# 统计代写|随机过程代写stochastic process代考|MXB334

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## 统计代写|随机过程代写stochastic process代考|Independence and Non-correlation

In general, non-correlation does not imply independence. However, this is nearly (see Example 3.2.8 below) true in the case of Gaussian vectors. We start with a definition in view of correctly stating the announced result.

Definition 3.2.6 Two random real vectors $X$ and $Y$ of respective dimensions $n$ and $q$ are said to be jointly Gaussian if the vector $Z$ defined by
$$Z^T=\left(X^T, Y^T\right)=\left(X_1, \ldots, X_n, Y_1, \ldots, Y_q\right)$$
is a Gaussian vector.

Theorem 3.2.7 Two jointly Gaussian random vectors $X$ and $Y$ of respective dimensions $n$ and $q$ are independent if and only if they are uncorrelated (that is, $\Gamma_{X Y}=0$ ).
Proof. Necessity: If $X$ and $Y$ are independent then, by the product formula for expectations,
$$\mathrm{E}\left[\left(X-m_X\right)\left(Y-m_Y\right)^T\right]=\mathrm{E}\left[X-m_X\right] \mathrm{E}\left[Y-m_Y\right]^T=0$$
Sufficiency: If $X$ and $Y$ are uncorrelated the vector $Z$ has for covariance matrix
$$\Gamma_Z=\left(\begin{array}{cc} \Gamma_X & 0 \ 0 & \Gamma_Y \end{array}\right)$$
and mean
$$m_Z=\left(\begin{array}{l} m_X \ m_Y \end{array}\right) \text {. }$$
It is a Gaussian vector by hypothesis and therefore, with $w=\left(u_1, \ldots, u_n, v_1, \ldots, v_q\right)^T$,
\begin{aligned} \mathrm{E}\left[\exp \left{i\left(u^T X+v^T Y\right)\right}\right] & =\mathrm{E}\left[\exp \left{i w^T Z\right}\right] \ & =\exp \left{i w^T m_Z-\frac{1}{2} w^T \Gamma_Z w\right} \ & =\exp \left{i\left(u^T m_X+v^T m_Y\right)-\frac{1}{2} u^T \Gamma_X u-\frac{1}{2} v^T \Gamma_Y v\right} \ & =\mathrm{E}\left[\exp \left{i u^T X\right}\right] \mathrm{E}\left[\exp \left{i v^T Y\right}\right], \end{aligned}
and the conclusion follows from the factorization theorem of characteristic functions (Theorem 3.1.52).

## 统计代写|随机过程代写stochastic process代考|The pdf of a Non-degenerate Gaussian Vector

If $\Gamma_X$ is degenerate, $X$ cannot have a probability density (see Theorem 3.1.34). However:
Theorem 3.2.9 Let $X$ be an n-dimensional Gaussian vector with mean $m$ and covariance matrix $\Gamma$, and assume that it is non-degenerate, that is, $\left(u^T \Gamma u=0\right)$ implies $u=0$. Then $X$ admits the probability density
$$f_X(x)=\frac{1}{(2 \pi)^{n / 2}(\operatorname{det} \Gamma)^{1 / 2}} \exp \left{-\frac{1}{2}(x-m)^T \Gamma^{-1}(x-m)\right} .$$
Proof. Since $\Gamma>0$ there exists a non-singular matrix $A$ of the same dimension as $\Gamma$ and such that $\Gamma=A A^T$. Define the $n$-vector $Z=A^{-1}(X-m)$. By Definition 3.2.3 it is a Gaussian vector, and furthermore $\mathrm{E}[Z]=0$ and
$$\Gamma_Z=A^{-1} \Gamma\left(A^T\right)^{-1}=A^{-1} A A^T\left(A^T\right)^{-1}=I .$$
Therefore
$$\sigma_Z(u)=\mathrm{E}\left[\exp \left{i u^T Z\right}\right]=\exp \left{-\frac{1}{2} \sum_{i=1}^n u_i^2\right} .$$
Since this is the characteristic function of a Gaussian vector of zero mean and having independent coordinates, we can assert that $Z_1, \ldots, Z_n$ are independent standard random variables. In particular, $Z$ admits the probability density
$$f_Z(z)=\prod_{i=1}^n \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\frac{1}{2} z_i^2 / 2}=\frac{1}{(2 \pi)^{n / 2}} \exp \left{-\frac{1}{2}|z|^2\right} .$$
Now, $X=A Z+m$, and therefore by the formula for the smooth change of variables
\begin{aligned} f_X(x) & =\frac{1}{|\operatorname{det} A|} f_Y\left(A^{-1}(x-m)\right) \ & =\frac{1}{(\operatorname{det} \Gamma)^{1 / 2}} \frac{1}{(2 \pi)^{n / 2}} \exp \left{-\frac{1}{2}\left|A^{-1}(x-m)\right|^2\right}, \end{aligned}
and this is precisely (3.37) since
\begin{aligned} \left|A^{-1}(x-m)\right|^2 & =\left(A^{-1}(x-m)\right)^T\left(A^{-1}(x-m)\right) \ & =(x-m)^T A^{-T} A^{-1}(x-m) \ & =(x-m)^T \Gamma^{-1}(x-m) . \end{aligned}

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Independence and Non-correlation

$$Z^T=\left(X^T, Y^T\right)=\left(X_1, \ldots, X_n, Y_1, \ldots, Y_q\right)$$

$$\mathrm{E}\left[\left(X-m_X\right)\left(Y-m_Y\right)^T\right]=\mathrm{E}\left[X-m_X\right] \mathrm{E}[Y$$

$$\Gamma_Z=\left(\begin{array}{llll} \Gamma_X & 0 & 0 & \Gamma_Y \end{array}\right)$$

$$m_Z=\left(m_X m_Y\right)$$

## 统计代写|随机过程代写stochastic process代考|The pdf of a Non-degenerate Gaussian Vector

3.1.34)。然而:

$f_{-} X(x)=$ Ifrac ${1}(2 \backslash$ |pi)^${n / 2}($ loperatorname{det $} \backslash G a m m a$

$\Gamma_Z=A^{-1} \Gamma\left(A^T\right)^{-1}=A^{-1} A A^T\left(A^T\right)^{-1}=I$

$$\left|A^{-1}(x-m)\right|^2=\left(A^{-1}(x-m)\right)^T\left(A^{-1}(x-m)\right)$$

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