问题

Definition (随机变量的线性组合)

Given a collection of $n$ random variables $X_1,\ldots,X_n$ and $n$ numerical constants $a_1,\ldots,a_n$, the rv
\[
Y=a_1 X_1+\cdots+a_n X_n=\sum_{i=1}^{n}a_i X_i
\]
is called a linear combination of the $X_i$'s.
 

Prove the following proposition.

Prop 1. Let $X_1,X_2,\ldots,X_n$ have mean values $\mu_1,\ldots,\mu_n$, respectively, and variances of $\sigma_1^2,\ldots,\sigma_n^2$, respectively.

(1) Whether or not the $X_i$'s are independent,
  \[
  \begin{split}
  E(a_1 X_1+a_2 X_2+\cdots+a_n X_n)&=a_1 E(X_1)+a_2 E(X_2)+\cdots+a_n E(X_n)\\
  &=a_1\mu_1+a_2\mu_2+\cdots+a_n\mu_n
  \end{split}
  \]

(2) If $X_1,\ldots,X_n$ are independent,
  \[
  \begin{split}
  V(a_1 X_1+a_2 X_2+\cdots+a_n X_n)&=a_1^2 V(X_1)+a_2^2 V(X_2)+\cdots+a_n^2 V(X_n)\\
  &=a_1^2\sigma_1^2+a_2^2\sigma_2^2+\cdots+a_n^2\sigma_n^2
  \end{split}
  \]
  and
  \[
  \sigma_{a_1 X_1+\cdots+a_n X_n}=\sqrt{a_1^2\sigma_1^2+\cdots+a_n^2\sigma_n^2}
  \]

(3) For any $X_1,\ldots,X_n$,
  \[
  V(a_1 X_1+\cdots+a_n X_n)=\sum_{i=1}^{n}\sum_{j=1}^{n}a_i a_j\mathrm{Cov}(X_i,X_j)
  \]

 

Then, we get the following proposition.

Prop 2. Let $X_1,X_2,\ldots,X_n$ be a random sample from a distribution with mean value $\mu$ and standard deviation $\sigma$. Then

• $E(\bar{X})=\mu_{\bar{X}}=\mu$.
• $V(\bar{X})=\sigma_{\bar{X}}^2=\frac{\sigma^2}{n}$ and $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$.

Here $\bar{X}=\frac{1}{n}(X_1+X_2+\cdots+X_n)$.

In addition, with $T_o=X_1+\cdots+X_n$ (the sample total), $E(T_o)=n\mu$, $V(T_o)=n\sigma^2$, and $\sigma_{T_o}=\sqrt{n}\sigma$.

注: 有时也记 $D(X)=V(X)$.

样本方差 $s^2$ 定义为

\[
s^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2
\]

其开方 $s$ 称为样本均方差或样本标准差.

\[
s=\sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2}
\]

样本 $k$ 阶原点矩 $m_k=\frac{1}{n}\sum_{i=1}^{n}X_i^k$, $k=1,2,\ldots$

样本 $k$ 阶中心矩 $M_k=\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^k$, $k=1,2,\ldots$

Prop. $E(s^2)=D(X)=\sigma^2$.