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We denote by <math>C_b(\R^d)</math> the space of bounded and continuous functions <math>\varphi:\R^d\to\R</math>. Moreover, we endow <math>C_b(\R^d)</math> with the supremums norm <math>\|\varphi\|_\infty=\sup_{x\in\R^d}\vert\varphi(x)\vert</math>. The space <math>(C_b(\R^d),\|\cdot\|_\infty)</math> forms a Banach space, i.e. it is a complete normed vector space. Next, we want to introduce the notion of law convergence in terms of probability measures. | |||
{{definitioncard|Weak and Law convergence|<ul style{{=}}"list-style-type:lower-roman"><li>Let <math>(\mu_n)_{n\geq 1}</math> be a sequence of probability measures on <math>\R^d</math>. We say that <math>(\mu_n)_{n\geq 1}</math> is converging weakly to a probability measure <math>\mu</math> on <math>\R^d</math>, and we write | |||
<math display="block"> | |||
\lim_{n\to\infty\atop w}\mu_n=\mu, | |||
</math> | |||
if for all <math>\varphi\in C_b(\R^d)</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d} \varphi d\mu_n=\int_{\R^d} \varphi d\mu. | |||
</math> | |||
</li> | |||
<li>Let <math>(\Omega,\A,\p)</math> be a probability space. A sequence of r.v.'s <math>(X_n)_{n\geq 1}</math>, taking values in <math>\R^d</math>, is said to converge in law to a r.v. <math>X</math> with values in <math>\R^d</math> and we write | |||
<math display="block"> | |||
\lim_{n\to\infty\atop law}X_n=X, | |||
</math> | |||
if <math>\lim_{n\to\infty\atop w}\p_{X_n}=\p_X</math>, or equivalently if for all <math>\varphi\in C_b(\R^d)</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)]\Longleftrightarrow \lim_{n\to\infty}\int_{\R^d}\varphi(x)d\p_{X_n}(x)=\int_{\R^d}\varphi(x)d\p_{X}(x). | |||
</math> | |||
</li> | |||
</ul>}} | |||
{{alert-info | One has to consider the following: | |||
<ul style{{=}}"list-style-type:lower-roman"><li>There is an abuse of language when we say that <math>\lim_{n\to\infty\atop law}X_n=X</math> because the r.v. <math>X</math> is not determined in a unique way, only <math>\p_X</math> is unique. | |||
</li> | |||
<li>Note also that the r.v.'s <math>X_n</math> and <math>X</math> need not be defined on the same probability space <math>(\Omega,\A,\p)</math>. | |||
</li> | |||
<li>The space of probability measures on <math>\R^d</math> can be viewed as a subspace of <math>C_b(\R^d)^*</math> (the dual space of <math>C_b(\R^d)</math>). The weak convergence then corresponds to convergence for the ''weak*''-topology. | |||
</li> | |||
<li>It is enough to show that <math>\lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)]</math> or <math>\lim_{n\to\infty}\int_{\R^d}\varphi(x)d\p_{X_n}(x)=\int_{\R^d} \varphi(x)d\p_X(x)</math> is satisfied for all <math>\varphi\in C_c(\R^d)</math>, where <math>C_c(\R^d)</math> is the space of continuous functions with compact support. That is, <math>\varphi\in C_c(\R^d)</math> if <math>supp(\varphi):=\overline{\{x\in\R^d\mid \varphi(x)\not=0\}}</math> is compact. | |||
</li> | |||
</ul> | |||
}} | |||
'''Example''' | |||
We got the following examples: | |||
<ul style{{=}}"list-style-type:lower-roman"><li>If <math>X_n</math> and <math>X\in\mathbb{Z}^d</math> for all <math>n\geq 1</math>, then <math>\lim_{n\to\infty\atop law}X_n=X</math> if and only if for all <math>X\in\mathbb{Z}^d</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\p[X_n=x]=\p[X=x]. | |||
</math> | |||
To see this, we use point (4) of the remark above. Let therefore <math>\varphi\in C_c(\R^d)</math>. Then | |||
<math display="block"> | |||
\E[\varphi(X_n)]=\sum_{k\in\mathbb{Z}^d}\varphi(k)\p[X_n=k]. | |||
</math> | |||
Since <math>\varphi</math> has compact support, i.e. <math>\varphi(x)=0</math> for <math>\vert x\vert \leq C</math> for some <math>C\geq 0</math>, we get | |||
<math display="block"> | |||
\E[\varphi(X_n)]=\sum_{k\in\mathbb{Z}^d\atop \vert k\vert\leq C}\varphi(k)\p[X_n=k]. | |||
</math> | |||
Hence we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\E[\varphi(X_n)]=\lim_{n\to\infty}\sum_{k\in\mathbb{Z}^d\atop \vert k\vert\leq C}\varphi(k)\p[X_n=k]=\sum_{k\in\mathbb{Z}^d\atop \vert k\vert \leq C}\varphi(k)\p[X=k]=\E[\varphi(X)]. | |||
</math> | |||
</li> | |||
<li>If <math>X_n</math> has density <math>\p_{X_n}(dx)=P_n(x)dx</math> for all <math>n\geq1</math> and if we assume that | |||
<math display="block"> | |||
\lim_{n\to\infty\atop a.e.}P_n(x)=P(x), | |||
</math> | |||
then there is a <math>q\geq 0</math> such that <math>\int_{\R^d}q(x)dx < \infty</math> and <math>P_n(x)\leq q(x)</math> a.e. Then an application of the dominated convergence theorem shows that | |||
<math display="block"> | |||
\int_{\R^d} P(x)dx=1, | |||
</math> | |||
and thus there exists a r.v. <math>X</math> with density <math>P</math> such that <math>\lim_{n\to\infty\atop law}X_n=X</math> and for <math>\varphi\in C_b(\R^d)</math> we get | |||
<math display="block"> | |||
\E[\varphi(X_n)]=\int_{\R^d}\varphi(x)P_n(x)dx, | |||
</math> | |||
and <math>\vert\varphi(x)P_n(x)\vert\leq \underbrace{\|\varphi\|_\infty q(x)}_{\in\mathcal{L}^1(\R^d)}</math>. So with the dominated convergence theorem we get | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d}\varphi(x)P_n(x)dx=\int_{\R^d}\varphi(x)P(x)dx=\E[\varphi(X)]. | |||
</math> | |||
</li> | |||
<li>Let <math>X_n\sim \mathcal{N}(0,\sigma_n^2)</math> such that <math>\lim_{n\to\infty}\sigma_n=0</math>. Then <math>\lim_{n\to\infty\atop law}X_n=0</math> and | |||
<math display="block"> | |||
\E[\varphi(X_n)]=\int_\R \varphi(x)e^{-\frac{x^2}{2\sigma_n^2}}\frac{1}{\sigma_n\sqrt{2\pi}}dx. | |||
</math> | |||
Now using that <math>u=\frac{x}{\sigma_n}</math>, we get <math>dx=\sigma_n du</math> and hence we have | |||
<math display="block"> | |||
\E[\varphi(X_n)]=\int_\R \varphi(x)e^{-\frac{x^2}{2\sigma_n^2}}\frac{1}{\sigma_n\sqrt{2\pi}}dx=\int_\R \varphi(\sigma_n u)e^{-\frac{u^2}{2}}\frac{1}{\sqrt{2\pi}}du. | |||
</math> | |||
Moreover, we have <math>\vert \varphi(\sigma_n u)e^{-\frac{u^2}{2}}\vert\leq \underbrace{\|\varphi\|_\infty e^{-\frac{u^2}{2}}}_{\in\mathcal{L}^1(\R)}</math>. Hence we get | |||
<math display="block"> | |||
\lim_{n\to\infty}\E[\varphi(X_n)]=\int_{\R}\varphi(0)e^{-\frac{u^2}{2}}\frac{1}{\sqrt{2\pi}}du. | |||
</math> | |||
</li> | |||
</ul> | |||
{{proofcard|Proposition|prop-1|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be a sequence of r.v.'s and assume that <math>\lim_{n\to\infty\atop \p}X_n=X</math>. Then <math>\lim_{n\to\infty\atop law}X_n=X</math>. | |||
|We first note that if <math>\lim_{n\to\infty\atop a.s.}X_n=X</math> then <math>\lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)]</math> for every <math>\varphi\in C_b(\R^d)</math>. Let us now assume that <math>(X_n)_{n\geq 1}</math> does not converge in law to <math>X</math>. Then there is a <math>\varphi\in C_b(\R^d)</math> such that <math>\E[\varphi(X_n)]</math> does not converge to <math>\E[\varphi(X)]</math>. We can hence extract a subsequence <math>(X_{n_k})_{k\geq 1}</math> from <math>(X_n)_{n\geq 1}</math> and find an <math>\epsilon > 0</math> such that | |||
<math display="block"> | |||
\vert\E[\varphi(X_{n_k})]-\E[\varphi(X)]\vert > \epsilon. | |||
</math> | |||
But this contradicts the fact that we can extract a further subsequence <math>(X_{n_{k_l}})_{l\geq 1}</math> from <math>(X_{n_k})_{k\geq 1}</math> such that | |||
<math display="block"> | |||
\lim_{l\to\infty\atop a.s.} X_{n_{k_{l}}}=X. | |||
</math>}} | |||
Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be a sequence of r.v.'s, A natural question would be to ask whether, under these condition, we have a <math>B\in\B(\R)</math> such that <math>\lim_{n\to\infty}\p[X_n\in B]=\p[X\in B]</math>. If we take <math>B=\{0\}</math> and use the previous example, we would get | |||
<math display="block"> | |||
\lim_{n\to\infty}\underbrace{\p[X_n=0]}_{=0}\not=\underbrace{\p[X=0]}_{=1}, | |||
</math> | |||
which shows that the answer to the question is negative. | |||
{{proofcard|Proposition|prop-2|Let <math>(\mu_n)_{n\geq 1}</math> be a sequence of probability measures on <math>\R^d</math> and <math>\mu</math> be a probability measure on <math>\R^d</math>. Then the following are equivalent. | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\lim_{n\to\infty\atop w}\mu_n=\mu</math>. | |||
</li> | |||
<li>For all open subsets <math>G\subset \R^d</math> we have | |||
<math display="block"> | |||
\limsup_n\mu_n(G)\geq \mu(G). | |||
</math> | |||
</li> | |||
<li>For all closed subsets <math>F\subset\R^d</math> we have | |||
<math display="block"> | |||
\limsup_n\mu_n(F)\leq \mu(F). | |||
</math> | |||
</li> | |||
<li>For all Borel measurable sets <math>B\in\B(\R^d)</math> with <math>\mu(\partial B)=0</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\mu_n(B)=\mu(B). | |||
</math> | |||
</li> | |||
</ul> | |||
|We immediately note that <math>(ii)\Longleftrightarrow (iii)</math> by taking complements. | |||
First we show <math>(i)\Longrightarrow (ii):</math> Let <math>G</math> be an open subset of <math>\R^d</math>. Define <math>\varphi_p(x):=p(d(x,G^C)\land 1)</math>. Then <math>\varphi_p</math> is continuous, bounded, <math>0\leq \varphi_p(x)\leq \one_{G}(x)</math> for all <math>x\in\R^d</math> and <math>\varphi_p\uparrow \one_{G}</math> (note that <math>d(x,F)=\inf_{y\in F}d(x,y)</math>) as <math>p\to\infty</math>. Moreover, <math>F</math> is closed if and only if <math>d(x,F)=0</math>. We also get that <math>\varphi_p(x)=0</math> on <math>G^C</math> and <math>0\leq \varphi_p(x)\leq 1\leq \one_{G}(x)</math> for all <math>x\in\R^d</math>. Therefore we get | |||
<math display="block"> | |||
\liminf_n\mu_n(G)\geq \sup_{p}\left(\liminf_nF\int_{\R^d}\varphi_pd\mu_n\right)=\sup_p\int_{\R^d} \varphi_pd\mu=\int \one_G d\mu=\mu(G). | |||
</math> | |||
Now we show that <math>(ii)</math> and <math>(iii)\Longrightarrow (iv):</math> For Borel measurable set <math>B\in\B(\R^d)</math> with <math>\mathring{B}\subset B\subset \bar B</math> we get | |||
<math display="block"> | |||
\limsup_n\mu_n(B)\leq \limsup_n\mu_n(\bar B)\leq \mu(\bar B), | |||
</math> | |||
<math display="block"> | |||
\liminf_n\mu_n(B)\geq \liminf_n\mu_n(\mathring{B})\geq \mu(\mathring{B}). | |||
</math> | |||
Therefore it follows that | |||
<math display="block"> | |||
\mu(\mathring{B})\leq \liminf_n \mu_n(B)\leq \limsup_n\mu_n(B)\leq \mu(B). | |||
</math> | |||
Moreover, if <math>\mu(\partial B)=0</math>, we get that <math>\mu(\bar B)=\mu(\mathring{B})=\mu(B)</math> and thus <math>\lim_{n\to\infty}\mu_n(B)=\mu(B)</math>. | |||
Now we show <math>(iv)\Longrightarrow (i):</math> Let therefore <math>\varphi\in C_b(\R^d)</math>. We can always use that <math>\varphi=\varphi^+-\varphi^-</math> and so, without loss of generality, we may assume that <math>\varphi\geq 0</math>. Let <math>\varphi\geq 0</math> and <math>K\geq 0</math> be such that <math>0\leq \varphi\leq K</math>. Then | |||
<math display="block"> | |||
\int_{\R^d}\varphi(x)d\mu(x)=\int_{\R^d}\underbrace{\left(\int_0^K\one_{\{t\leq \varphi(x)\}}dt\right)}_{K\land \varphi(x)=\varphi(x)}d\mu(x)=\int_0^K\mu(E_t^\varphi)dt, | |||
</math> | |||
where <math>E_t^\varphi:=\{x\in\R^d\mid \varphi(x)\geq t\}</math>. Similarly, we have | |||
<math display="block"> | |||
\int_{\R^d}\varphi(x)d\mu_n(x)=\int_0^K\mu_n(E_t^\varphi)dt. | |||
</math> | |||
Now we can note that <math>\partial E_t^\varphi\subset\{x\in\R^d\mid \varphi(x)=t\}</math>. Moreover, there are at most countably many values of <math>t</math> for which <math>\mu\left(\{x\in\R^d\mid \varphi(x)=t\}\right) > 0</math>. Indeed, for an integer<math>K\geq 1</math> we get that <math>\mu\left(\{x\in\R^d\mid \varphi(x)=t\}\right)\geq \frac{1}{K}</math>. This can happen for at most <math>K</math> distinct values of <math>t</math>. Thence we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\mu_n(E_t^\varphi)=\mu(E_t^\varphi)dt\text{a.e.}, | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d} \varphi(x)d\mu_n(x)=\int_0^K\mu_n(E_t^\varphi)dt\xrightarrow{n\to\infty} \int_0^K\mu(E_t^\varphi)dt=\int_{\R^d}\varphi(x)d\mu(x). | |||
</math>}} | |||
''Consequences:'' We look at the case of <math>d=1</math>. Let <math>(X_n)_{n\geq 1}</math> be a sequence of r.v.'s with values in <math>\R</math> and let <math>X</math> be a r.v. with values in <math>\R</math>. One can show that | |||
<math display="block"> | |||
\lim_{n\to\infty\atop law}X_n=X\Longleftrightarrow \lim_{n\to\infty}F_{X_n}(t)=F_X(t). | |||
</math> | |||
{{proofcard|Proposition|prop-3|Let <math>(\mu_n)_{n\geq 1}</math> and <math>\mu</math> be probability measures on <math>\R^d</math>. Let <math>H\subset C_b(\R^d)</math> such that <math>\bar H\supset C_c(\R^d)</math>. Then the following are equivalent. | |||
<ul style{{=}}"list-style-type:lower-roman"><li><math>\lim_{n\to\infty\atop w}\mu_n=\mu.</math> | |||
</li> | |||
<li>For all <math>\varphi\in C_c(\R^d)</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d} \varphi d\mu_n=\int_{\R^d} \varphi d\mu. | |||
</math> | |||
</li> | |||
<li>For all <math>\varphi\in H</math> we have | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d}\varphi d\mu_n=\int_{\R^d} \varphi d\mu. | |||
</math> | |||
</li> | |||
</ul> | |||
|It is obvious that <math>(i)\Longrightarrow (ii)</math> and <math>(i)\Longrightarrow (iii)</math>. | |||
Therefore we first show <math>(ii)\Longrightarrow (i):</math> Let therefore <math>\varphi\in C_b(\R^d)</math> and let <math>(f_k)_{k\geq 1}\in C_c(\R^d)</math> with <math>0\leq f_k\leq 1</math> and <math>f_k\uparrow 1</math> as <math>k\to\infty</math>. Then for all <math>k\geq 1</math> we get that <math>\varphi f_k\in C_c(\R^d)</math> and hence | |||
<math display="block"> | |||
\lim_{n\to\infty}\int_{\R^d}\varphi f_kd\mu_n=\int_{\R^d} \varphi f_k d\mu. | |||
</math> | |||
Moreover, we have | |||
<math display="block"> | |||
\left\vert \int_{\R^d} \varphi d\mu-\int_{\R^d}\varphi f_k d\mu\right\vert\leq \sup_{x\in\R^d}\vert\varphi(x)\vert\left(1-\int_{\R^d} f_kd\mu\right) | |||
</math> | |||
and also | |||
<math display="block"> | |||
\left\vert \int_{\R^d} \varphi d\mu_n-\int_{\R^d} \varphi f_k d\mu_n\right\vert\leq \sup_{x\in\R^d}\vert \varphi(x)\vert \left(1-\int_{\R^d} f_k d\mu_n\right). | |||
</math> | |||
Hence, for all <math>k\geq 1</math> we get | |||
<math display="block"> | |||
\begin{align*} | |||
\limsup_n \left\vert\int_{\R^d}\varphi d\mu-\int_{\R^d} \varphi d\mu_n\right\vert&\leq \sup_{x\in \R^d}\vert \varphi(x)\vert \limsup_n\left[\left(1-\int_{\R^d} f_k d\mu_n\right)+\left(1-\int_{\R^d} f_k d\mu\right)\right]\\ | |||
&=2\sup_{x\in\R^d}\vert\varphi(x)\vert\left(1-\int f_kd\mu\right)\xrightarrow{k\to\infty}0. | |||
\end{align*} | |||
</math> | |||
Now we show <math>(iii)\Longrightarrow (ii):</math> Let therefore <math>\varphi\in C_c(\R^d)</math>. Then there is a sequence <math>(\varphi_k)_{k\geq 1}\subset H</math> such that <math>\|\varphi-\varphi_k\|_\infty\leq \frac{1}{k}</math> for all <math>k\geq 1</math>. This implies that | |||
<math display="block"> | |||
\begin{multline*} | |||
\limsup_n\left\vert\int \varphi d\mu_n-\int \varphi d\mu\right\vert\\\leq \limsup_n\left(\left\vert\int \varphi d\mu_n-\int\varphi_k d\mu_n\right\vert+\underbrace{\left\vert\int \varphi_k d\mu_n-\int\varphi_k d\mu\right\vert}_{\xrightarrow{n\to\infty}0} +\left\vert \int \varphi_k d\mu-\int\varphi d\mu\right\vert\right)\leq \frac{2}{k}. | |||
\end{multline*} | |||
</math> | |||
The claim follows now for <math>k\to\infty</math>.}} | |||
{{proofcard|Theorem (Lèvy)|thm-1|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>(\mu_n)_{n\geq 1}</math> be a sequence of probability measures on <math>\R^d</math> associated to a sequence of real r.v.'s <math>(X_n)_{n\geq 1}</math>. Moreover, let <math>\hat{\mu}_n(\xi)=\int_{\R^d}e^{i\xi x}d\mu_n(x)</math> and <math>\Phi_X(\xi)=\E[e^{i\xi x}]</math>. Then for all <math>\xi\in\R^d</math> we get | |||
<math display="block"> | |||
\lim_{n\to\infty\atop w}\mu_n=\mu \Longleftrightarrow\lim_{n\to\infty}\hat\mu_n(\xi)=\hat\mu(\xi). | |||
</math> | |||
Equivalently, for all <math>\xi\in\R^d</math> we get | |||
<math display="block"> | |||
\lim_{n\to\infty\atop law}X_n=X\Longleftrightarrow\lim_{n\to\infty}\Phi_{X_n}(\xi)=\Phi_X(\xi). | |||
</math> | |||
|It is obvious that <math>\lim_{n\to\infty\atop w}\mu_n=\mu</math> implies that <math>\lim_{n\to\infty}\hat{\mu}_n(\xi)=\hat{\mu}(\xi)</math>. Therefore <math>e^{i\xi X}</math> is continuous and bounded. For notation conventions we deal with the case <math>d=1</math>. Let therefore <math>f\in C_c(\R^d)</math>. For <math>\sigma > 0</math> we also note <math>g_\sigma(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}</math>. | |||
\begin{exer} | |||
Show that <math>g_\sigma *f\xrightarrow{\sigma\to 0}f</math> uniformly on <math>\R</math>. | |||
\end{exer} | |||
\begin{exer} | |||
Show that if <math>\nu</math> is a probability measure, then | |||
<math display="block"> | |||
\int_\R g_\sigma * fd\nu=\int_\R f(x)(g_\sigma *\nu)(x)dx=\int_\R f(x)\frac{1}{\sigma\sqrt{2\pi}}\int_\R e^{i\xi x}g_{\frac{1}{\sigma}}(\xi)\hat{\nu}(\xi)d\xi dx. | |||
</math> | |||
\end{exer} | |||
Since <math>\lim_{n\to\infty}\hat{\mu}_n(\xi)=\hat{\mu}(\xi)</math>, we get by the dominated convergence theorem that | |||
<math display="block"> | |||
\int_\R e^{i\xi x}g_{\frac{1}{\sigma}}(\xi)\hat{\mu}_n(\xi)d\xi\xrightarrow{n\to\infty}\int_\R e^{i\xi x}g_{\frac{1}{\sigma}}(\xi)\hat{\mu}(\xi)d\xi. | |||
</math> | |||
These quantities are bounded by 1, and hence we can apply the dominated convergence theorem to obtain | |||
<math display="block"> | |||
\int_\R g_\sigma * fd\mu_n\xrightarrow{n\to\infty} \int_\R g_\sigma * fd\mu. | |||
</math> | |||
Let now <math>H:=\{\varphi=g_\sigma *f\mid \sigma > 0,f\in C_c(\R^d)\}\subset C_b(\R^d)</math>. Since <math>f\in C_c(\R^d)</math> we get that <math>\|g_\sigma *f-f\|_{\infty}\xrightarrow{n\to\infty}0</math> and thus <math>\bar H\supset C_c(\R^d)</math>. The result now follows from the previous proposition.}} | |||
{{proofcard|Theorem (Lévy)|thm-2|Let <math>(\mu_n)_{n\geq 1}</math> be a sequence of probability measures on <math>\R^d</math> with characteristic functions <math>(\Phi_n)_{n\geq 1}</math>. If <math>\Phi_n</math> converges pointwise to a function <math>\Phi</math> which is continuous at 0, then | |||
<math display="block"> | |||
\lim_{n\to\infty\atop w}\mu_n=\mu | |||
</math> | |||
for some probability measure <math>\mu</math> on <math>\R^d</math>. | |||
|No proof here.}} | |||
'''Example''' | |||
Let <math>(X_n)_{n\geq 1}</math> be a sequence of poisson r.v.'s with parameter <math>\lambda</math>. Moreover, consider the sequence <math>Z_n=\frac{X_n-1}{\sqrt{n}}</math>. Then we have | |||
<math display="block"> | |||
\begin{multline*} | |||
E\left[e^{iu Z_n}\right]=\E\left[e^{iu\left(\frac{X_n-u}{\sqrt{n}}\right)}\right]=e^{-i\frac{u}{\sqrt{n}}}\E\left[e^{iu\frac{X_n}{\sqrt{n}}}\right]=e^{-i\frac{u}{\sqrt{n}}}e^{\lambda\left(e^{i \frac{u}{\sqrt{n}}}-1\right)}\\=e^{-i\frac{u}{\sqrt{n}}}e^{\lambda\left(i\frac{u}{\sqrt{n}}-\frac{u^2}{2n}+O\left(\frac{1}{n}\right)\right)}\xrightarrow{n\to\infty}e^{-\frac{u^2}{2}}. | |||
\end{multline*} | |||
</math> | |||
Since <math>\E\left[e^{iu\mathcal{N}(0,1)} \right]=e^{-\frac{u^2}{2}}</math>, we deduce that <math>\lim_{n\to\infty\atop law}Z_n=\lim_{n\to\infty\atop law}\frac{X_n-u}{\sqrt{n}}=\mathcal{N}(0,1)</math>. Before stating and proving the central limit theorem, we give two extra results on convergence in law. | |||
{{proofcard|Theorem|thm-3|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be a sequence of r.v.'s and <math>X</math> a r.v. and assume that <math>\lim_{n\to\infty\atop law}X_n=X</math> and that <math>X</math> is a.s. equal to a constant <math>a</math>. Then | |||
<math display="block"> | |||
\lim_{n\to\infty\atop\p}X_n=X. | |||
</math> | |||
|Let <math>f(x):=\vert x-a\vert\land 1</math>. Then <math>f</math> is a continuous and bounded map and therefore <math>\lim_{n\to\infty}\E[f(X_n)]=\E[f(X)]=0</math>, i.e. <math>\lim_{n\to\infty}\E[\vert X_n-a\vert \land 1]=0</math> which implies that <math>\lim_{n\to\infty\atop \p}X_n=X</math>.}} | |||
{{proofcard|Theorem|thm-4|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>(X_n)_{n\geq 1}</math> be a sequence of r.v.'s and <math>X</math> be r.v. in <math>\R^d</math>. Assume that <math>X_n</math> has density <math>f_n</math> for all <math>n\geq 1</math> and <math>X</math> has density <math>f</math>. Moreover, assume that <math>\lim_{n\to\infty}f_n(x)=f(x)</math>, a.e. Then | |||
<math display="block"> | |||
\lim_{n\to\infty\atop law}X_n=X. | |||
</math> | |||
|We need to show that <math>\E[h(X_n)]\xrightarrow{n\to\infty}\E[h(X)]</math>, where <math>h:\R^d\to\R</math> is a bounded and measurable map and | |||
<math display="block"> | |||
\E[h(X_n)]=\int_{\R^d}h(x)f_n(x)dx, | |||
</math> | |||
<math display="block"> | |||
\E[h(X)]=\int_{\R^d}h(x)f(x)dx. | |||
</math> | |||
Let <math>h:\R^d\to\R</math> be a bounded and measurable map. Moreover, let <math>\alpha=\sup_{x\in\R^d}\vert h(x)\vert.</math> Set <math>h_1(x)=h(x)+\alpha\geq 0</math> and <math>h_2(x)=\alpha-h(x)\geq 0</math>. So it follows that <math>h_1f_n\geq 0</math> and <math>h_2f_n\geq 0</math>. Moreover, we also get that | |||
<math display="block"> | |||
h_1f_n\xrightarrow{n\to\infty}h_1f\text{a.e.} | |||
</math> | |||
<math display="block"> | |||
h_2f_n\xrightarrow{n\to\infty}h_2f\text{a.e.} | |||
</math> | |||
With Fatou's lemma we get | |||
<math display="block"> | |||
\E[h_1(X)]=\int_{\R^d}h_1(x)f(x)dx\leq \liminf_n\int_{\R^d}h_1(x)f_n(x)dx=\liminf_n\E[h_1(X_n)]. | |||
</math> | |||
Similarly we get <math>\E[h_2(X)]\leq \liminf_n\E[h_2(X_n)]</math>. Now substitute <math>h_1(x)=h(x)+\alpha</math> and <math>h_2(x)=\alpha-h(x)</math>, where we use <math>\liminf_n(-a_n)=\liminf_n(a_n)</math> to obtain | |||
<math display="block"> | |||
\limsup_n\E[h(X_n)]\leq \E[h(X)]\leq \liminf_n\E[h(X_n)], | |||
</math> | |||
which implies that | |||
<math display="block"> | |||
\liminf_n\E[h(X_n)]=\limsup_n\E[h(X_n)]=\E[h(X)]. | |||
</math>}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Revision as of 01:53, 8 May 2024
We denote by [math]C_b(\R^d)[/math] the space of bounded and continuous functions [math]\varphi:\R^d\to\R[/math]. Moreover, we endow [math]C_b(\R^d)[/math] with the supremums norm [math]\|\varphi\|_\infty=\sup_{x\in\R^d}\vert\varphi(x)\vert[/math]. The space [math](C_b(\R^d),\|\cdot\|_\infty)[/math] forms a Banach space, i.e. it is a complete normed vector space. Next, we want to introduce the notion of law convergence in terms of probability measures.
- Let [math](\mu_n)_{n\geq 1}[/math] be a sequence of probability measures on [math]\R^d[/math]. We say that [math](\mu_n)_{n\geq 1}[/math] is converging weakly to a probability measure [math]\mu[/math] on [math]\R^d[/math], and we write
[[math]] \lim_{n\to\infty\atop w}\mu_n=\mu, [[/math]]if for all [math]\varphi\in C_b(\R^d)[/math] we have[[math]] \lim_{n\to\infty}\int_{\R^d} \varphi d\mu_n=\int_{\R^d} \varphi d\mu. [[/math]]
- Let [math](\Omega,\A,\p)[/math] be a probability space. A sequence of r.v.'s [math](X_n)_{n\geq 1}[/math], taking values in [math]\R^d[/math], is said to converge in law to a r.v. [math]X[/math] with values in [math]\R^d[/math] and we write
[[math]] \lim_{n\to\infty\atop law}X_n=X, [[/math]]if [math]\lim_{n\to\infty\atop w}\p_{X_n}=\p_X[/math], or equivalently if for all [math]\varphi\in C_b(\R^d)[/math] we have[[math]] \lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)]\Longleftrightarrow \lim_{n\to\infty}\int_{\R^d}\varphi(x)d\p_{X_n}(x)=\int_{\R^d}\varphi(x)d\p_{X}(x). [[/math]]
- There is an abuse of language when we say that [math]\lim_{n\to\infty\atop law}X_n=X[/math] because the r.v. [math]X[/math] is not determined in a unique way, only [math]\p_X[/math] is unique.
- Note also that the r.v.'s [math]X_n[/math] and [math]X[/math] need not be defined on the same probability space [math](\Omega,\A,\p)[/math].
- The space of probability measures on [math]\R^d[/math] can be viewed as a subspace of [math]C_b(\R^d)^*[/math] (the dual space of [math]C_b(\R^d)[/math]). The weak convergence then corresponds to convergence for the weak*-topology.
- It is enough to show that [math]\lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)][/math] or [math]\lim_{n\to\infty}\int_{\R^d}\varphi(x)d\p_{X_n}(x)=\int_{\R^d} \varphi(x)d\p_X(x)[/math] is satisfied for all [math]\varphi\in C_c(\R^d)[/math], where [math]C_c(\R^d)[/math] is the space of continuous functions with compact support. That is, [math]\varphi\in C_c(\R^d)[/math] if [math]supp(\varphi):=\overline{\{x\in\R^d\mid \varphi(x)\not=0\}}[/math] is compact.
Example
We got the following examples:
- If [math]X_n[/math] and [math]X\in\mathbb{Z}^d[/math] for all [math]n\geq 1[/math], then [math]\lim_{n\to\infty\atop law}X_n=X[/math] if and only if for all [math]X\in\mathbb{Z}^d[/math] we have
[[math]] \lim_{n\to\infty}\p[X_n=x]=\p[X=x]. [[/math]]To see this, we use point (4) of the remark above. Let therefore [math]\varphi\in C_c(\R^d)[/math]. Then[[math]] \E[\varphi(X_n)]=\sum_{k\in\mathbb{Z}^d}\varphi(k)\p[X_n=k]. [[/math]]Since [math]\varphi[/math] has compact support, i.e. [math]\varphi(x)=0[/math] for [math]\vert x\vert \leq C[/math] for some [math]C\geq 0[/math], we get[[math]] \E[\varphi(X_n)]=\sum_{k\in\mathbb{Z}^d\atop \vert k\vert\leq C}\varphi(k)\p[X_n=k]. [[/math]]Hence we have[[math]] \lim_{n\to\infty}\E[\varphi(X_n)]=\lim_{n\to\infty}\sum_{k\in\mathbb{Z}^d\atop \vert k\vert\leq C}\varphi(k)\p[X_n=k]=\sum_{k\in\mathbb{Z}^d\atop \vert k\vert \leq C}\varphi(k)\p[X=k]=\E[\varphi(X)]. [[/math]]
- If [math]X_n[/math] has density [math]\p_{X_n}(dx)=P_n(x)dx[/math] for all [math]n\geq1[/math] and if we assume that
[[math]] \lim_{n\to\infty\atop a.e.}P_n(x)=P(x), [[/math]]then there is a [math]q\geq 0[/math] such that [math]\int_{\R^d}q(x)dx \lt \infty[/math] and [math]P_n(x)\leq q(x)[/math] a.e. Then an application of the dominated convergence theorem shows that[[math]] \int_{\R^d} P(x)dx=1, [[/math]]and thus there exists a r.v. [math]X[/math] with density [math]P[/math] such that [math]\lim_{n\to\infty\atop law}X_n=X[/math] and for [math]\varphi\in C_b(\R^d)[/math] we get[[math]] \E[\varphi(X_n)]=\int_{\R^d}\varphi(x)P_n(x)dx, [[/math]]and [math]\vert\varphi(x)P_n(x)\vert\leq \underbrace{\|\varphi\|_\infty q(x)}_{\in\mathcal{L}^1(\R^d)}[/math]. So with the dominated convergence theorem we get[[math]] \lim_{n\to\infty}\int_{\R^d}\varphi(x)P_n(x)dx=\int_{\R^d}\varphi(x)P(x)dx=\E[\varphi(X)]. [[/math]]
- Let [math]X_n\sim \mathcal{N}(0,\sigma_n^2)[/math] such that [math]\lim_{n\to\infty}\sigma_n=0[/math]. Then [math]\lim_{n\to\infty\atop law}X_n=0[/math] and
[[math]] \E[\varphi(X_n)]=\int_\R \varphi(x)e^{-\frac{x^2}{2\sigma_n^2}}\frac{1}{\sigma_n\sqrt{2\pi}}dx. [[/math]]Now using that [math]u=\frac{x}{\sigma_n}[/math], we get [math]dx=\sigma_n du[/math] and hence we have[[math]] \E[\varphi(X_n)]=\int_\R \varphi(x)e^{-\frac{x^2}{2\sigma_n^2}}\frac{1}{\sigma_n\sqrt{2\pi}}dx=\int_\R \varphi(\sigma_n u)e^{-\frac{u^2}{2}}\frac{1}{\sqrt{2\pi}}du. [[/math]]Moreover, we have [math]\vert \varphi(\sigma_n u)e^{-\frac{u^2}{2}}\vert\leq \underbrace{\|\varphi\|_\infty e^{-\frac{u^2}{2}}}_{\in\mathcal{L}^1(\R)}[/math]. Hence we get[[math]] \lim_{n\to\infty}\E[\varphi(X_n)]=\int_{\R}\varphi(0)e^{-\frac{u^2}{2}}\frac{1}{\sqrt{2\pi}}du. [[/math]]
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s and assume that [math]\lim_{n\to\infty\atop \p}X_n=X[/math]. Then [math]\lim_{n\to\infty\atop law}X_n=X[/math].
We first note that if [math]\lim_{n\to\infty\atop a.s.}X_n=X[/math] then [math]\lim_{n\to\infty}\E[\varphi(X_n)]=\E[\varphi(X)][/math] for every [math]\varphi\in C_b(\R^d)[/math]. Let us now assume that [math](X_n)_{n\geq 1}[/math] does not converge in law to [math]X[/math]. Then there is a [math]\varphi\in C_b(\R^d)[/math] such that [math]\E[\varphi(X_n)][/math] does not converge to [math]\E[\varphi(X)][/math]. We can hence extract a subsequence [math](X_{n_k})_{k\geq 1}[/math] from [math](X_n)_{n\geq 1}[/math] and find an [math]\epsilon \gt 0[/math] such that
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s, A natural question would be to ask whether, under these condition, we have a [math]B\in\B(\R)[/math] such that [math]\lim_{n\to\infty}\p[X_n\in B]=\p[X\in B][/math]. If we take [math]B=\{0\}[/math] and use the previous example, we would get
which shows that the answer to the question is negative.
Let [math](\mu_n)_{n\geq 1}[/math] be a sequence of probability measures on [math]\R^d[/math] and [math]\mu[/math] be a probability measure on [math]\R^d[/math]. Then the following are equivalent.
- [math]\lim_{n\to\infty\atop w}\mu_n=\mu[/math].
- For all open subsets [math]G\subset \R^d[/math] we have
[[math]] \limsup_n\mu_n(G)\geq \mu(G). [[/math]]
- For all closed subsets [math]F\subset\R^d[/math] we have
[[math]] \limsup_n\mu_n(F)\leq \mu(F). [[/math]]
- For all Borel measurable sets [math]B\in\B(\R^d)[/math] with [math]\mu(\partial B)=0[/math] we have
[[math]] \lim_{n\to\infty}\mu_n(B)=\mu(B). [[/math]]
We immediately note that [math](ii)\Longleftrightarrow (iii)[/math] by taking complements. First we show [math](i)\Longrightarrow (ii):[/math] Let [math]G[/math] be an open subset of [math]\R^d[/math]. Define [math]\varphi_p(x):=p(d(x,G^C)\land 1)[/math]. Then [math]\varphi_p[/math] is continuous, bounded, [math]0\leq \varphi_p(x)\leq \one_{G}(x)[/math] for all [math]x\in\R^d[/math] and [math]\varphi_p\uparrow \one_{G}[/math] (note that [math]d(x,F)=\inf_{y\in F}d(x,y)[/math]) as [math]p\to\infty[/math]. Moreover, [math]F[/math] is closed if and only if [math]d(x,F)=0[/math]. We also get that [math]\varphi_p(x)=0[/math] on [math]G^C[/math] and [math]0\leq \varphi_p(x)\leq 1\leq \one_{G}(x)[/math] for all [math]x\in\R^d[/math]. Therefore we get
Consequences: We look at the case of [math]d=1[/math]. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s with values in [math]\R[/math] and let [math]X[/math] be a r.v. with values in [math]\R[/math]. One can show that
Let [math](\mu_n)_{n\geq 1}[/math] and [math]\mu[/math] be probability measures on [math]\R^d[/math]. Let [math]H\subset C_b(\R^d)[/math] such that [math]\bar H\supset C_c(\R^d)[/math]. Then the following are equivalent.
- [math]\lim_{n\to\infty\atop w}\mu_n=\mu.[/math]
- For all [math]\varphi\in C_c(\R^d)[/math] we have
[[math]] \lim_{n\to\infty}\int_{\R^d} \varphi d\mu_n=\int_{\R^d} \varphi d\mu. [[/math]]
- For all [math]\varphi\in H[/math] we have
[[math]] \lim_{n\to\infty}\int_{\R^d}\varphi d\mu_n=\int_{\R^d} \varphi d\mu. [[/math]]
It is obvious that [math](i)\Longrightarrow (ii)[/math] and [math](i)\Longrightarrow (iii)[/math]. Therefore we first show [math](ii)\Longrightarrow (i):[/math] Let therefore [math]\varphi\in C_b(\R^d)[/math] and let [math](f_k)_{k\geq 1}\in C_c(\R^d)[/math] with [math]0\leq f_k\leq 1[/math] and [math]f_k\uparrow 1[/math] as [math]k\to\infty[/math]. Then for all [math]k\geq 1[/math] we get that [math]\varphi f_k\in C_c(\R^d)[/math] and hence
Hence, for all [math]k\geq 1[/math] we get
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](\mu_n)_{n\geq 1}[/math] be a sequence of probability measures on [math]\R^d[/math] associated to a sequence of real r.v.'s [math](X_n)_{n\geq 1}[/math]. Moreover, let [math]\hat{\mu}_n(\xi)=\int_{\R^d}e^{i\xi x}d\mu_n(x)[/math] and [math]\Phi_X(\xi)=\E[e^{i\xi x}][/math]. Then for all [math]\xi\in\R^d[/math] we get
Equivalently, for all [math]\xi\in\R^d[/math] we get
It is obvious that [math]\lim_{n\to\infty\atop w}\mu_n=\mu[/math] implies that [math]\lim_{n\to\infty}\hat{\mu}_n(\xi)=\hat{\mu}(\xi)[/math]. Therefore [math]e^{i\xi X}[/math] is continuous and bounded. For notation conventions we deal with the case [math]d=1[/math]. Let therefore [math]f\in C_c(\R^d)[/math]. For [math]\sigma \gt 0[/math] we also note [math]g_\sigma(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{x^2}{2\sigma^2}}[/math]. \begin{exer} Show that [math]g_\sigma *f\xrightarrow{\sigma\to 0}f[/math] uniformly on [math]\R[/math]. \end{exer} \begin{exer} Show that if [math]\nu[/math] is a probability measure, then
Let [math](\mu_n)_{n\geq 1}[/math] be a sequence of probability measures on [math]\R^d[/math] with characteristic functions [math](\Phi_n)_{n\geq 1}[/math]. If [math]\Phi_n[/math] converges pointwise to a function [math]\Phi[/math] which is continuous at 0, then
No proof here.
Example
Let [math](X_n)_{n\geq 1}[/math] be a sequence of poisson r.v.'s with parameter [math]\lambda[/math]. Moreover, consider the sequence [math]Z_n=\frac{X_n-1}{\sqrt{n}}[/math]. Then we have
Since [math]\E\left[e^{iu\mathcal{N}(0,1)} \right]=e^{-\frac{u^2}{2}}[/math], we deduce that [math]\lim_{n\to\infty\atop law}Z_n=\lim_{n\to\infty\atop law}\frac{X_n-u}{\sqrt{n}}=\mathcal{N}(0,1)[/math]. Before stating and proving the central limit theorem, we give two extra results on convergence in law.
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s and [math]X[/math] a r.v. and assume that [math]\lim_{n\to\infty\atop law}X_n=X[/math] and that [math]X[/math] is a.s. equal to a constant [math]a[/math]. Then
Let [math]f(x):=\vert x-a\vert\land 1[/math]. Then [math]f[/math] is a continuous and bounded map and therefore [math]\lim_{n\to\infty}\E[f(X_n)]=\E[f(X)]=0[/math], i.e. [math]\lim_{n\to\infty}\E[\vert X_n-a\vert \land 1]=0[/math] which implies that [math]\lim_{n\to\infty\atop \p}X_n=X[/math].
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math](X_n)_{n\geq 1}[/math] be a sequence of r.v.'s and [math]X[/math] be r.v. in [math]\R^d[/math]. Assume that [math]X_n[/math] has density [math]f_n[/math] for all [math]n\geq 1[/math] and [math]X[/math] has density [math]f[/math]. Moreover, assume that [math]\lim_{n\to\infty}f_n(x)=f(x)[/math], a.e. Then
We need to show that [math]\E[h(X_n)]\xrightarrow{n\to\infty}\E[h(X)][/math], where [math]h:\R^d\to\R[/math] is a bounded and measurable map and
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].