【Multivariate Data Analysis 02】Multivariate Normal and Related Distributions 多元正态及其相关分布

Chapter 2 Multivariate Normal and Related Distributions

General Assumption

the underlying multivariate distribution is multivariate normal

2.1 Multivariate Normal Distribution

Definition

A random vector $x$ is said to have a multivariate normal distribution if every linear combination of its components has an univariate normal distribution.

Suppose $a=(a_1{a}_2{)}^{\prime }$ and $x=(x_1{x}_2{)}^{\prime }$ . The multinormality of $x$ requires that $a^{\prime }x=a_1{x}_1+a_2{x}_2$ is univariate normal for all $a_1$ and $a_2$.

• If $a_1=0$, then $a^{\prime }x=x_2$
• If $a_1=a_2=0$, then $a^{\prime }x=0$, we don’t want this.

Properties

These ten properties are absolutely fundamental, I proved them all.

Property 1

If $x$ is multinormal, for any constant vector $a$, $a^{\prime }x\sim N(a^{\prime }\mu ,a^{\prime }\Sigma a)$

---

Proof
$$\begin{array}{rl}& \because E(a^{\prime }x)=a^{\prime }\mu andVar(a^{\prime }x)=a\Sigma a^{\prime }\\ & \because a^{\prime }x\sim N,根据multinormal的定义\\ & \therefore a^{\prime }x\sim N(a^{\prime }\mu ,a^{\prime }\Sigma a)\end{array}$$

where $a$ is of $p\times 1$, $\mu$ is of $p\times 1$, and $\Sigma$ is a square matrix of p dimensions.

---

Property 2

The moment generating function (m.g.f. 矩母函数 ) of a multinormal random vector $x$ with mean vector $\mu$ and covariance matrix $\Sigma$ is given by
$$M_x(t)=exp(t^{\prime }\mu +\frac{1}{2}{t}^{\prime }\Sigma t)$$
Thus, a multinormal distribution is **identified by its means $\mu$ and covariances $\Sigma$ **, We use the notation $x\sim N_p(\mu ,\Sigma )$

---

Proof

Follow the Hint 对式子进行一些拆解就好了
∵ y = t ′ x ∼ N ( t ′ μ , t ′ Σ t ) ∴ M x ( t ) = E ( e t ′ x ) = E ( e y ) = ∫ − ∞ ∞ e y p ( y ) d y = ∫ − ∞ ∞ e y 1 2 π ∣ t ′ Σ t ∣ 1 2 e x p { − ( y − t ′ μ ) 2 2 t ′ Σ t } d y = ∫ − ∞ ∞ 1 2 π ∣ t ′ Σ t ∣ 1 2 e x p { − y 2 − 2 y t ′ μ + ( t ′ μ ) 2 − y 2 t ′ Σ t 2 t ′ Σ t } d y = ∫ − ∞ ∞ 1 2 π ∣ t ′ Σ t ∣ 1 2 e x p { − y 2 − 2 y ( t ′ μ + t ′ Σ t ) 2 t ′ Σ t } d y = ∫ − ∞ ∞ 1 2 π ∣ t ′ Σ t ∣ 1 2 e x p { − ( y − ( t ′ μ + t ′ Σ t ) ) 2 − ( t ′ Σ t ) 2 − 2 t ′ μ t ′ Σ t 2 t ′ Σ t } d y = e x p { t ′ μ + 1 2 t ′ Σ t } ∫ − ∞ ∞ 1 2 π ∣ t ′ Σ t ∣ 1 2 e x p { − ( y − ( t ′ μ + t ′ Σ t ) ) 2 2 t ′ Σ t } d y = e x p ( t ′ μ + 1 2 t ′ Σ t ) \begin{aligned} \because y &= t'x\sim N(t'\mu, t'\Sigma t) \\ \therefore M_x(t) &= E(e^{t'x})=E(e^y) \\ &=\int_{-\infty}^{\infty}e^yp(y)dy \\ &=\int_{-\infty}^{\infty}e^y\frac{1}{\sqrt{2\pi}|t'\Sigma t|^{\frac{1}{2}}}exp\{-\frac{(y-t'\mu)^2}{2t'\Sigma t}\}dy \\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}|t'\Sigma t|^{\frac{1}{2}}}exp\{-\frac{y^2-2yt'\mu+(t'\mu)^2-y2t'\Sigma t}{2t'\Sigma t}\}dy \\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}|t'\Sigma t|^{\frac{1}{2}}}exp\{-\frac{y^2-2y(t'\mu+t'\Sigma t)}{2t'\Sigma t}\}dy \\ &=\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}|t'\Sigma t|^{\frac{1}{2}}}exp\{-\frac{(y-(t'\mu+t'\Sigma t))^2-(t'\Sigma t)^2-2t'\mu t'\Sigma t}{2t'\Sigma t}\}dy \\ &=exp\{t'\mu+\frac{1}{2}t'\Sigma t\}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}|t'\Sigma t|^{\frac{1}{2}}}exp\{-\frac{(y-(t'\mu+t'\Sigma t))^2}{2t'\Sigma t}\}dy \\ &= exp(t'\mu+\frac{1}{2}t'\Sigma t) \\\\ \end{aligned} ∵y∴Mx​(t)​=t′x∼N(t′μ,t′Σt)=E(et′x)=E(ey)=∫−∞∞​eyp(y)dy=∫−∞∞​ey2π ​∣t′Σt∣21​1​exp{ −2t′Σt(y−t′μ)2​}dy=∫−∞∞​2π ​∣t′Σt∣21​1​exp{ −2t′Σty2−2yt′μ+(t′μ)2−y2t′Σt​}dy=∫−∞∞​2π ​∣t′Σt∣21​1​exp{ −2t′Σty2−2y(t′μ+t′Σt)​}dy=∫−∞∞​2π ​∣t′Σt∣21​1​exp{ −2t′Σt(y−(t′μ<