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Month: November 2020

Sub-Gaussian tail bound for local martingales

Herman Chernoff
Herman Chernoff (1923 – ) One of the first to play with exponential Markov inequalities in the 1950s. He was not aware of the work of Harald Cramér in the 1930s!

This post is devoted to a sub-Gaussian tail bound and exponential square integrability for local martingales, taken from my master course on stochastic calculus.

Sub-Gaussian tail bound and exponential square integrability for local martingales. Let \( {M={(M_t)}_{t\geq0}} \) be a continuous local martingale issued from the origin. Then for all \( {t,K,r\geq0} \),

\[ \mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r, \langle M\rangle_t\leq K\Bigr) \leq2\mathrm{e}^{-\frac{r^2}{2K}}, \]

and in particular, if \( {\langle M\rangle_t\leq Ct} \) then

\[ \mathbb{P}\Bigr(\sup_{s\in[0,t]}|M_s|\geq r\Bigr) \leq2\mathrm{e}^{-\frac{r^2}{2Ct}} \]

and, for all \( {\alpha<\frac{1}{2Ct}} \),

\[ \mathbb{E}\Bigr(\mathrm{e}^{\alpha\sup_{s\in[0,t]}|M_s|^2}\Bigr)<\infty. \]

The condition \( {\langle M\rangle_t\leq Ct} \) is a comparison to Brownian motion for which equality holds.

Proof. For all \( {\lambda,t\geq0} \), the Doléans-Dade exponential

\[ X^\lambda ={\Bigr(\mathrm{e}^{\lambda M_t-\frac{\lambda^2}{2}\langle M\rangle_t}\Bigr)}_{t\geq0} \]

is a positive super-martingale with \( {X^\lambda_0=1} \) and \( {\mathbb{E}X^\lambda_t\leq1} \) for all \( {t\geq0} \). For all \( {t,\lambda,r,K\geq0} \), by using the maximal inequality for the super-martingale \( {X^\lambda} \) in the last step,

\[ \begin{array}{rcl} \mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}M_s\geq r\Bigr) &\leq&\mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}X^\lambda_s\geq\mathrm{e}^{\lambda r-\frac{\lambda^2}{2}K}\Bigr) \\ &\leq&\mathbb{P}\Bigr(\sup_{0\leq s\leq t}X^\lambda_s\geq\mathrm{e}^{\lambda r-\frac{\lambda^2}{2}K}\Bigr)\\ &\leq&\mathbb{E}(X^\lambda_0)\mathrm{e}^{-\lambda r+\frac{\lambda^2}{2}K} =\mathrm{e}^{-\lambda r+\frac{\lambda^2}{2}K}. \end{array} \]

Taking \( {\lambda=r/K} \) gives

\[ \mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}M_s\geq r\Bigr) \leq\mathrm{e}^{-\frac{r^2}{2K}}. \]

The same reasoning for \( {-M} \) instead of \( {M} \) provides (note that \( {\langle -M\rangle=\langle M\rangle} \) obviously)

\[ \mathbb{P}\Bigr(\langle M\rangle_t\leq K,\sup_{0\leq s\leq t}(-M_s)\geq r\Bigr) \leq\mathrm{e}^{-\frac{r^2}{2K}}. \]

The union bound (hence the prefactor \( {2} \)) gives then the first desired inequality. The exponential square integrability comes from the usual link between tail bound and integrability, namely if \( {X=\sup_{s\in[0,t]}|M_s|} \), \( {U(x)=\mathrm{e}^{\alpha x^2}} \), \( {\alpha<\frac{1}{2Kt}} \), then, by Fubini-Tonelli,

\[ \begin{array}{rcl} \mathbb{E}(U(X)) &=&\mathbb{E}\Bigr(\int_0^XU'(x)\mathrm{d}x\Bigr)\\ &=&\mathbb{E}\Bigr(\int_0^\infty\mathbf{1}_{x\leq X}U'(x)\mathrm{d}x\Bigr)\\ &=&\int_0^\infty U'(x)\mathbb{P}(X\geq x)\mathrm{d}x\  &\leq&\int_0^\infty2\alpha x\mathrm{e}^{\alpha x^2}\mathrm{e}^{-\frac{x^2}{2Kt}}\mathrm{d}x <\infty. \end{array} \]

Doob maximal inequality for super-martingales. If \( {M} \) is a continuous super-martingale, then for all \( {t\geq0} \) and \( {\lambda>0} \), denoting \( {M^-=\max(0,-M)} \),

\[ \mathbb{P}\Bigr(\max_{s\in[0,t]}|M_s|\geq\lambda\Bigr) \leq\frac{\mathbb{E}(M_0)+2\mathbb{E}(M^-_t)}{\lambda}. \]

In particular when \( {M} \) is non-negative then \( {\mathbb{E}(M^-)=0} \) and the upper bound is \( {\frac{\mathbb{E}(M_0)}{\lambda}} \).

Proof. Let us define the bounded stopping time

\[ T=t\wedge \inf\{s\in[0,t]:M_s\geq \lambda\}. \]

We have \( {M_T\in\mathrm{L}^1} \) since \( {|M_T|\leq\max(|M_0|,|M_t|,\lambda)} \). By the Doob stopping theorem for the sub-martingale \( {-M} \) and the bounded stopping times \( {0} \) and \( {T} \) that satisfy \( {M_0\in\mathrm{L}^1} \) and \( {M_T\in\mathrm{L}^1} \), we get

\[ \mathbb{E}(M_0) \geq\mathbb{E}(M_T) \geq \lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) +\mathbb{E}(M_t\mathbf{1}_{\max_{s\in[0,t]}M_s<\lambda}) \]

hence, recalling that \( {M^-=\max(-M,0)} \),

\[ \lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) \leq \mathbb{E}(M_0)+\mathbb{E}(M^-_t). \]

This produces the desired inequality when \( {M} \) is non-negative. For the general case, we observe that the Jensen inequality for the nondecreasing convex function \( {u\in\mathbb{R}\mapsto\max(u,0)} \) and the sub-martingale \( {-M} \) shows that \( {M^-} \) is a sub-martingale. Thus, by the Doop maximal inequality for non-negative sub-martingales,

\[ \lambda\mathbb{P}(\max_{s\in[0,t]}M^-_s\geq \lambda) \leq\mathbb{E}(M^-_t). \]

Finally, putting both inequalities together gives

\[ \lambda\mathbb{P}(\max_{s\in[0,t]}|M_s|\geq \lambda) \leq \lambda\mathbb{P}(\max_{s\in[0,t]}M_s\geq \lambda) +\lambda\mathbb{P}(\max_{s\in[0,t]}M^-_s\geq \lambda) \leq\mathbb{E}(M_0)+2\mathbb{E}(M^-_t). \]

Doob maximal inequalities. Let \( {M} \) be a continuous process.

  1. If \( {M} \) is a martingale or a non-negative sub-martingale then for all \( {p\geq1} \), \( {t\geq0} \), \( {\lambda>0} \),

    \[ \mathbb{P}\Bigr(\max_{s\in[0,t]}|M_s|\geq\lambda\Bigr) \leq\frac{\mathbb{E}(|M_t|^p)}{\lambda^p}. \]

  2. If \( {M} \) is a martingale then for all \( {p>1} \) and \( {t\geq0} \),

    \[ \mathbb{E}\Bigr(\max_{s\in[0,t]}|M_s|^p\Bigr) \leq\Bigr(\frac{p}{p-1}\Bigr)^p\mathbb{E}(|M_t|^p) \]

    in other words

    \[ \Bigr\|\max_{s\in[0,t]}|M_s|\Bigr\|_p\leq\frac{p}{p-1}\|M_t\|_p. \]

    In particular if \( {M_t\in\mathrm{L}^p} \) then \( {M^*_t=\max_{s\in[0,t]}M_s\in\mathrm{L}^p} \).

Comments. This inequality allows to control the tail of the supremum by the moment at the terminal time. It is a continuous time martingale version of the simpler Kolmogorov maximal inequality for sums of independent and identically distributed random variables. Note that \( {q=1/(1-1/p)=p/(p-1)} \) is the Hölder conjugate of \( {p} \) namely \( {1/p+1/q=1} \). The inequality is often used with \( {p=2} \), for which \( {(p/(p-1))^p=4} \).

Proof. We can always assume that the right hand side is finite, otherwise the inequalities are trivial.

  1. If \( {M} \) is a martingale, then by the Jensen inequality for the convex function \( {u\in\mathbb{R}\mapsto |u|^p} \), the process \( {|M|^p} \) is a sub-martingale. Similarly, If \( {M} \) is a non-negative sub-martingale then, since \( {u\in[0,+\infty)\mapsto u^p} \) is convex and non-decreasing it follows that \( {M^p=|M|^p} \) is a sub-martingale. Therefore in all cases \( {{(|M_s|^p)}_{s\in[0,t]}} \) is a sub-martingale. Let us define the bounded stopping time

    \[ T=t\wedge \inf\{s\geq0:|M_s|\geq\lambda\}. \]

    Note that \( {|M_T|\leq\max(|M_0|,\lambda)} \) and thus \( {M_T\in\mathrm{L}^1} \). The Doob stopping theorem for the sub-martingale \( {|M|^p} \) and the bounded stopping times \( {T} \) and \( {t} \) that satisfy \( {T\leq t} \) gives

    \[ \mathbb{E}(|M_T|^p)\leq\mathbb{E}(|M_t|^p). \]

    On the other hand the definition of \( {T} \) gives

    \[ |M_T|^p \geq\lambda^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda} +|M_t|^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|<\lambda}\\ \geq\lambda^p\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda}. \]

    It remains to combine these inequalities to get the desired result.

  2. If we introduce for all \( {n\geq1} \) the “localization” stopping time

    \[ T_n=t\wedge\inf\{s\geq0:|M_s|\geq n\}, \]

    then the desired inequality for the bounded sub-martingale \( {{(|M_{s\wedge T_n}|)}_{s\in[0,t]}} \) would give

    \[ \mathbb{E}(\max_{s\in[0,t]}|M_{s\wedge T_n}|^p) \leq\left(\frac{p}{p-1}\right)^p\mathbb{E}(|M_t|^p), \]

    and the desired result for \( {{(M_s)}_{s\in[0,t]}} \) would then follow by monotone convergence theorem. Thus this shows that we can assume without loss of generality that \( {{(|M_s|)}_{s\in[0,t]}} \) is bounded, in particular that \( {\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)<\infty} \). This a martingale localization argument! The previous proof gives

    \[ \mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda) \leq\frac{\mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda})}{\lambda} \]

    for all \( {\lambda>0} \), and thus

    \[ \int_0^\infty\lambda^{p-1} \mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda)\mathrm{d}\lambda \leq \int_0^\infty\lambda^{p-2} \mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda})\mathrm{d}\lambda. \]

    Now the Fubini-Tonelli theorem gives

    \[ \int_0^\infty\lambda^{p-1}\mathbb{P}(\max_{s\in[0,t]}|M_s|\geq\lambda)\mathrm{d}\lambda =\mathbb{E}\int_0^{\max_{s\in[0,t]}|M_s|}\lambda^{p-1}\mathrm{d}\lambda =\frac{1}{p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p). \]

    and similarly (here we need \( {p>1} \))

    \[ \int_0^\infty\lambda^{p-2}\mathbb{E}(|M_t|\mathbf{1}_{\max_{s\in[0,t]}|M_s|\geq\lambda)})\mathrm{d}\lambda =\frac{1}{p-1}\mathbb{E}(|M_t|\max_{s\in[0,t]}|M_s|^{p-1}). \]

    Combining all this gives

    \[ \mathbb{E}(\max_{s\in[0,t]}|M_s|^p) \leq\frac{p}{p-1} \mathbb{E}(M_t\max_{s\in[0,t]}|M_s|^{p-1}). \]

    But since the Hölder inequality gives

    \[ \mathbb{E}(|M_t|\max_{s\in[0,t]}|M_s|^{p-1}) \leq\mathbb{E}(|M_t|^p)^{1/p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)^{\frac{p-1}{p}}, \]

    we obtain

    \[ \mathbb{E}(\max_{s\in[0,t]}|M_s|^p) \leq\frac{p}{p-1}\mathbb{E}(|M_t|^p)^{1/p}\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)^{\frac{p-1}{p}}. \]

    Consequently, since \( {\mathbb{E}(\max_{s\in[0,t]}|M_s|^p)<\infty} \), we obtain the desired inequality.

Doob stopping theorem for sub-martingales. If \( {M} \) is a continuous sub-martingale and \( {S} \) and \( {T} \) are bounded stopping times such that \( {S\leq T,\quad M_S\in\mathrm{L}^1,\quad M_T\in\mathrm{L}^1} \), then

\[ \mathbb{E}(M_S)\leq\mathbb{E}(M_T). \]

Proof. We proceed as in the proof of the Doob stopping theorem for martingales, by assuming first that \( {S} \) and \( {T} \) take their values in the finite set \( {\{t_1,\ldots,t_n\}} \) where \( {t_1<\cdots<t_n} \). In this case \( {M_T} \) and \( {M_S} \) are in \( {\mathrm{L}^1} \) automatically. The inequality \( {S\leq T} \) gives \( {\mathbf{1}_{S\geq t}\leq\mathbf{1}_{T\geq t}} \) for all \( {t} \). Using this fact and the sub-martingale property of \( {M} \), we get

\[ \begin{array}{rcl} \mathbb{E}(M_S) &=&\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\overbrace{\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})}^{\geq0}\mathbf{1}_{S\geq t_k}\Bigr)\\ &\leq&\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})\mathbf{1}_{T\geq t_k}\Bigr)\\ &=&\mathbb{E}(M_T). \end{array} \]

More generally, when \( {S} \) and \( {T} \) are arbitrary bounded stopping times satisfying \( {S\leq T} \), we proceed by approximation as in the proof of the Doob stopping for martingales, using again the sub-martingale nature of \( {M} \) to get uniform integrability.

Doob stopping theorem for martingales. If \( {M} \) is a continuous martingale and \( {T:\Omega\rightarrow[0,+\infty]} \) is a stopping time then \( {{(M_{t\wedge T})}_{t\geq0}} \) is a martingale: for all \( {t\geq0} \) and \( {s\in[0,t]} \), we have

\[ M_{t\wedge T}\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_{t\wedge T}\mid\mathcal{F}_s)=M_{s\wedge T}. \]

Moreover, if \( {T} \) is bounded, or if \( {T} \) is almost surely finite and \( {{(M_{t\wedge T})}_{t\geq0}} \) is uniformly integrable (for instance dominated by an integrable random variable), then

\[ M_T\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_T)=\mathbb{E}(M_0). \]

Comments. The most important is that \( {{(M_{t\wedge T})}_{t\geq0}} \) is a martingale. We have always \( {\lim_{t\rightarrow\infty}M_{T\wedge t}\mathbf{1}_{T<\infty}=M_T\mathbf{1}_{T<\infty}} \) almost surely. When \( {T<\infty} \) almost surely we could use what we know on \( {M} \) and \( {T} \) to deduce by monotone or dominated convergence that this holds in \( {\mathrm{L}^1} \), giving \( {\mathbb{E}(M_T)=\mathbb{E}(\lim_{t\rightarrow\infty}M_{t\wedge T})=\lim_{t\rightarrow\infty}\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_0)} \).

The theorem states that this is automatically the case when \( {T} \) is bounded or when \( {M^T} \) is uniformly integrable. Furthermore, if \( {M^T} \) is uniformly integrable then it can be shown that \( {M_\infty} \) exists, giving a sense to \( {M_T} \) even on \( {\{T=\infty\}} \), and then \( {\mathbb{E}(M_T)=\mathbb{E}(M_0)} \).

Proof. Let assume first that \( {T} \) takes a finite number of values \( {t_1<\cdots<t_n} \). Let us show that \( {M_T\in\mathrm{L}^1} \) and \( {\mathbb{E}(M_T)=\mathbb{E}(M_0)} \). We have \( {M_T=\sum_{k=1}^nM_{t_k}\mathbf{1}_{T=t_k}\in\mathrm{L}^1} \), and moreover, using

\[ \{T\geq t_k\}=(\cup_{i=1}^{k-1}\{T=t_i\})^c\in\mathcal{F}_{t_{k-1}}, \]

and the martingale property \( {\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})=0} \), for all \( {k} \), we get

\[ \mathbb{E}(M_T) =\mathbb{E}(M_0) +\mathbb{E}\Big(\sum_{k=1}^n\mathbb{E}(M_{t_k}-M_{t_{k-1}}\mid\mathcal{F}_{t_{k-1}})\mathbf{1}_{T\geq t_k}\Bigr) =\mathbb{E}(M_0). \]

Suppose now that \( {T} \) takes an infinite number of values but is bounded by some constant \( {C} \). For all \( {n\geq0} \), we approximate \( {T} \) by the piecewise constant random variable (discretization of \( {[0,C]} \)).

\[ T_n=C\mathbf{1}_{T=C}+\sum_{k=1}^{n}t_{k}\mathbf{1}_{t_{k-1}\leq T<t_{k}} \quad\text{where}\quad t_k=t_{n,k}=C\frac{k}{n}. \]

This is a stopping time since for all \( {t\geq0} \), \( {\{T_n\leq t\}=\{T_n\leq\lfloor t\rfloor\}\in\mathcal{F}_{\lfloor t\rfloor}} \), which reduces the problem to a discrete \( {t} \), and then for all integer \( {m\geq0} \), we have that \( {\{T_n=m\} =\varnothing\in\mathcal{F}_0} \) if \( {m\not\in\{t_{k}:1\leq k\leq n\}} \), while \( {\{T_n=m\}=\{T=C\}\in\mathcal{F}_C} \) if \( {m=C} \), and

\[ \{T_n=m\} =\{T<t_{k-1}\}^c\cap\{T<t_{k}\}\in\mathcal{F}_{t_{k}} \]

if \( {m=t_{k}} \), \( {1\leq k\leq n} \), where we used the fact that for all \( {t\geq0} \),

\[ \{T=t\}=\{T\leq t\}\cap\{T<t\}^c=\{T\leq t\}\cap\cap_{r=1}^\infty\{T>t-1/r\}\in\mathcal{F}_t. \]

Since \( {T_n} \) takes a finite number of values, the previous step gives \( {\mathbb{E}(M_{T_n})=\mathbb{E}(M_0)} \). On the other hand, almost surely, \( {T_n\rightarrow T} \) as \( {n\rightarrow\infty} \). Since \( {M} \) is continuous, it follows that almost surely \( {M_{T_n}\rightarrow M_T} \) as \( {n\rightarrow\infty} \). Let us show now that \( {{(M_{T_n})}_{n\geq1}} \) is uniformly integrable. Since for all \( {n\geq0} \), \( {T_n} \) takes its values in a finite set \( {t_1<\cdots<t_{m_n}\leq C} \), the martingale property and the Jensen inequality give, for all \( {R>0} \),

\[ \begin{array}{rcl} \mathbb{E}(|M_C|\mathbf{1}_{|M_{T_n}|\geq R}) &=&\sum_k\mathbb{E}(|M_C|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\sum_k\mathbb{E}(\mathbb{E}(|M_C|\mid\mathcal{F}_{t_k})\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &\geq&\sum_k\mathbb{E}(|\mathbb{E}(M_C\mid\mathcal{F}_{t_k})|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\sum_k\mathbb{E}(|M_{t_k}|\mathbf{1}_{|M_{t_k}|\geq R,T_n=t_k})\\ &=&\mathbb{E}(|M_{T_n}|\mathbf{1}_{|M_{T_n}|\geq R}). \end{array} \]

Now \( {M} \) is continuous and thus locally bounded, and \( {M_C\in\mathrm{L}^1} \), thus, by dominated convergence,

\[ \sup_n\mathbb{E}(|M_{T_n}|\mathbf{1}_{|M_{T_n}|>R}) \leq\mathbb{E}(|M_C|\mathbf{1}_{\sup_{s\in[0,C]}|M_s|\geq R}) \underset{R\rightarrow\infty}{\longrightarrow}0. \]

Therefore \( {{(M_{T_n})}_{n\geq0}} \) is uniformly integrable. As a consequence

\[ \overset{\mathrm{a.s.}}{\lim_{n\rightarrow\infty}}M_{T_n}=M_T\in\mathrm{L}^1 \quad\text{and}\quad \mathbb{E}(M_T)=\lim_{n\rightarrow\infty}\mathbb{E}(M_{T_n})=\mathbb{E}(M_0). \]

Let us suppose now that \( {T} \) is an arbitrary stopping time. For all \( {0\leq s\leq t} \) and \( {A\in\mathcal{F}_s} \), the random variable \( {S=s\mathbf{1}_A+t\mathbf{1}_{A^c}} \) is a (finite) stopping time, and what precedes for the finite stopping time \( {t\wedge T\wedge S} \) gives \( {M_{t\wedge T\wedge S}\in\mathrm{L}^1} \) and \( {\mathbb{E}(M_{t\wedge T\wedge S})=\mathbb{E}(M_0)} \). Now, using the definition of \( {S} \), we have

\[ \mathbb{E}(M_0) =\mathbb{E}(M_{t\wedge T\wedge S}) =\mathbb{E}(\mathbf{1}_AM_{s\wedge T}) +\mathbb{E}(\mathbf{1}_{A^c}M_{t\wedge T}) =\mathbb{E}(\mathbf{1}_A(M_{s\wedge T}-M_{t\wedge T}))+\mathbb{E}(M_{t\wedge T}). \]

Since \( {\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_0)} \), we get the martingale property for \( {{(M_{t\wedge T})}_{t\geq0}} \), namely

\[ \mathbb{E}((M_{t\wedge T}-M_{s\wedge T})\mathbf{1}_A)=0. \]

Finally, suppose that \( {T<\infty} \) almost surely and \( {{(M_{t\wedge T})}_{t\geq0}} \) is uniformly integrable. The random variable \( {M_T} \) is well defined and \( {\lim_{t\rightarrow\infty}M_{t\wedge T}=M_T} \) almost surely because \( {M} \) is continuous. Furthermore, since \( {{(M_{t\wedge T})}_{t\geq0}} \) is uniformly integrable, it follows that \( {M_T\in\mathrm{L}^1} \) and \( {\lim_{t\rightarrow\infty}M_{t\wedge T}=M_T} \) in \( {\mathrm{L}^1} \). In particular \( {\mathbb{E}(M_0)\underset{\forall t}{=}\mathbb{E}(M_{t\wedge T})=\lim_{t\rightarrow\infty}\mathbb{E}(M_{t\wedge T})=\mathbb{E}(M_T)} \). Further reading in the same spirit.

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