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Davide Giraudo
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The following is contained in Durrett's book Probability theory and examples.

First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$.

Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.

First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$.

Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.

The following is contained in Durrett's book Probability theory and examples.

First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$.

Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.

Source Link
Davide Giraudo
  • 4k
  • 2
  • 30
  • 42

First remark: the limit always exists. Consider indeed $U_n$ the number of upcrossings of $[a,b]$ by $M_0,\dots,M_n$. Then $(b-a)\mathbb E(U_n)\leqslant \mathbb E(M_0-a)^+$.

Using towering property of conditional expectation, we obtain $\mathbb E(M_0\mid\mathcal F_n)=M_n$. If $M_0\in\mathbb L^1$, then the family $\{\mathbb E(M_0\mid \mathcal G),\mathcal G\subset\mathcal F \}$ is uniformly integrable and we conclude convergence in $\mathbb L^1$.