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===Backward Martingales and the law of large numbers=== | |||
{{definitioncard|Backward Filtration| | |||
Let <math>(\Omega,\F,\p)</math> be a probability space. A backward filtration is a family <math>(\F_n)_{n\in-\N}</math> of <math>\sigma</math>-Algebras indexed by the negative integers, which we will denote by <math>(\F_n)_{n\leq 0}</math>, such that for all <math>n\leq m\leq 0</math> we have | |||
<math display="block"> | |||
\F_n\subset\F_m. | |||
</math> | |||
}} | |||
{{alert-info | | |||
We will write | |||
<math display="block">\F_{-\infty}:=\bigcap_{n\leq 0}\F_n. | |||
</math> | |||
It is clear that <math>\F_{-\infty}</math> is also a <math>\sigma</math>-Algebra included in <math>\F</math>. A stochastic process <math>(X_n)_{n\leq 0}</math>, indexed by the negative integers, is called a backwards martingale (resp. backwards sub- or supermartingale) if for all <math>n\leq 0</math>, <math>X_n</math> is <math>\F_n</math>-measurable, <math>\E[\vert X_n\vert] < \infty</math> and for all <math>n\leq m</math> we have | |||
<math display="block"> | |||
\E[X_m\mid \F_n]=X_n\text{(resp. $\E[X_m\mid \F_n] < X_n$ or $\E[X_m\mid \F_n]\geq X_n$).} | |||
</math> | |||
}} | |||
{{proofcard|Theorem (Backward convergence theorem)|thm-1|Let <math>(\Omega,\F,(\F_n)_{n\leq 0},\p)</math> be a backward filtered probability space. Let <math>(X_n)_{n\leq 0}</math> be a backward supermartingale. Assume that | |||
<math display="block"> | |||
\begin{equation} | |||
\sup_{n\leq 0}\E[\vert X_n\vert] < \infty. | |||
\end{equation} | |||
</math> | |||
Then <math>(X_n)_{n\leq 0}</math> is u.i. and converges a.s. and in <math>L^1</math> to <math>X_{-\infty}</math> as <math>n\to-\infty</math>. Moreover, for all <math>n\leq 0</math>, we have | |||
<math display="block"> | |||
\E[X_n\mid \F_{-\infty}]\leq X_{-\infty}a.s. | |||
</math> | |||
|First we show a.s. convergence. Let therefore <math>k\geq 1</math> be a fixed integer. For <math>n\in\{1,...,k\}</math>, let <math>Y_n^k=X_{n-k}</math> and <math>\mathcal{G}_n^k=\F_{n-k}</math>. For <math>n > k</math>, we take <math>Y_n^k=X_0</math> and <math>\mathcal{G}_n^k=\F_0</math>. Then <math>(Y_n^k)_{n\geq 0}</math> is a supermartingale with respect to <math>(\mathcal{G}_n^k)_{n\geq 0}</math>. We now apply Doob's upcrossing inequality to the submartingale <math>(-Y_n^k)_{n\geq 0}</math> to obtain that for <math>a < b</math> | |||
<math display="block"> | |||
(b-a)\E[N_k([a,b],-Y^k)]\leq \E[(-Y_n^k-a)^+]=\E[(-X_0-a)^+]\leq \vert a\vert +\E[\vert X_0\vert]. | |||
</math> | |||
We note that when <math>k\uparrow \infty</math>, <math>N_k([a,b],-Y^k)</math> increases and | |||
<math display="block"> | |||
\begin{multline*} | |||
N([a,b],-X):=\sup\{k\in\N\mid \exists m_1 < n_1 < \dotsm < m_k < n_k\leq 0;\\ | |||
-X_{m_1}\leq a,-X_{n_1}\geq b,...,-X_{m_k}\leq a,-X_{n_k}\geq b\}. | |||
\end{multline*} | |||
</math> | |||
With monotone convergence we get | |||
<math display="block"> | |||
(b-a)\E[N([a,b],-X)]\leq \vert a\vert +\E[\vert X_0\vert] < \infty. | |||
</math> | |||
One can easily show that <math>(X_n)_{n\leq 0}</math> converges a.s. to <math>X_\infty</math> as <math>n\to-\infty</math> and Fatou implies then that <math>\E[\vert X_{-\infty}\vert] < \infty</math>. We want to show that <math>(X_n)_{n\leq 0}</math> is u.i. Thus, let <math>\varepsilon > 0</math>. The sequence <math>(\E[X_n])_{n\geq 0}</math> is increasing and bounded; we can take <math>k\leq 0</math> small enough to get for <math>n\leq k</math>, | |||
<math display="block"> | |||
\E[X_n]\leq \E[X_k]+\frac{\varepsilon}{2}. | |||
</math> | |||
Moreover, the finite family <math>\{X_k,X_{k+1},...,X_{-1},X_0\}</math> is u.i. and one can then choose <math>\alpha > 0</math> large enough such that for all <math>k\leq n\leq 0</math> | |||
<math display="block"> | |||
\E[\vert X_n\vert\one_{\{\vert X_n\vert > \alpha}] < \varepsilon. | |||
</math> | |||
We can also choose <math>\delta > 0</math> sufficiently small such that for all <math>A\in\F</math>, <math>\p[A] < \delta</math> implies that <math>\E[\vert X_n\mid \one_A] < \frac{\varepsilon}{2}</math>. Now if <math>n < k</math>, we get | |||
<math display="block"> | |||
\begin{align*} | |||
\E[\vert X_n\vert \one_{\{ \vert X_n\vert > \alpha}]&=\E[-X_n\one_{\{ X_n < -\alpha\}}]+\E[X_n\one_{\{ X_n > \alpha\}}]=-\E[X_n\one_{\{ X_n < -\alpha\}}]+\E[X_n]-\E[X_n\one_{\{X_n\leq \alpha\}}]\\ | |||
&\leq -\E[\E[X_k\mid \F_n]\one_{\{ X_n < -\alpha\}}]+\E[X_k]+\frac{\varepsilon}{2}-\E[\E[X_k\mid \F_n]\one_{X_\leq \alpha}]\\ | |||
&=-\E[X_k\one{\{ X_n < -\alpha\}}]+\E[X_k]+\frac{\varepsilon}{2}-\E[X_k\one_{\{ X_n\leq \alpha\}}]\\ | |||
&=-\E[X_n\one_{\{ X_n < -\alpha\}}]+\E[X_n\one_{X_n > \alpha\}}]+\frac{\varepsilon}{2}\leq \E[\vert X_k\vert \one_{\{ \vert X_n\vert > \alpha\}}]+\frac{\varepsilon}{2}. | |||
\end{align*} | |||
</math> | |||
Next, we observe that | |||
<math display="block"> | |||
\p[\vert X_n\vert > \alpha]\leq \frac{1}{\alpha}\E[X_n]\leq \frac{C}{\alpha}, | |||
</math> | |||
where <math>C=\sup_{n\leq 0}\E[\vert X_n\vert] < \infty</math>. Choose <math>\alpha</math> such that <math>\frac{C}{\alpha} < \delta</math>. Consequently, we get | |||
<math display="block"> | |||
\E[\vert X_k\vert\one_{\{\vert X_n\vert > \alpha\}}] < \frac{\varepsilon}{2}. | |||
</math> | |||
Hence, for all <math>n < k</math>, <math>\E[\vert X_n\vert \one_{\{ \vert X_n\vert > \alpha\}}] < \varepsilon</math>. This inequality is also true for <math>k\leq n\leq 0</math> and thus we have that <math>(X_n)_{n\leq 0}</math> is u.i. To conclude, we note that u.i. and a.s. convergence implies <math>L^1</math> convergence. Then, for <math>m\leq n</math> and <math>A\in\F_{-\infty}\subset \F_m</math>, we have | |||
<math display="block"> | |||
\E[X_n\one_A]\leq \E[X_m\one_A]\xrightarrow{m\to-\infty}\E[X_{-\infty}\one_A]. | |||
</math> | |||
Therefore, <math>\E[\E[X_n\mid \F_{-\infty}]\one_A]\leq \E[X_{-\infty}\one_A]</math> and hence | |||
<math display="block"> | |||
\E[X_n\mid \F_{-\infty}]\leq X_{-\infty}. | |||
</math>}} | |||
{{alert-info | | |||
Equation <math>(1)</math> is always satisfied for backward martingales. Indeed, for all <math>n\leq 0</math> we get | |||
<math display="block"> | |||
\E[X_0\mid \F_n]=X_n, | |||
</math> | |||
which implies that <math>\E[\vert X_n\vert]\leq \E[\vert X_0\vert]</math> and thus | |||
<math display="block"> | |||
\sup_{n\leq 0}\E[\vert X_n\vert] < \infty. | |||
</math> | |||
Backward martingales are therefore always u.i. | |||
}} | |||
{{proofcard|Corollary|cor-1|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>Z</math> be a r.v. in <math>L^1(\Omega,\F,\p)</math> and let <math>(\mathcal{G}_n)_{n\geq 0}</math> be a decreasing family of <math>\sigma</math>-Algebras. Then | |||
<math display="block"> | |||
\E[Z\mid \mathcal{G}_n]\xrightarrow{n\to\infty\atop\text{a.s. and $L^1$}}\E[Z\mid \mathcal{G}_\infty], | |||
</math> | |||
where | |||
<math display="block"> | |||
\mathcal{G}_\infty:=\bigcap_{n\geq 0}\mathcal{G}_n. | |||
</math> | |||
|For <math>n\geq 0</math> define <math>X_{-n}:=\E[Z\mid \F_n]</math>, where <math>\F_{-n}=\mathcal{G}_n</math>. Then <math>(X_n)_{n\leq 0}</math> is a backward martingale with respect to <math>(\F_n)_{n\leq 0}</math>. Hence theorem 14.1. implies that <math>(X_n)_{n\leq 0}</math> converges a.s. and in <math>L^1</math> for <math>n\to\infty</math>. Moreover{{efn|This follows from the last part of theorem 14.1.}}, | |||
<math display="block"> | |||
X_\infty=\E[X_0\mid \F_{-\infty}]=\E[\E[Z\mid \F_0]\mid \F_{-\infty}]=\E[Z\mid \F_{-\infty}]=\E[Z\mid \mathcal{G}_\infty]. | |||
</math>}} | |||
{{proofcard|Theorem (Kolmogorov's 0-1 law)|thm-2|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be a sequence of independent r.v.'s with values in arbitrary measure spaces. For <math>n\geq 1</math>, define the <math>\sigma</math>-Algebra | |||
<math display="block"> | |||
\B_n:=\sigma(\{X_k\mid k\geq n\}). | |||
</math> | |||
The tail <math>\sigma</math>-Algebra <math>\B_\infty</math> is defined as | |||
<math display="block"> | |||
\B_\infty:=\bigcap_{n=1}^\infty\B_n. | |||
</math> | |||
Then <math>\B_\infty</math> is trivial in the sense that for all <math>B\in \B_\infty</math> we have <math>\p[B]\in\{0,1\}</math>. | |||
|This proof can be found in the stochastics I notes.}} | |||
{{proofcard|Lemma|lem5|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>Z\in L^1(\Omega,\F,\p)</math> and <math>\mathcal{H}_1</math> and <math>\mathcal{H}_2</math> two <math>\sigma</math>-Algebras included in <math>\F</math>. Assume that <math>\mathcal{H}_2</math> is independent of <math>\sigma(Z)\lor \mathcal{H}_1</math>. Then | |||
<math display="block"> | |||
\E[Z\mid \mathcal{H}_1\lor \mathcal{H}_2]=\E[Z\mid \mathcal{H}_1]. | |||
</math> | |||
|Let <math>A\in \mathcal{H}_1\lor \mathcal{H}_2</math> such that <math>A=B\cap C</math>, where <math>B\in \mathcal{H}_1</math> and <math>C\in\mathcal{H}_2</math>. Then | |||
<math display="block"> | |||
\begin{multline*} | |||
\E[Z\one_A]=\E[Z\one_B\one_C]=\E[Z\one_B]\E[\one_C]=\E[\one_A]\E[\E[Z\mid \mathcal{H}_1]\one_B]\\=\E[\E[Z\mid \mathcal{H}_1]\one_B\one_C]=\E[\E[Z\mid \mathcal{H}_1]\mid \one_A]. | |||
\end{multline*} | |||
</math> | |||
Now we note that | |||
<math display="block"> | |||
\sigma(W=\{ B\cap C\mid B\in\mathcal{H}_1,C\in \mathcal{H}_2\})=\mathcal{H}_1\lor \mathcal{H}_2 | |||
</math> | |||
and <math>W</math> is stable under finite intersections. Thus the monotone class theorem implies that for all <math>A\in \mathcal{H}_1\lor \mathcal{H}_2</math> we have | |||
<math display="block"> | |||
\E[Z\one_A]=\E[\E[Z\mid\mathcal{H}_1]\one_A] | |||
</math>}} | |||
{{proofcard|Theorem (Strong law of large numbers)|thm-3|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>(\xi_n)_{n\geq 1}</math> be a sequence of iid r.v.'s such that for all <math>n\geq 1</math> we have <math>\E[\vert \xi_n\vert] < \infty</math>. Moreover, let <math>S_0=0</math> and <math>S_n=\sum_{j=1}^n\xi_j</math>. Then | |||
<math display="block"> | |||
\frac{S_n}{n}\xrightarrow{n\to\infty\atop\text{a.s. and $L^1$}}\E[\xi_1]. | |||
</math> | |||
|At first, we want to show that <math>\E[\xi_1\mid S_n]=\frac{S_n}{n}</math>. Indeed, we know that there is a measurable map <math>g</math> such that | |||
<math display="block"> | |||
\E[\xi_1\mid S_n]=g(S_n). | |||
</math> | |||
Moreover, we know that for <math>k\in\{1,...,n\}</math> we have <math>(\xi_1,S_n)</math> and <math>(\xi_k,S_n)</math> have the same law. Now for all bounded and Borel measurable maps <math>h</math> we have | |||
<math display="block"> | |||
\E[\xi_k h(S_n)]=\E[\xi_1 h(S_n)]=\E[g(S_n)h(S_n)]. | |||
</math> | |||
Thus <math>\E[\xi_k\mid S_n]=g(S_n)</math>. We have | |||
<math display="block"> | |||
\sum_{j=1}^n\E[\xi_j\mid S_n]=\E\left[\sum_{j=1}^n \xi_j\mid S_n\right]=S_n, | |||
</math> | |||
but on the other hand we have that | |||
<math display="block"> | |||
\sum_{j=1}^n\E[\xi_j\mid S_n]=ng(s_n). | |||
</math> | |||
Hence <math>g(S_n)=\frac{S_n}{n}</math>. Now take <math>\mathcal{H}_1=\sigma(S_n)</math> and <math>\mathcal{H}_2=\sigma(S_n,\xi_{n+1},\xi_{n+2},...)</math>. Thus, by [[#lem5 |lemma]], we get | |||
<math display="block"> | |||
\E[\xi_1\mid S_n,\xi_{n+1},\xi_{n+2},...]=\E[\xi_1\mid S_n]. | |||
</math> | |||
Now define <math>\mathcal{G}:=\sigma(S_n,\xi_{n+1},\xi_{n+2},...)</math>. Then we have <math>\mathcal{G}_{n+1}\subset\mathcal{G}_n</math> because <math>S_{n+1}=S_n+\xi_{n+1}</math>. Hence, it follows that <math>\E[\xi_1\mid \mathcal{G}_n]=\E[\xi_1\mid S_n]=\frac{S_n}{n}</math> converges a.s. and in <math>L^1</math> to some r.v., but Kolmogorov's 0-1 law implies that this limit is a.s. constant. In particular, <math>\E\left[\frac{S_n}{n}\right]=\E[\xi_1]</math> converges in <math>L^1</math> to this limit, which is thus <math>\E[\xi_1]</math>.}} | |||
===Martingales bounded in <math>L^2</math> and random series=== | |||
Let <math>(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> be a filtered probability space. Let <math>(M_n)_{n\geq 0}</math> be a martingale in <math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, i.e. <math>\E[M_n^2] < \infty</math> for all <math>n\geq 0</math>. We say that <math>(M_n)_{n\geq 0}</math> is bounded in <math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> if <math>\sup_{n\geq 0}\E[M_n^2] < \infty</math>. For <math>n\leq \nu</math>, we have that <math>\E[M_\nu\mid \F_n]=M_n</math> implies that <math>(M_\nu-M_n)</math> is orthogonal to <math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. Hence, for all <math>s\leq t\leq n\leq \nu</math>, <math>(M_\nu-M_n)</math> is orthogonal to <math>(M_t-M_s)</math>. | |||
<math display="block"> | |||
\left\langle M_\nu-M_n,M_t-M_s\right\rangle=0\Longleftrightarrow \E[(M_\nu-M_n)(M_t-M_s)]=0. | |||
</math> | |||
Now write <math>M_n=M_0+\sum_{k=1}^n(M_k-M_{k-1})</math>. <math>M_n</math> is then a sum of orthogonal terms and therefore | |||
<math display="block"> | |||
\E[M_n^2]=\E[M_0^2]+\sum_{k=1}^n\E[(M_k-M_{k-1})^2]. | |||
</math> | |||
{{proofcard|Theorem|thm-4|Let <math>(M_n)_{n\geq 0}</math> be a martingale in <math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>. Then <math>(M_n)_{n\geq 0}</math> is bounded in <math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> if and only if <math>\sum_{k\geq 1}\E[(M_k-M_{k-1})^2] < \infty</math> and in this case | |||
<math display="block"> | |||
M_n\xrightarrow{n\to\infty\atop \text{a.s. and $L^1$}}M_\infty. | |||
</math>|}} | |||
{{proofcard|Theorem|thm-5|Suppose that <math>(X_n)_{n\geq 1}</math> is a sequence of independent r.v.'s such that for all <math>k\geq 1</math>, <math>\E[X_k]=0</math> and <math>\sigma_k^2=Var(X_k) < \infty</math>. Then | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\sum_{k\geq1}\sigma^2_k < \infty</math> implies that <math>\sum_{k\geq 1}X_k</math> converges a.s. | |||
</li> | |||
<li>If there is a <math>C > 0</math> such that for all <math>\omega\in\Omega</math> and <math>k\geq1</math>, <math>\vert X_k(\omega)\vert\leq C</math>, then <math>\sum_{k\geq 1}X_k</math> converges a.s. implies that <math>\sum_{k\geq 1}\sigma_k^2 < \infty</math>. | |||
</li> | |||
</ul> | |||
|Consider <math>\F_n=\sigma(X_1,...,x_n)</math> with <math>F_0=\{\varnothing,\Omega\}</math>, <math>S_n=\sum_{j=1}^nX_j</math> with <math>S_0=0</math> and <math>A_n=\sum_{k=1}^n\sigma_k^2</math> with <math>A_0=0</math>. Moreover, set <math>M_n=S_n^2-A_n</math>. Then <math>(S_n)_{n\geq 0}</math> is a martingale and | |||
<math display="block"> | |||
\E[(S_n-S_{n-1})^2]=\E[X_n^2]=\sigma_n^2. | |||
</math> | |||
Thus <math>\sum_{n\geq 1}\sigma_n^2 < \infty</math> inplies <math>\sum_{n\geq 1}\E[(S_n-S_{n-1})^2] < \infty</math> and hence <math>(S_n)_{n\geq 0}</math> | |||
is bounded in | |||
<math>L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)</math>, which means that <math>S_n</math> converges a.s. Next we show that <math>(M_n)_{n\geq 0}</math> is a martingale. We have | |||
<math display="block"> | |||
\E[(S_n-S_{n-1})^2\mid \F_{n-1}]=\E[X_n^2\mid \F_{n-1}]=\E[X_n^2]=\sigma_n^2. | |||
</math> | |||
Hence we get | |||
<math display="block"> | |||
\begin{multline*} | |||
\sigma_n^2=\E[(S_n-S_{n-1})^2\mid \F_{n-1}]=\E[S_n^2-2S_{n-1}S_n+S_{n-1}^2\mid F_{n-1}]=\\ | |||
=\E[S_n^2\mid\F_{n-1}]-2S_{n-1}^2+S_{n-1}^2=\E[S_n^2\mid \F_{n-1}]-S_{n-1}^2, | |||
\end{multline*} | |||
</math> | |||
which implies that <math>(M_n)_{n\geq 0}</math> is a martingale. Let <math>T:=\inf\{n\in\N\mid \vert S_n\vert > \alpha\}</math> for some constant <math>\alpha</math>. Then <math>T</math> is a stopping time. <math>(M_{n\land T})_{n\geq 1}</math> is a martingale and hence | |||
<math display="block"> | |||
\E[M_{n\land T}]=\E[S_{n\land T}^2]-\E[A_{n\land T}]=0. | |||
</math> | |||
Therefore <math>\E[S_{n\land T}^2]=\E[A_{n\land T}]</math> and if <math>T</math> is finite, <math>\vert S_T-S_{T-1}\vert < \vert X_T\vert\leq C</math> for some constant <math>C</math>, thus <math>\vert S_{n\land T}\vert \leq C+\alpha</math> and hence <math>\E[A_{n\land T}]\leq (C+\alpha)^2</math> for all <math>n\geq 0</math>. Now since <math>A_n</math> is increasing we get that <math>\E[A_{n\land T}]\leq (C+\alpha)^2 < \infty.</math> Since <math>\sum_{n\geq 0}X_n</math> converges a.s., <math>\sum_{k=1}^n X_k</math> is bounded and there exists <math>\alpha > 0</math> such that <math>\p[T=\alpha] > 0</math>. Choosing{{efn|Note that <math>\E\left[\sum_{k\geq 1}\sigma_k^2\one_{\{ T=\infty\}}\right]=\sum_{k\geq 1}\sigma_k^2\p[T=\infty]</math>}} <math>\alpha</math> right yields <math>\sum_{k\geq 1}\sigma_k^2 < \infty</math>.}} | |||
'''Example''' | |||
Let <math>(a_n)_{n\geq 1}</math> be a sequence of real numbers and let <math>(\xi_n)_{n\geq 1}</math> be iid r.v.'s with <math>\p[\xi=\pm 1]=\frac{1}{2}</math>. Then <math>\sum_{n\geq 1}a_n\xi_n</math> converges a.s. if and only if <math>\sum_{n\geq 1}a_n^2 < \infty</math>. Indeed, we get <math>\vert a_n\xi_n\vert=\vert a_n\vert\xrightarrow{n\to\infty}0</math> and therefore there exists a <math>C > 0</math> such that for all <math>n\geq 1</math>, <math>\vert a_n\vert \leq C</math>. Now for a r.v. <math>X</math> recall that we write <math>\Phi_X(t)=\E\left[e^{itX}\right]</math>. We also know | |||
<math display="block"> | |||
e^{itx}=\sum_{n\geq 0}\frac{i^nt^nx^n}{n!},e^{itX}=\sum_{n\geq 0}\frac{i^nt^nX^n}{n!}. | |||
</math> | |||
Moreover, define <math>R_n(x)=e^{ix}-\sum_{k=0}^n\frac{i^kx^k}{k!}</math>. Therefore we get | |||
<math display="block"> | |||
\vert R_n(x)\vert\leq \min \left(2\frac{\vert x\vert^n}{n!},\frac{\vert x\vert^{n+1}}{(n+1)!}\right). | |||
</math> | |||
Indeed, <math>\vert R_0(x)\vert=\vert e^{ix}-1\vert=\left\vert\int_0^x ie^{iy}dy\right\vert\leq \min(2,\vert x\vert).</math> Moreover, we have <math>\vert R_n(x)\vert=\left\vert \int_0^x iR_{n-1}(y)dy\right\vert</math>. Hence the claim follows by a simple induction on <math>n</math>. If <math>X</math> is such that <math>\E[X]=0</math>, <math>\E[X^2]=\sigma^2 < \infty</math> and <math>e^{itX}-\left( 1+itX-\frac{t^2X^2}{2}\right)=R_2(tX)</math> we get | |||
<math display="block"> | |||
\E\left[e^{itX}\right]=1-\frac{\sigma^2t^2}{2}+\E[R_2(tX)] | |||
</math> | |||
and <math>\E[R_2(tX)]\leq t^2\E[\vert X\vert^2\land tX^3]</math>. With dominated convergence it follows that <math>\Phi(t)=1-\frac{t^2\sigma^2}{2}+o(t^2)</math> as <math>t\to 0</math>. | |||
{{proofcard|Lemma|lem-1|Let <math>(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> be a filtered probability space. Let <math>(X_n)_{n\geq 0}</math> be a sequence of independent r.v.'s bounded by <math>k > 0</math>. Then if <math>\sum_{n\geq 0}X_n</math> converges a.s., <math>\sum_{n\geq 0}\E[X_n]</math> and <math>\sum_{n\geq 0}Var(X_n)</math> both converge. | |||
|If <math>Z</math> is a r.v. such that <math>\vert Z\vert < k</math>, <math>\E[Z]=0</math> and <math>\sigma^2=Var(Z) < \infty</math>, then for <math>\vert t\vert\leq \frac{1}{k}</math> we get | |||
<math display="block"> | |||
\vert \Phi_Z(t)\vert\leq 1-\frac{t^2\sigma^2}{2}+\frac{t^3k\E[Z^2]}{6}\leq 1-\frac{t^2\sigma^2}{2}+\frac{t^2\sigma^2}{6}=1-\frac{t^2\sigma^2}{3}\leq \exp\left(-\frac{t^2\sigma^2}{3}\right). | |||
</math> | |||
Let <math>Z_n=X_n-\E[X_n]</math>. Then <math>\vert \Phi_{Z_n}(t)\vert=\vert \Phi_{X_n}(t)\vert</math> and <math>\vert Z_n\vert\leq 2k</math>. If <math>\sum_{n\geq 0}X_n=\infty</math>, we get | |||
<math display="block"> | |||
\prod_{n\geq0}\vert\Phi_{X_n}(t)\vert=\prod_{n\geq 0}\vert\Phi_{Z_n}(t)\vert\leq \exp\left(-\frac{1}{3}t^2\sum_{n\geq 0}Var(X_n)\right)=0. | |||
</math> | |||
This is a contradiction, since <math>\vert \Phi_{\sum_{n\geq 0}X_n}(t)\vert\xrightarrow{n\to\infty}\vert\Phi(t)\vert</math> with <math>\Phi</math> continuous and <math>\Phi(0)=1</math>. Hence <math>\sum_{n\geq 0}Var(X_n)=\sum_{n\geq 0}Var(Z_n) < \infty</math>. Since <math>\E[Z_n]=0</math> and <math>\sum_{n\geq 0}Var(Z_n) < \infty</math>, we have <math>\sum_{n\geq 0}Z_n</math> converges a.s., but <math>\sum_{n\geq 0}-Z_n=\sum_{n\geq 0}X_n-\sum_{n\geq 0}\E[X_n]</math> and thus since <math>\sum_{n\geq 0}X_n</math> converges a.s. it follows that <math>\sum_{n\geq 0}\E[X_n]</math> converges.}} | |||
{{proofcard|Theorem (Kolmogorov's three series theorem)|thm-6|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 0}</math> be a sequence of independent r.v.'s. Then <math>\sum_{n\geq 0}X_n</math> converges a.s. if and only if for some <math>k > 0</math> (then for every <math>k > 0</math>) the following properties hold. | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\sum_{n\geq 0}\p[\vert X_n\vert > k] < \infty</math> | |||
</li> | |||
<li><math>\sum_{n\geq 0}\E\left[X_n^{(k)}\right]</math> converges, where <math>X_n^{(k)}=X_n\one_{\{\vert X_n\vert\leq k\}}</math> | |||
</li> | |||
<li><math>\sum_{n\geq 0}Var\left(X_n^{(k)}\right) < \infty</math> | |||
</li> | |||
</ul> | |||
|Suppose that for some <math>k > 0</math>, <math>(i)</math>, <math>(ii)</math> and <math>(iii)</math> hold. Then | |||
<math display="block"> | |||
\sum_{n\geq 0}\p\left[X_n\not=X_n^{(k)}\right]=\sum_{n\geq 0}\p[\vert X_n\vert > k] < \infty. | |||
</math> | |||
It follows from the Borel-Cantelli lemma that <math>\p\left[X_n=X_n^{(k)}\text{for all but finitely many $n$}\right]=1</math>. Hence we only need to show that <math>\sum_{n\geq 0}X_n^{k}</math> converges a.s. Because of <math>(ii)</math> it is enough to show that <math>\sum_{n\geq 0}Y_n^{(k)}</math> converges, where <math>Y_n^{(k)}=X_n^{(k)}-\E\left[X_n^{(k)}\right]</math>. The convergence of <math>\sum_{n\geq 0}Y_n^{(k)}</math> follows then from <math>(iii)</math>. Conversely assume that <math>\sum_{n\geq 0}X_n</math> converges a.s. and that <math>k\in(0,\infty)</math>. Since <math>X\xrightarrow{n\to\infty}0</math> a.s., we have that <math>\vert X_n\vert > k</math> for only finitely many <math>n</math>. Therefore, the Borel-Cantelli lemma implies <math>(i)</math>. Since <math>X_n=X_n^{(k)}</math> for all but finitely many <math>n</math>, <math>\sum_{n\geq 0}X_n^{(k)}</math> converges and it follows from lemma 14.8. that <math>(ii)</math> and <math>(iii)</math> have to hold.}} | |||
{{proofcard|Lemma (Ces\‘aro)|lem-2|Suppose <math>(b_n)_{n\geq 1}</math> is a sequence of strictly positive real numbers with <math>b_n\uparrow\infty</math> as <math>n\to\infty</math>. Let <math>(v_n)_{n\geq 1}</math> be a sequence of real numbers such that <math>v_n\xrightarrow{n\to\infty}v</math>. Then | |||
<math display="block"> | |||
\frac{1}{b_n}\sum_{k=1}^n(b_k-b_{k-1})v_k\xrightarrow{n\to\infty}v(b_0=0) | |||
</math> | |||
|Note that | |||
<math display="block"> | |||
\begin{multline*} | |||
\left\vert \frac{1}{b_n}\sum_{k=1}^n(b_k-b_{k-1})v_k-v\right\vert=\left\vert \frac{1}{b_n}\sum_{k=1}^n(b_k-b_{k-1})(v_k-v)\right\vert\leq \frac{1}{b_n}\sum_{k=1}^N(b_k-b_{k-1})\vert v_k-v\vert\\+\frac{1}{b_n}\sum_{k=N+1}^n(b_k-b_{k-1})\vert v_k-v\vert. | |||
\end{multline*} | |||
</math> | |||
Now we only have to choose <math>N</math> such that <math>n\geq N</math> and <math>\vert v_k-v\vert < \varepsilon</math> for any <math>\varepsilon > 0</math>.}} | |||
{{proofcard|Lemma (Kronecker)|lem-3|Let <math>(b_n)_{n\geq 1}</math> be a sequence of real numbers, strictly positive with <math>b_n\uparrow \infty</math> as <math>n\to\infty</math>. Let <math>(x_n)_{n\geq 1}</math> be a sequence of real numbers. Then if <math>\sum_{n\geq 1}\frac{x_n}{b_n}</math> converges, we get that | |||
<math display="block"> | |||
\frac{x_1+\dotsm +x_n}{b_n}\xrightarrow{n\to\infty}0. | |||
</math> | |||
|Let <math>v_n=\sum_{k=1}^n\frac{x_n}{b_n}</math> and <math>v=\lim_{n\to\infty}v_n</math>. Then <math>v_n-v_{n-1}=\frac{x_n}{b_n}</math>. Moreover, we note that | |||
<math display="block"> | |||
\sum_{k=1}^nx_k=\sum_{k=1}^nb_k(v_k-v_{k-1})=b_nv_n-\sum_{k=1}^n(b_k-b_{k-1})v_k, | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
\frac{x_1+\dotsm+x_n}{b_n}=v_n-\frac{1}{b_n}\sum_{k=1}^n(b_k-b_{k-1})v_k\xrightarrow{n\to\infty}v-v=0. | |||
</math>}} | |||
{{proofcard|Proposition|prop-1|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>(w_n)_{n\geq 1}</math> be a sequence of r.v.’s such that <math>\E[w_n]=0</math> for all <math>n\geq 1</math> and <math>\sum_{n\geq 1}\frac{Var(w_n)}{n^2} < \infty</math>. Then | |||
<math display="block"> | |||
\frac{1}{n}\sum_{n\geq 1}w_n\xrightarrow{n\to\infty \atop a.s.}0. | |||
</math> | |||
|Exercise.{{efn|From Kronecker's lemma it is enough to prove that <math>\sum_{k\geq 1}\frac{w_k}{k}</math> converges a.s.}}}} | |||
{{proofcard|Theorem|thm-7|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be independent and non-negative r.v.'s such that <math>\E[X_n]=1</math> for all <math>n\geq 1</math>. Define <math>M_0=1</math> and for <math>n\in\N</math>, let | |||
<math display="block"> | |||
M_n=\prod_{j=1}^nX_j. | |||
</math> | |||
Then <math>(M_n)_{n\geq 1}</math> is a non-negative martingale, so that <math>M_\infty:=\lim_{n\to\infty}M_n</math> exists a.s. Then the following are equivalent. | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\E[M_\infty]=1</math>. | |||
</li> | |||
<li><math>M_n\xrightarrow{n\to\infty\atop L^1}M_\infty</math>. | |||
</li> | |||
<li><math>(M_n)_{n\geq 1}</math> is u.i. | |||
</li> | |||
<li><math>\prod_{n}a_n > 0</math>, where <math>0 < a_n=\E[X_n^{1/2}]\leq 1</math>. | |||
</li> | |||
<li><math>\sum_{n}(1-a_n) < \infty</math>. | |||
</li> | |||
</ul> | |||
Moreover, if one of the following (then every one) statements hold, then | |||
<math display="block"> | |||
\p[M_\infty=0]=1. | |||
</math> | |||
|Exercise.}} | |||
===A martingale central limit theorem=== | |||
{{proofcard|Theorem|thm-8|Let <math>(\Omega,\F,(\F_n)_{n\geq 0},\p)</math> be a filtered probability space. Let <math>(X_n)_{n\geq 0}</math> be a sequence of real valued r.v.'s such that for all <math>n\geq 1</math> | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\E[X_n\mid \F_{n-1}]=0</math>. | |||
</li> | |||
<li><math>\E[X_n^2\mid \F_{n-1}]=1</math>. | |||
</li> | |||
<li><math>\E[\vert X_n\vert^3\mid \F_{n-1}]\leq k < \infty</math>. | |||
</li> | |||
</ul> | |||
Let <math>S_n=\sum_{j=1}^nX_j</math>. Then | |||
<math display="block"> | |||
\frac{S_n}{\sqrt{n}}\xrightarrow{n\to\infty\atop law}\mathcal{N}(0,1). | |||
</math> | |||
|Define <math>\Phi_{n,j}(u)=\E\left[e^{iu\frac{X_j}{\sqrt{n}}}\mid \F_{j-1}\right]</math>. A Taylor expansion yields | |||
<math display="block"> | |||
\exp\left(iu\frac{X_j}{\sqrt{n}}\right)=1+iu\frac{X_j}{\sqrt{n}}-\frac{u^2}{2n}X_j^2-\frac{iu^3}{6n^{3/2}}\bar X_j^3, | |||
</math> | |||
where <math>\bar X_j</math> is a random number between 0 and <math>X_j</math>. Therefore we get | |||
<math display="block"> | |||
\Phi_{n,j}(u)=1+iu\frac{1}{\sqrt{n}}\E[X_j\mid \F_{j-1}]-\frac{u^2}{2n}\E[X_j^2\mid \F_{j-1}]-\frac{iu^3}{6n^{3/2}}\E[\bar X_j^3\mid \F_{j-1}] | |||
</math> | |||
and thus | |||
<math display="block"> | |||
\Phi_{n,j}(u)-1+\frac{u^2}{2n}=-\frac{iu^3}{6n^{3/2}}\E[\bar X_j^2\mid \F_{j-1}]. | |||
</math> | |||
Hence we get | |||
<math display="block"> | |||
\E\left[e^{iu\frac{S_p}{\sqrt{n}}}\right]=\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}e^{iu\frac{X_p}{\sqrt{n}}}\right]=\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\E\left[e^{iu\frac{X_p}{\sqrt{n}}}\mid \F_{p-1}\right]\right]=\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\Phi_{n,p}(u)\right]. | |||
</math> | |||
Consequently, we get | |||
<math display="block"> | |||
\E\left[e^{iu\frac{S_p}{\sqrt{n}}}\right]=\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\left(1+\frac{u^2}{2n}-\frac{iu^3}{6n^{3/2}}\bar X_p^3\right)\right]. | |||
</math> | |||
Thus we get that | |||
<math display="block"> | |||
\E\left[e^{iu\frac{S_p}{\sqrt{n}}}-\left(1-\frac{u^2}{2n}\right)e^{iu\frac{S_{p-1}}{\sqrt{n}}}\right]=\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\frac{iu^3}{6n^{3/2}}\bar X_p^3\right], | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
\left\vert \E\left[e^{iu \frac{S_p}{\sqrt{n}}}-\left(1-\frac{u^2}{2n}\right)e^{iu\frac{S_{p-1}}{\sqrt{n}}}\right]\right\vert\leq \frac{K\vert u\vert ^3}{6n^{3/2}}.(\star) | |||
</math> | |||
Let us fix <math>n\in\N</math>. For <math>n</math> large enough, we have <math>0\leq 1-\frac{u^2}{2n}\leq 1</math>. Multiplying both sides of <math>(\star)</math> by <math>\left(1-\frac{u^2}{2n}\right)^{n-p}</math>, we get | |||
<math display="block"> | |||
\left\vert \left(1-\frac{u^2}{2n}\right)^{n-p}\E\left[e^{iu\frac{S_p}{\sqrt{n}}}\right]-\left(1-\frac{u^2}{2n}\right)^{n-p+1}\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\right]\right\vert\leq \frac{K\vert u\vert^3}{6n^{3/2}}.(\star\star) | |||
</math> | |||
By taking <math>K</math> sufficiently large, we can assume that <math>(\star\star)</math> holds for all <math>n</math>. Now we note that | |||
<math display="block"> | |||
\E\left[e^{iu\frac{S_n}{\sqrt{n}}}\right]-\left(1-\frac{u^2}{2n}\right)^n=\sum_{p=1}^n\left\{\left(1-\frac{u^2}{2n}\right)^{n-p}\E\left[e^{iu\frac{S_p}{\sqrt{n}}}\right]-\left(1-\frac{u^2}{2n}\right)^{n-p+1}\E\left[e^{iu\frac{S_{p-1}}{\sqrt{n}}}\right]\right\}. | |||
</math> | |||
Therefore we get | |||
<math display="block"> | |||
\left\vert\E\left[e^{iu\frac{S_n}{\sqrt{n}}}\right]-\left(1-\frac{u^2}{2n}\right)^n\right\vert\leq n\frac{K\vert u\vert^3}{n^{3/2}}=\frac{K\vert u\vert^3}{6\sqrt{n}}, | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
\lim_{n\to\infty}\E\left[e^{iu\frac{S_n}{\sqrt{n}}}\right]=e^{-\frac{u^2}{2}}. | |||
</math>}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | |||
==Notes== | |||
{{notelist}} |
Latest revision as of 23:55, 8 May 2024
Backward Martingales and the law of large numbers
Let [math](\Omega,\F,\p)[/math] be a probability space. A backward filtration is a family [math](\F_n)_{n\in-\N}[/math] of [math]\sigma[/math]-Algebras indexed by the negative integers, which we will denote by [math](\F_n)_{n\leq 0}[/math], such that for all [math]n\leq m\leq 0[/math] we have
We will write
Let [math](\Omega,\F,(\F_n)_{n\leq 0},\p)[/math] be a backward filtered probability space. Let [math](X_n)_{n\leq 0}[/math] be a backward supermartingale. Assume that
Then [math](X_n)_{n\leq 0}[/math] is u.i. and converges a.s. and in [math]L^1[/math] to [math]X_{-\infty}[/math] as [math]n\to-\infty[/math]. Moreover, for all [math]n\leq 0[/math], we have
First we show a.s. convergence. Let therefore [math]k\geq 1[/math] be a fixed integer. For [math]n\in\{1,...,k\}[/math], let [math]Y_n^k=X_{n-k}[/math] and [math]\mathcal{G}_n^k=\F_{n-k}[/math]. For [math]n \gt k[/math], we take [math]Y_n^k=X_0[/math] and [math]\mathcal{G}_n^k=\F_0[/math]. Then [math](Y_n^k)_{n\geq 0}[/math] is a supermartingale with respect to [math](\mathcal{G}_n^k)_{n\geq 0}[/math]. We now apply Doob's upcrossing inequality to the submartingale [math](-Y_n^k)_{n\geq 0}[/math] to obtain that for [math]a \lt b[/math]
With monotone convergence we get
Next, we observe that
Equation [math](1)[/math] is always satisfied for backward martingales. Indeed, for all [math]n\leq 0[/math] we get
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math]Z[/math] be a r.v. in [math]L^1(\Omega,\F,\p)[/math] and let [math](\mathcal{G}_n)_{n\geq 0}[/math] be a decreasing family of [math]\sigma[/math]-Algebras. Then
For [math]n\geq 0[/math] define [math]X_{-n}:=\E[Z\mid \F_n][/math], where [math]\F_{-n}=\mathcal{G}_n[/math]. Then [math](X_n)_{n\leq 0}[/math] is a backward martingale with respect to [math](\F_n)_{n\leq 0}[/math]. Hence theorem 14.1. implies that [math](X_n)_{n\leq 0}[/math] converges a.s. and in [math]L^1[/math] for [math]n\to\infty[/math]. Moreover[a],
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of independent r.v.'s with values in arbitrary measure spaces. For [math]n\geq 1[/math], define the [math]\sigma[/math]-Algebra
The tail [math]\sigma[/math]-Algebra [math]\B_\infty[/math] is defined as
Then [math]\B_\infty[/math] is trivial in the sense that for all [math]B\in \B_\infty[/math] we have [math]\p[B]\in\{0,1\}[/math].
This proof can be found in the stochastics I notes.
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math]Z\in L^1(\Omega,\F,\p)[/math] and [math]\mathcal{H}_1[/math] and [math]\mathcal{H}_2[/math] two [math]\sigma[/math]-Algebras included in [math]\F[/math]. Assume that [math]\mathcal{H}_2[/math] is independent of [math]\sigma(Z)\lor \mathcal{H}_1[/math]. Then
Let [math]A\in \mathcal{H}_1\lor \mathcal{H}_2[/math] such that [math]A=B\cap C[/math], where [math]B\in \mathcal{H}_1[/math] and [math]C\in\mathcal{H}_2[/math]. Then
Now we note that
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math](\xi_n)_{n\geq 1}[/math] be a sequence of iid r.v.'s such that for all [math]n\geq 1[/math] we have [math]\E[\vert \xi_n\vert] \lt \infty[/math]. Moreover, let [math]S_0=0[/math] and [math]S_n=\sum_{j=1}^n\xi_j[/math]. Then
At first, we want to show that [math]\E[\xi_1\mid S_n]=\frac{S_n}{n}[/math]. Indeed, we know that there is a measurable map [math]g[/math] such that
Martingales bounded in [math]L^2[/math] and random series
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Let [math](M_n)_{n\geq 0}[/math] be a martingale in [math]L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math], i.e. [math]\E[M_n^2] \lt \infty[/math] for all [math]n\geq 0[/math]. We say that [math](M_n)_{n\geq 0}[/math] is bounded in [math]L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] if [math]\sup_{n\geq 0}\E[M_n^2] \lt \infty[/math]. For [math]n\leq \nu[/math], we have that [math]\E[M_\nu\mid \F_n]=M_n[/math] implies that [math](M_\nu-M_n)[/math] is orthogonal to [math]L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. Hence, for all [math]s\leq t\leq n\leq \nu[/math], [math](M_\nu-M_n)[/math] is orthogonal to [math](M_t-M_s)[/math].
Now write [math]M_n=M_0+\sum_{k=1}^n(M_k-M_{k-1})[/math]. [math]M_n[/math] is then a sum of orthogonal terms and therefore
Let [math](M_n)_{n\geq 0}[/math] be a martingale in [math]L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math]. Then [math](M_n)_{n\geq 0}[/math] is bounded in [math]L^2(\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] if and only if [math]\sum_{k\geq 1}\E[(M_k-M_{k-1})^2] \lt \infty[/math] and in this case
Suppose that [math](X_n)_{n\geq 1}[/math] is a sequence of independent r.v.'s such that for all [math]k\geq 1[/math], [math]\E[X_k]=0[/math] and [math]\sigma_k^2=Var(X_k) \lt \infty[/math]. Then
- [math]\sum_{k\geq1}\sigma^2_k \lt \infty[/math] implies that [math]\sum_{k\geq 1}X_k[/math] converges a.s.
- If there is a [math]C \gt 0[/math] such that for all [math]\omega\in\Omega[/math] and [math]k\geq1[/math], [math]\vert X_k(\omega)\vert\leq C[/math], then [math]\sum_{k\geq 1}X_k[/math] converges a.s. implies that [math]\sum_{k\geq 1}\sigma_k^2 \lt \infty[/math].
Consider [math]\F_n=\sigma(X_1,...,x_n)[/math] with [math]F_0=\{\varnothing,\Omega\}[/math], [math]S_n=\sum_{j=1}^nX_j[/math] with [math]S_0=0[/math] and [math]A_n=\sum_{k=1}^n\sigma_k^2[/math] with [math]A_0=0[/math]. Moreover, set [math]M_n=S_n^2-A_n[/math]. Then [math](S_n)_{n\geq 0}[/math] is a martingale and
which implies that [math](M_n)_{n\geq 0}[/math] is a martingale. Let [math]T:=\inf\{n\in\N\mid \vert S_n\vert \gt \alpha\}[/math] for some constant [math]\alpha[/math]. Then [math]T[/math] is a stopping time. [math](M_{n\land T})_{n\geq 1}[/math] is a martingale and hence
Example
Let [math](a_n)_{n\geq 1}[/math] be a sequence of real numbers and let [math](\xi_n)_{n\geq 1}[/math] be iid r.v.'s with [math]\p[\xi=\pm 1]=\frac{1}{2}[/math]. Then [math]\sum_{n\geq 1}a_n\xi_n[/math] converges a.s. if and only if [math]\sum_{n\geq 1}a_n^2 \lt \infty[/math]. Indeed, we get [math]\vert a_n\xi_n\vert=\vert a_n\vert\xrightarrow{n\to\infty}0[/math] and therefore there exists a [math]C \gt 0[/math] such that for all [math]n\geq 1[/math], [math]\vert a_n\vert \leq C[/math]. Now for a r.v. [math]X[/math] recall that we write [math]\Phi_X(t)=\E\left[e^{itX}\right][/math]. We also know
Moreover, define [math]R_n(x)=e^{ix}-\sum_{k=0}^n\frac{i^kx^k}{k!}[/math]. Therefore we get
Indeed, [math]\vert R_0(x)\vert=\vert e^{ix}-1\vert=\left\vert\int_0^x ie^{iy}dy\right\vert\leq \min(2,\vert x\vert).[/math] Moreover, we have [math]\vert R_n(x)\vert=\left\vert \int_0^x iR_{n-1}(y)dy\right\vert[/math]. Hence the claim follows by a simple induction on [math]n[/math]. If [math]X[/math] is such that [math]\E[X]=0[/math], [math]\E[X^2]=\sigma^2 \lt \infty[/math] and [math]e^{itX}-\left( 1+itX-\frac{t^2X^2}{2}\right)=R_2(tX)[/math] we get
and [math]\E[R_2(tX)]\leq t^2\E[\vert X\vert^2\land tX^3][/math]. With dominated convergence it follows that [math]\Phi(t)=1-\frac{t^2\sigma^2}{2}+o(t^2)[/math] as [math]t\to 0[/math].
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a sequence of independent r.v.'s bounded by [math]k \gt 0[/math]. Then if [math]\sum_{n\geq 0}X_n[/math] converges a.s., [math]\sum_{n\geq 0}\E[X_n][/math] and [math]\sum_{n\geq 0}Var(X_n)[/math] both converge.
If [math]Z[/math] is a r.v. such that [math]\vert Z\vert \lt k[/math], [math]\E[Z]=0[/math] and [math]\sigma^2=Var(Z) \lt \infty[/math], then for [math]\vert t\vert\leq \frac{1}{k}[/math] we get
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 0}[/math] be a sequence of independent r.v.'s. Then [math]\sum_{n\geq 0}X_n[/math] converges a.s. if and only if for some [math]k \gt 0[/math] (then for every [math]k \gt 0[/math]) the following properties hold.
- [math]\sum_{n\geq 0}\p[\vert X_n\vert \gt k] \lt \infty[/math]
- [math]\sum_{n\geq 0}\E\left[X_n^{(k)}\right][/math] converges, where [math]X_n^{(k)}=X_n\one_{\{\vert X_n\vert\leq k\}}[/math]
- [math]\sum_{n\geq 0}Var\left(X_n^{(k)}\right) \lt \infty[/math]
Suppose that for some [math]k \gt 0[/math], [math](i)[/math], [math](ii)[/math] and [math](iii)[/math] hold. Then
Suppose [math](b_n)_{n\geq 1}[/math] is a sequence of strictly positive real numbers with [math]b_n\uparrow\infty[/math] as [math]n\to\infty[/math]. Let [math](v_n)_{n\geq 1}[/math] be a sequence of real numbers such that [math]v_n\xrightarrow{n\to\infty}v[/math]. Then
Note that
Now we only have to choose [math]N[/math] such that [math]n\geq N[/math] and [math]\vert v_k-v\vert \lt \varepsilon[/math] for any [math]\varepsilon \gt 0[/math].
Let [math](b_n)_{n\geq 1}[/math] be a sequence of real numbers, strictly positive with [math]b_n\uparrow \infty[/math] as [math]n\to\infty[/math]. Let [math](x_n)_{n\geq 1}[/math] be a sequence of real numbers. Then if [math]\sum_{n\geq 1}\frac{x_n}{b_n}[/math] converges, we get that
Let [math]v_n=\sum_{k=1}^n\frac{x_n}{b_n}[/math] and [math]v=\lim_{n\to\infty}v_n[/math]. Then [math]v_n-v_{n-1}=\frac{x_n}{b_n}[/math]. Moreover, we note that
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math](w_n)_{n\geq 1}[/math] be a sequence of r.v.’s such that [math]\E[w_n]=0[/math] for all [math]n\geq 1[/math] and [math]\sum_{n\geq 1}\frac{Var(w_n)}{n^2} \lt \infty[/math]. Then
Exercise.[c]
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be independent and non-negative r.v.'s such that [math]\E[X_n]=1[/math] for all [math]n\geq 1[/math]. Define [math]M_0=1[/math] and for [math]n\in\N[/math], let
- [math]\E[M_\infty]=1[/math].
- [math]M_n\xrightarrow{n\to\infty\atop L^1}M_\infty[/math].
- [math](M_n)_{n\geq 1}[/math] is u.i.
- [math]\prod_{n}a_n \gt 0[/math], where [math]0 \lt a_n=\E[X_n^{1/2}]\leq 1[/math].
- [math]\sum_{n}(1-a_n) \lt \infty[/math].
Moreover, if one of the following (then every one) statements hold, then
Exercise.
A martingale central limit theorem
Let [math](\Omega,\F,(\F_n)_{n\geq 0},\p)[/math] be a filtered probability space. Let [math](X_n)_{n\geq 0}[/math] be a sequence of real valued r.v.'s such that for all [math]n\geq 1[/math]
- [math]\E[X_n\mid \F_{n-1}]=0[/math].
- [math]\E[X_n^2\mid \F_{n-1}]=1[/math].
- [math]\E[\vert X_n\vert^3\mid \F_{n-1}]\leq k \lt \infty[/math].
Let [math]S_n=\sum_{j=1}^nX_j[/math]. Then
Define [math]\Phi_{n,j}(u)=\E\left[e^{iu\frac{X_j}{\sqrt{n}}}\mid \F_{j-1}\right][/math]. A Taylor expansion yields
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].