Proposition

Proposition
設 $X_i, i = 1,2,\ldots ,n$ 為獨立隨機變數, 且都具有常態分布, 其參數分別為 $\mu _i, \sigma _i^2, i=1,2,\ldots ,n$ 則 $ \displaystyle\sum_{i=1}^n X_i$為參數是 $\displaystyle\sum_{i=1}^n \mu _i $ 和 $ \displaystyle\sum _{i=1}^n \sigma _i^2 $ 的 常態隨機變數.

Example (Sums of Independent Poisson Random Variables)
設 X 和 Y 為獨立卜瓦松隨機變數, 其參數分別為 $\lambda _1$ 和 $\lambda _2$, 試求 X+Y 的分布.
Solution:
因為事件 $\{ X+Y = n\} $ 可寫成互斥事件 $ \{ X=k, Y = n-k \} $ 的聯集, 其中 $ 0 \leq k \leq n $, 故得
$\begin{array}{rcl} P\{X+Y=n \} & = & \displaystyle\sum_{k=0}^n P\{X=k, Y=n-k\} ... ...{e^{- (\lambda _1+\lambda _2)}}{n!}(\lambda _1+\lambda _2)^n \\ \\ \end{array}$
也就是說, X+Y 為參數是 $\lambda _1+\lambda _2$ 的卜瓦松隨機變數.          $\rule[0.02em]{1.0mm}{1.5mm}$

Example (Sums of Independent Binomial Random Variables) Let X and Y be independent binomial random variables with respective parameters (n,p) and (m,p). Calculate the distribution of X+Y. [Solution:
Without any computation at all we can immediately conclude, by recalling the interpretation of a binomial random variable, that X+Y is binomial with parameters (n+m,p). This follows because X represents the number of successes in n independent trials, each of which results in a success with probability p; similarly, Y represents the number of successes in mindependent trials, each trial being a success with probability p. Hence, as X and Y are assumed independent, it follows that X+Yrepresents the number of successes in n+m independent trials when each trial has a probability p of being a success. Therefore, X+Y is a binomial random variable with parameters (n+m, p). To check this result analytically, note that
$\begin{array}{rcl} P\{X+Y=k\}&=&\displaystyle\sum_{i=0}^n P\{X=i,Y=k-i\} \\ \\ ... ...yle\sum_{i=0}^n{n\choose i}p^iq^{n-i}{m\choose k-i}p^{k-i}q^{m-k+i} \end{array}$
where q=1-p and where ${r\choose j}=0$ when j>r. Hence
$P\{X+Y=k\}=p^kq^{n+m-k}\displaystyle\sum_{i=0}^n{n\choose i}{m\choose k-i}$
and the result follows upon application of the combinatiorial identity
${n+m\choose k}=\displaystyle\sum_{i=0}^n{n\choose i}{m\choose k-i}\qquad\rule[0.02em]{1.0mm}{1.5mm}$