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Proof of Independent Gaussian Random Variables are Jointly Gaussian

lemmalem:independent-gaussians-jointly-gaussian-2026a
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Reason: Proof of lem:independent-gaussians-jointly-gaussian-2026a via scalar standardizations and an assembled joint Gaussian representation. Approved by Aaron.

Proof

Throughout, expectations, variances, and covariances of the Gaussian random variables involved are defined and finite by Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector. Write ΞΌi=E[Xi]\mu_i=\mathbb{E}[X_i] and Οƒi2=Var⁑(Xi)\sigma_i^2=\operatorname{Var}(X_i) for 1≀i≀p1\le i\le p, with the variance of square-integrable random variables, and let

J={ i∈{1,…,p}Β :Β Οƒi2>0 },J=\{\,i\in\{1,\dots,p\}\ :\ \sigma_i^2>0\,\},

let mm be the number of elements of JJ, and enumerate J={i1,…,im}J=\{i_1,\dots,i_m\} with i1<β‹―<imi_1<\dots<i_m; the case m=0m=0 (empty JJ) is allowed. For iβˆ‰Ji\notin J we have Οƒi2=0\sigma_i^2=0: by the moment formulas (claim 2) of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to any Gaussian representation of the one-term tuple (Xi)(X_i), the variance is a sum of squares, hence nonnegative.

Step 1: Standardized variables. Suppose mβ‰₯1m\ge1 and fix l∈{1,…,m}l\in\{1,\dots,m\}. Let Οƒil\sigma_{i_l} be the positive square root of Οƒil2\sigma_{i_l}^2 and define

Zl=Xilβˆ’ΞΌilΟƒil.Z_l=\frac{X_{i_l}-\mu_{i_l}}{\sigma_{i_l}}.

By the standardization claim (claim 2) of Standardization and Cumulative Distribution Function of a Gaussian Random Variable, ZlZ_l is a standard normal random variable.

Step 2: Z1,…,ZmZ_1,\dots,Z_m are independent. Fix l∈{1,…,m}l\in\{1,\dots,m\} and define gl:Rβ†’Rg_l:\mathbb{R}\to\mathbb{R} by gl(x)=(xβˆ’ΞΌil)/Οƒilg_l(x)=(x-\mu_{i_l})/\sigma_{i_l}, so that Zl=gl(Xil)Z_l=g_l(X_{i_l}) pointwise on Ξ©\Omega. For every real aa, since Οƒil>0\sigma_{i_l}>0,

{x∈R:gl(x)>a}=(ΞΌil+Οƒila, ∞),\{x\in\mathbb{R}: g_l(x)>a\}=(\mu_{i_l}+\sigma_{i_l}a,\ \infty),

an open interval and hence a Borel set; by the generator criterion of Measurable Function and Real-Valued Measurable Function, glg_l is measurable from R\mathbb{R} with the Borel Οƒ\sigma-algebra to itself. Hence for every Borel set BB the set glβˆ’1(B)g_l^{-1}(B) is Borel, and

{Zl∈B}={Xil∈glβˆ’1(B)}.\{Z_l\in B\}=\{X_{i_l}\in g_l^{-1}(B)\}.

Now fix Borel sets B1,…,BmB_1,\dots,B_m and put Cl=glβˆ’1(Bl)C_l=g_l^{-1}(B_l) for 1≀l≀m1\le l\le m. Since X1,…,XpX_1,\dots,X_p are independent, the events {X1∈E1},…,{Xp∈Ep}\{X_1\in E_1\},\dots,\{X_p\in E_p\} are independent for the Borel choice Eil=ClE_{i_l}=C_l (1≀l≀m1\le l\le m) and Ei=RE_i=\mathbb{R} for iβˆ‰Ji\notin J. Independence of events requires the product identity for every nonempty subfamily, in particular for every nonempty subfamily of the events {Xil∈Cl}={Zl∈Bl}\{X_{i_l}\in C_l\}=\{Z_l\in B_l\} (1≀l≀m1\le l\le m). Hence for every nonempty SβŠ†{1,…,m}S\subseteq\{1,\dots,m\},

P(β‹‚l∈S{Zl∈Bl})=∏l∈SP(Zl∈Bl),P\Bigl(\bigcap_{l\in S}\{Z_l\in B_l\}\Bigr)=\prod_{l\in S}P(Z_l\in B_l),

and, the Borel sets B1,…,BmB_1,\dots,B_m being arbitrary, the random variables Z1,…,ZmZ_1,\dots,Z_m are independent.

Step 3: A Gaussian representation. Define real numbers aija_{ij} for 1≀i≀p1\le i\le p, 1≀j≀m1\le j\le m by: aij=Οƒila_{ij}=\sigma_{i_l} if i=ili=i_l and j=lj=l for some l∈{1,…,m}l\in\{1,\dots,m\}, and aij=0a_{ij}=0 otherwise.

If i=il∈Ji=i_l\in J, then pointwise on Ω\Omega,

ΞΌi+βˆ‘j=1maijZj=ΞΌil+ΟƒilZl=Xil,\mu_i+\sum_{j=1}^{m}a_{ij}Z_j=\mu_{i_l}+\sigma_{i_l}Z_l=X_{i_l},

directly from the definition of ZlZ_l in Step 1; in particular the equality holds with probability one. If iβˆ‰Ji\notin J, then Οƒi2=0\sigma_i^2=0, so P(Xi=ΞΌi)=1P(X_i=\mu_i)=1 by the degenerate case (claim 1) of Standardization and Cumulative Distribution Function of a Gaussian Random Variable, while the ii-th row of (aij)(a_{ij}) is zero, so that

ΞΌi+βˆ‘j=1maijZj=ΞΌi.\mu_i+\sum_{j=1}^{m}a_{ij}Z_j=\mu_i .

Hence P(Xi=ΞΌi+βˆ‘j=1maijZj)=1P\bigl(X_i=\mu_i+\sum_{j=1}^{m}a_{ij}Z_j\bigr)=1 for every i∈{1,…,p}i\in\{1,\dots,p\}, and by Steps 1 and 2 the random variables Z1,…,ZmZ_1,\dots,Z_m are independent standard normal. Therefore (m,(ΞΌi),(aij),(Zj))\bigl(m,(\mu_i),(a_{ij}),(Z_j)\bigr) is a Gaussian representation of (X1,…,Xp)(X_1,\dots,X_p); the case m=0m=0, in which every XiX_i is almost surely constant, is expressly allowed by Gaussian Random Vectors and Jointly Gaussian Random Variables. Thus (X1,…,Xp)(X_1,\dots,X_p) is a Gaussian random vector.

Step 4: Distinct components are uncorrelated. By the moment formulas (claim 2) of Square-Integrability, Moments, and Covariance Matrix of a Gaussian Random Vector applied to the representation of Step 3, for 1≀i<k≀p1\le i<k\le p,

Cov⁑(Xi,Xk)=βˆ‘j=1maij akj.\operatorname{Cov}(X_i,X_k)=\sum_{j=1}^{m}a_{ij}\,a_{kj}.

Fix j=l∈{1,…,m}j=l\in\{1,\dots,m\}. By construction ailβ‰ 0a_{il}\ne0 only if i=ili=i_l, and aklβ‰ 0a_{kl}\ne0 only if k=ilk=i_l; since iβ‰ ki\ne k, at least one of the two factors vanishes, so every summand is 00. Hence Cov⁑(Xi,Xk)=0\operatorname{Cov}(X_i,X_k)=0. β– \blacksquare

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