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A random vector <math>Z=(Z_1,...,Z_n)</math> is said to be Gaussian, if for all <math>\lambda_1,...,\lambda_n\in\R</math> | |||
<math display="block"> | |||
\lambda_1 Z_1+\dotsm +\lambda_n Z_n | |||
</math> | |||
is Gaussian.Moreover, <math>Z</math> is called centered, if <math>\E[Z_j]=0</math> for all <math>1\leq j\leq n</math>. Let <math>Z</math> be a Gaussian vector. Then for all <math>\xi\in\R^n</math> we get | |||
<math display="block"> | |||
\E\left[e^{i\langle\xi, Z\rangle}\right]=\exp\left(-\frac{1}{2}\xi^t C_Z\xi\right), | |||
</math> | |||
where <math>C_Z:=(C_{ij})</math> and <math>C_{ij}=\E[Z_iZ_j]</math>. If <math>Cov(Z_i,Z_j)=0</math>, then <math>Z_i</math> and <math>Z_j</math> are independent. More generally, we have the Gaussian vectors | |||
<math display="block"> | |||
(\underbrace{X_1,...,X_{i_1}}_{Y_1},\underbrace{X_{i_1+1},...,X_{i_2}}_{Y_2},...,\underbrace{X_{i_{n-1}+1},...,X_{i_n}}_{Y_n}). | |||
</math> | |||
<math>Y_1</math> and <math>Y_2</math> are independent if and only if <math>Cov(X_j,X_n)=0</math>, where <math>1\leq j\leq i_1</math> and <math>i_1+1\leq k\leq i_2</math>. If <math>Z_1,...,Z_n</math> are independent Gaussian r.v.'s, we have that | |||
<math display="block"> | |||
Z=(Z_1,...,Z_n) | |||
</math> | |||
is a Gaussian vector. If <math>Z</math> is a Gaussian vector and <math>A\in \mathcal{M}(m\times n,\R)</math> , we get that <math>AZ</math> is again a Gaussian vector. | |||
{{proofcard|Theorem|thm1|Let <math>(\Omega,\F,\p)</math> be a probability space. Let <math>X\in L^1(\Omega,\F,\p)</math> and <math>Y_1,...,Y_p\in L^1(\Omega,\F,\p)</math> and let <math>(X,Y_1,...,Y_p)</math> be a centered Gaussian vector. Then | |||
<math display="block"> | |||
\E[X\mid Y_1,...,Y_p] | |||
</math> | |||
is the orthogonal projection of <math>X</math> on the vector space | |||
<math display="block"> | |||
span\{Y_1,...,Y_p\}. | |||
</math> | |||
Consequently, there exists real numbers <math>\lambda_1,...,\lambda_p</math> such that | |||
<math display="block"> | |||
\E[X\mid Y_1,...,Y_p]=\lambda_1 Y_1+\dotsm +\lambda_p Y_p. | |||
</math> | |||
{{alert-info | | |||
\label{rem4} | |||
Moreover, for a measurable map <math>h:\R\to\R_+</math> we get | |||
<math display="block"> | |||
\E[h(X)\mid Y_1,...,Y_p]=\int_\R h(x)Q_{\sum_{j=1}^n\lambda_j Y_j,\sigma^2}(x)dx, | |||
</math> | |||
where <math>\sigma^2=\E\left[\left( X-\sum_{j=1}^n\lambda_j Y_j\right)\right]</math> and | |||
<math display="block"> | |||
Q_{n,\sigma^2}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-m)^2}{2\sigma^2}\right). | |||
</math> | |||
}} | |||
|[Proof of [[#rem4 |Remark]]] | |||
Exercise.{{efn|This is done similarly to the proof of [[#thm1 |theorem]]}}}} | |||
\begin{proof}[Proof of [[#thm1 |Theorem]]] | |||
Let <math>\tilde X=\lambda_1 Y_1+\dotsm +\lambda_pY_p</math> be the orthogonal projection of <math>X</math> onto <math>span\{Y_1,...,Y_p\}</math>, meaning that for all <math>1\leq j\leq p</math> | |||
<math display="block"> | |||
\E[(X-\tilde X)Y_j]=0. | |||
</math> | |||
Note that this condition gives us explicitly the <math>\lambda_j's</math>. We obtain therefore that <math>(Y_1,...,Y_p,(X-\tilde X))</math> | |||
is a Gaussian vector. Moreover, we get <math>\E[(X-\tilde X)Y_j]=Cov(X-\tilde X,Y_j)=0</math> and thus <math>X-\tilde X</math> is independent of <math>Y_1,...,Y_p</math>. Hence | |||
<math display="block"> | |||
\E[X\mid Y_1,...,Y_p]=\E[X-\tilde X+\tilde X\mid Y_1,...,Y_p]=\E[X-\tilde X\mid Y_1,...,Y_p]+\E[\tilde X\mid Y_1,...,Y_p]=\E[X-\tilde X]+\tilde X=\tilde X. | |||
</math> | |||
\end{proof} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} | |||
==Notes== | |||
{{notelist}} |
Revision as of 01:53, 8 May 2024
A random vector [math]Z=(Z_1,...,Z_n)[/math] is said to be Gaussian, if for all [math]\lambda_1,...,\lambda_n\in\R[/math]
is Gaussian.Moreover, [math]Z[/math] is called centered, if [math]\E[Z_j]=0[/math] for all [math]1\leq j\leq n[/math]. Let [math]Z[/math] be a Gaussian vector. Then for all [math]\xi\in\R^n[/math] we get
where [math]C_Z:=(C_{ij})[/math] and [math]C_{ij}=\E[Z_iZ_j][/math]. If [math]Cov(Z_i,Z_j)=0[/math], then [math]Z_i[/math] and [math]Z_j[/math] are independent. More generally, we have the Gaussian vectors
[math]Y_1[/math] and [math]Y_2[/math] are independent if and only if [math]Cov(X_j,X_n)=0[/math], where [math]1\leq j\leq i_1[/math] and [math]i_1+1\leq k\leq i_2[/math]. If [math]Z_1,...,Z_n[/math] are independent Gaussian r.v.'s, we have that
is a Gaussian vector. If [math]Z[/math] is a Gaussian vector and [math]A\in \mathcal{M}(m\times n,\R)[/math] , we get that [math]AZ[/math] is again a Gaussian vector.
Let [math](\Omega,\F,\p)[/math] be a probability space. Let [math]X\in L^1(\Omega,\F,\p)[/math] and [math]Y_1,...,Y_p\in L^1(\Omega,\F,\p)[/math] and let [math](X,Y_1,...,Y_p)[/math] be a centered Gaussian vector. Then
\label{rem4} Moreover, for a measurable map [math]h:\R\to\R_+[/math] we get
\begin{proof}[Proof of Theorem] Let [math]\tilde X=\lambda_1 Y_1+\dotsm +\lambda_pY_p[/math] be the orthogonal projection of [math]X[/math] onto [math]span\{Y_1,...,Y_p\}[/math], meaning that for all [math]1\leq j\leq p[/math]
Note that this condition gives us explicitly the [math]\lambda_j's[/math]. We obtain therefore that [math](Y_1,...,Y_p,(X-\tilde X))[/math] is a Gaussian vector. Moreover, we get [math]\E[(X-\tilde X)Y_j]=Cov(X-\tilde X,Y_j)=0[/math] and thus [math]X-\tilde X[/math] is independent of [math]Y_1,...,Y_p[/math]. Hence
\end{proof}
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].
Notes
- This is done similarly to the proof of theorem