PropertiesOfPoissonRandomVariables 2009-02-25 19:22:42 2009-02-25 19:22:42 Derivation Poisson random variable \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} % define commands here \newcommand*{\abs}[1]{\left\lvert #1\right\rvert} \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} \begin{prop} If $X_1,X_2$ are independent Poisson random variables with parameters $\lambda_1,\lambda_2$, then $X_1+X_2$ is a Poisson random variable with parameter $\lambda_1+\lambda_2$. \end{prop} \begin{proof} Let $X:=X_1+X_2$ and $\lambda:=\lambda_1+\lambda_2$, let us calculate the distribution function of $X$: \begin{eqnarray*} F_X(x) &=& P(X\le x) = P(X_1+X_2\le x) = \sum_{i=0}^x P(X_1+X_2=i) \\ &=& \sum_{i=0}^x \sum_{j=0}^i P(X_1=j \mbox{ and } X_2=i-j) = \sum_{i=0}^x \sum_{j=0}^i P(X_1=j)P(X_2=i-j) \\ &=& \sum_{i=0}^x \sum_{j=0}^i \frac{e^{-\lambda_1} \lambda_1^j}{j!} \frac{e^{-\lambda_2} \lambda_2^{i-j}}{(i-j)!} = \sum_{i=0}^x \sum_{j=0}^i \frac{e^{-\lambda}}{i!} \binom{i}{j} \lambda_1^j \lambda_2^{i-j} \\ &=& \sum_{i=0}^x \frac{e^{-\lambda}}{i!} \sum_{j=0}^i \binom{i}{j} \lambda_1^j \lambda_2^{i-j} = \sum_{i=0}^x \frac{e^{-\lambda}}{i!} (\lambda_1+\lambda_2)^i = \sum_{i=0}^x \frac{e^{-\lambda}}{i!} \lambda^i. \end{eqnarray*} As a result, $X$ is a Poisson random variable with parameter $\lambda$. Notice that in the fifth equation, we used the assumption that $X_1$ and $X_2$ are independent. \end{proof} As a corollary, any sum of independent Poisson random variables is Poisson, with parameter the sum of the parameters from the independent random variables. \begin{prop} A Poisson random variable is infinitely divisible. \end{prop} \begin{proof} Let $X$ be a Poisson random variable with parameter $\lambda$. Let $n$ be any positive integer. Let $X_1,\ldots, X_n$ be independent identically distributed Poisson random variables with parameter $\frac{\lambda}{n}$. Then the sum of these random variables is easily seen to be Poisson, with parameter $\lambda$, and is therefore identically distributed as $X$. \end{proof}