Hence

[[math]] \lambda\p\left[\max_{0\leq k\leq n}X_k^+\geq \lambda\right]\geq\E\left[X_n^+\one_{\{\max_{0\leq k\leq n}X_k^+\geq \lambda\}}\right]. [[/math]]

Now ler [math]Y_n:=\max_{0\leq k\leq n}X_k^+[/math]. Then for any [math]k \gt 0[/math], we have

[[math]] \begin{align*} \E[(Y_n\land k)^p]&=\int_0^\infty p\lambda^{p-1}\p[Y_n\land k\geq \lambda]d\lambda=\int_0^np\lambda^{p-1}\p[Y_n\geq\lambda]d\lambda\\ &\leq \int_0^n p\lambda^{p-1}\left(\frac{1}{\lambda}\E[X_n^+\one_{\{ Y_n\geq \lambda\}}]\right)d\lambda\\ &=\E\left[\int_0^n p\lambda^{p-1}X_n^+\one_{\{Y_n\geq\lambda\}}d\lambda\right]=\E\left[\int_0^{Y_n\land k}p\lambda^{p-2}X_n^+d\lambda\right]\\ &=\E\left[\frac{p}{p-1}(Y_n\land k)^{p-1}X_n^+\right]\\ &\leq q\E[(X_n^+)^p]^{1/p}\E[(Y_n\land k)^p]^{1/q}, \end{align*} [[/math]]

where we have used that [math]q=\frac{p}{p-1}[/math] and Markov's inequality. Therefore we obtain

[[math]] \E[(Y_n\land k)^p]\leq q\E[(X_n^+)^p]^{1/p}\E[(Y_n\land k)^p]^{1/q}. [[/math]]

Since [math]\E[(Y_n\land k)^p]\not=0[/math], we can divide by it to get

[[math]] \E[(Y_n\land k)^p]^{1-1/q=1/p}\leq q\E[(X_n^+)^p]^{1/p} [[/math]]

and thus

[[math]] \| (Y_k\land n)\|_p\leq q\| X_n^+\|_p. [[/math]]

Now for [math]k\to\infty[/math], monotone convergence implies that

[[math]] \|Y_n\|_p\leq q\|X_n^+\|_p. [[/math]]

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Corollary

Let [math](\Omega,\F,(\F_n)_{n\geq0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a martingale and [math]p \gt 1[/math], [math]q \gt 1[/math] such that [math]\frac{1}{p}+\frac{1}{q}=1[/math]. Then

[[math]] \left\| \sup_{n\geq 0}\vert X_n\vert\right\|_p\leq q\sup_{n\geq 0}\| X_n\|_p. [[/math]]

Show Proof

Exercise

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General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].