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Exercise 2.29 (Gaussian distribution)

A random vector $(X_{1}, \dots, X_{n})$ taking values in $R^{n}$ is said to be a gaussian random vector if there exists $μ = (μ_{1}, \dots, μ_{n}) \in R^{n}$ and an $n \times n$ positive definite real symmetric matrix $Σ : = {(σ_{ij})}_{1 \leq i, j \leq n}$ such that

P ((X_{1}, \dots, X_{n}) \in S) = \frac{1}{{(2 π)}^{n ∕ 2} {(\det Σ)}^{1 ∕ 2}} \int_{S} e^{- \frac{1}{2} {(x - μ)}^{T} Σ^{- 1} (x - μ)} dx

for all Borel sets $S \subset R^{n}$ (where we identify elements of $R^{n}$ with column vectors). The distribution of $(X_{1}, \dots, X_{n})$ is called a multivariate normal distribution.

1.: If $(X_{1}, \dots, X_{n})$ is a gaussian random vector with the indicated parameters $μ, Σ$ , show that $E X_{i} = μ_{i}$ and $Cov (X_{i}, X_{j}) = σ_{ij}$ for $1 \leq i, j \leq n$ . In particular $Var (X_{i}) = σ_{ii}$ . Thus we see that the parameters of a gaussian random variable can be recovered from the mean and covariances.
2.: If $(X_{1}, \dots, X_{n})$ is a gaussian random vector and $1 \leq i, j \leq n$ , show that $X_{i}$ and $X_{j}$ are independent if and only if the covariance $Cov (X_{i}, X_{j})$ vanishes. Furthermore, show that $(X_{1}, \dots, X_{n})$ are jointly independent if and only if all the covariances $Cov (X_{i}, X_{j})$ for $1 \leq i < j \leq n$ vanish. In particular, for gaussian random vectors, joint independence is equivalent to pairwise independence. (Contrast this with Exercise 23.)
3.: Give an example of two real random variables $X, Y$ , each of which is gaussian, and for which $Cov (X, Y) = 0$ , but such that $X$ and $Y$ are not independent. Why does this not contradict (ii)?

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A random vector $(X_1,\dots,X_n)$ taking values in ${\bf R}^n$ is said to be a gaussian random vector if there exists $\mu = (\mu_1,\dots,\mu_n) \in {\bf R}^n$ and an $n \times n$ positive definite real symmetric matrix $\Sigma := (\sigma_{ij})_{1 \leq i,j \leq n}$ such that

$$\displaystyle \mathop{\bf P}( (X_1,\dots,X_n) \in S) = \frac{1}{(2\pi)^{n/2} (\det \Sigma)^{1/2}} \int_S e^{-\frac{1}{2} (x-\mu)^T \Sigma^{-1} (x-\mu)}\ dx$$

for all Borel sets $S \subset {\bf R}^n$ (where we identify elements of ${\bf R}^n$ with column vectors). The distribution of $(X_1,\dots,X_n)$ is called a multivariate normal distribution.

\begin{enumerate}
\item If $(X_1,\dots,X_n)$ is a gaussian random vector with the indicated parameters $\mu, \Sigma$, show that ${\bf E} X_i = \mu_i$ and $\hbox{Cov}(X_i,X_j) = \sigma_{ij}$ for $1 \leq i,j \leq n$. In particular $\hbox{Var}(X_i) = \sigma_{ii}$. Thus we see that the parameters of a gaussian random variable can be recovered from the mean and covariances.
\item If $(X_1,\dots,X_n)$ is a gaussian random vector and $1 \leq i, j \leq n$, show that $X_i$ and $X_j$ are independent if and only if the covariance $\hbox{Cov}(X_i,X_j)$ vanishes. Furthermore, show that $(X_1,\dots,X_n)$ are jointly independent if and only if all the covariances $\hbox{Cov}(X_i,X_j)$ for $1 \leq i < j \leq n$ vanish. In particular, for gaussian random vectors, joint independence is equivalent to pairwise independence. (Contrast this with Exercise 23.)
\item Give an example of two real random variables $X,Y$, each of which is gaussian, and for which $\hbox{Cov}(X,Y)=0$, but such that $X$ and $Y$ are not independent. Why does this not contradict (ii)?
\end{enumerate}

Exercise 2.29 (Gaussian distribution)

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