> [!definition] > > Let $\alpha > 0$, then the standard Poisson [[Probability Distribution|distribution]] with rate $\alpha$ is > $ > \pi_\alpha(\bracs{k}) = e^{-\alpha}\frac{\alpha^k}{k!} \quad \pi_\alpha = e^{-\alpha}\sum_{k = 0}^\infty \frac{\alpha^k}{k!}\delta_k = e^{-\alpha}\sum_{k = 0}^\infty \frac{\alpha^k}{k!}\delta_{1}^{*k} > $ > If $\nu$ is another [[Probability|probability]] measure, then the compound Poisson distribution with rate $\alpha$ and intensity $\nu$ is > $ > \pi_\alpha = e^{-\alpha}\sum_{k = 0}^\infty \frac{\alpha^k}{k!}\nu^{*k} > $ > [!theorem] > > Let $(\Omega, \cf, \bp)$ be a probability space, $N$ be a Poisson random variable with rate $\alpha$, and $\seq{X_n}$ be a family of [[Probabilistic Independence|independent]], identically distributed random variables with distribution $\nu$. Let $S_n = \sum_{k \le n}X_k$ be the $n$-th partial sum, then the random variable $S_N$ has distribution equal to the compound Poisson distribution with rate $\alpha$ and intensity $\nu$. > [!theorem] > > Let $\pi_{\alpha, \nu}$ be the Poisson distribution with rate $\alpha$ and intensity $\nu$, then > $ > \wh \pi_{\alpha, \nu} = e^{\alpha \wh \nu(\xi) - \alpha} = e^{\alpha(\wh \nu - 1)} > $ > therefore $\pi_{\alpha, \nu} \in \mathcal I(\real^d)$ is [[Infinitely Divisible Distribution|infinitely divisible]]. Importantly, > $ > e^{\alpha \wh \nu - \alpha}(\xi) = \exp\paren{\alpha \int e^{\braks{i\angles{x, \xi}}} - 1 d\nu(x)} > $ > the mass $\nu$ assigns at $0$ is irrelevant. Let $M = \alpha \nu$, then we can assume without loss of generality that $M(\bracs{0}) = 0$, and $\pi_{\alpha, \nu} = \pi_{M}$. > > *Proof*. > $ > \wh \pi_{\alpha, \nu} = \sum_{k = 0}^\infty e^{-\alpha}\frac{\alpha^k \cdot \wh \nu^k}{k!} = e^{\alpha \wh \nu(\xi) - \alpha} > $ > [!theorem] Derivation > > $ > P(p, n) = {n \choose k}p^{k}(1-p)^{n -k} > $ > Divide a time interval evenly into $n$ small time intervals, and let $p$ be the [[Probability|probability]] that the event happens in one of those intervals. The number of successes $k$ as a [[Random Variable|random variable]] over the total interval follows a [[Binomial Random Variable|binomial distribution]]. Let a parameter $\lambda = np$, which means that $p = \frac{\lambda}{n}$. > > $ > \begin{align*} > P(p, n) &= {n \choose k}\frac{\lambda}{n}^{k}\paren{1-\frac{\lambda}{n}}^{n -k} \\ > &= \frac{n!}{k!(n-k)!}\frac{\lambda}{n}^{k}\paren{1-\frac{\lambda}{n}}^{n -k} \\ > &= \frac{\lambda^k}{k!}\frac{n!}{(n-k)!}\paren{\frac{1}{n}}^{k}\paren{1-\frac{\lambda}{n}}^{n -k} > \end{align*} > $ > > Take a [[Limit|limit]] such that the number of subintervals approach infinity over smaller and smaller intervals: > $ > \begin{align*} > Po(\lambda) &= \frac{\lambda^k}{k!}\limv{n}\frac{n!}{(n-k)!}\paren{\frac{1}{n}}^{k}\paren{1-\frac{\lambda}{n}}^{n -k}\\ > &= \frac{\lambda^k}{k!}\limv{n} > \frac{\prod_{i=n-k+1}^{n}i}{n^k}\paren{1-\frac{\lambda}{n}}^{n -k}\\ > &= \frac{\lambda^k}{k!}\limv{n} > \prod_{i=n-k+1}^{n}\frac{i}{n^k} > \cdot\limv{n}\paren{1-\frac{\lambda}{n}}^{n -k}\\ > &= \frac{\lambda^k}{k!}\limv{n}\paren{1-\frac{\lambda}{n}}^{n -k}\\ > &= \frac{\lambda^k}{k!}\lim_{-n/\lambda \to \infty}\paren{1 + \paren{\frac{1}{-\frac{\lambda}{n}}}}^{-\lambda/n\cdot-\lambda} \cdot > \limv{n}\paren{1-\frac{\lambda}{n}}^{-k}\\ > &= \frac{\lambda^k}{k!}e^{-\lambda} \cdot > \limv{n}\paren{1-\frac{\lambda}{n}}^{-k}\\ > &= \frac{\lambda^k}{k!}e^{-\lambda} > \end{align*} > $ > [!theorem] > > The parameter $\lambda$ is the [[Expectation|expected value]] and the [[Variance|variance]] of the distribution, and is proportional to the length of the interval. > $ > \begin{align*} > \ev(Po(\lambda)) &= \var{Po(\lambda)} = \lambda \\ > \text{Expected Occurances in Interval} &= \text{Frequency} \times \text{Interval Length} \\ > \lambda &\propto \text{Interval Length} > \end{align*} > $ > > *Proof*. Since the Poisson random variable is a limit of the binomial variable, which is a [[Random Sample|random sample]] of [[Bernoulli Random Variable|Bernoulil]] random variables. Taking $\lambda = \frac{p}{n}$: > $ > Po(\lambda) = S_\lambda(\infty) = \sum_{i = 1}^{\infty}X_i \quad > \text{where}\ X \sim B(1, \frac{\lambda}{\infty}) > $ > Evaluating this as a limit for [[Expectation|expected value]] and [[Variance|variance]], > $ > \begin{align*} > \ev{(Po(\lambda))} > &= \ev\paren{\sum_{i = 1}^{\infty}B\paren{1, \frac{\lambda}{\infty}}} \\ > &= \ev\paren{\limv{n}\sum_{i = 1}^{n}B\paren{1, \frac{\lambda}{n}}} \\ > &= \limv{n}\sum_{i = 1}^{n}\ev\paren{B\paren{1, \frac{\lambda}{n}}} \\ > &= \limv{n}\sum_{i = 1}^{n}\frac{\lambda}{n}\\ > &= \limv{n}\frac{n\lambda}{n}\\ > &= \lambda > \end{align*} > $ > $ > \begin{align*} > \var{Po(\lambda)} &= > \var{\sum_{i = 1}^{\infty}B\paren{1, \frac{\lambda}{\infty}}} \\ > &= \var{\limv{n}\sum_{i = 1}^{n}B\paren{1, \frac{\lambda}{n}}}\\ > &= \limv{n}\sum_{i = 1}^{n}\var{B\paren{1, \frac{\lambda}{n}}}\\ > &= \limv{n}\sum_{i = 1}^{n}\frac{\lambda}{n}\paren{1 - \frac{\lambda}{n}}\\ > &= \limv{n}\frac{n\lambda}{n}\paren{1 - \frac{\lambda}{n}}\\ > &= \limv{n}\lambda\paren{1 - \frac{\lambda}{n}} \\ > &= \lambda > \end{align*} > $ > [!theorem] Cumulative Distribution > > $ > P(Po(\lambda) \le x) = e^{-\lambda}\paren{1 + \lambda + \frac{\lambda^2}{2!} + \cdots + \frac{\lambda^n}{x}} > $ > *Proof*. Since the range of a Poisson random variable starts at 0, start the sum there. > > $ > \begin{align*} > P(Po(\lambda) \le x) &= \sum_{j = 0}^{x}\frac{\lambda^j}{j!}e^{-\lambda} \\ > &= e^{-\lambda}\sum_{j = 0}^{x}\frac{\lambda^j}{j!} \\ > &= e^{-\lambda}\paren{1 + \lambda + \frac{\lambda^2}{2!} + \cdots + \frac{\lambda^x}{x}} > \end{align*} > $ > > Notably, if $x \to \infty$, then the sum becomes the [[Power Series|power series]] of the [[Exponential Function|exponential function]]: > $ > \begin{align*} > \lim_{x \to \infty}P(Po(\lambda) \le x) &= e^{-\lambda}\lim_{x \to \infty}\sum_{j = 0}^{x}\frac{\lambda^j}{j!} \\ > &= e^{-\lambda}e^{\lambda} = 1 > \end{align*} > $ > Which lands a nice proof that the Poisson distribution is a [[Probability Distribution|probability distribution]]. > [!theorem] [[Moment Generating Function]] > > $ > M_X(t) = \exp(\lambda(e^{t} - 1)) > $ > > *Proof*. Using the [[Power Series|power series]] of the [[Exponential Function|exponential function]]: > $ > \begin{align*} > M_X(t) &= \ev(e^{tX}) \\ > &= \sum_{j = 0}^{\infty} e^{tx_j}p_j \\ > &= \sum_{j = 0}^{\infty} e^{tj}\frac{\lambda^j}{j!}e^{-\lambda} \\ > &= e^{-\lambda}\sum_{j = 0}^{\infty} e^{tj}\frac{\lambda^j}{j!} \\ > &= e^{-\lambda}\sum_{j = 0}^{\infty}\frac{(e^{t})^j\lambda^j}{j!} \\ > &= e^{-\lambda}\sum_{j = 0}^{\infty}\frac{(e^{t}\lambda)^j}{j!} \\ > &= e^{-\lambda}e^{e^{t}\lambda} \\ > &= \exp(e^{t}\lambda - \lambda) \\ > &= \exp(\lambda(e^t - 1)) > \end{align*} > $ > > The first moment of the Poisson distribution can be used to show that its expected value and variance is $\lambda$. > > *Proof*. Using the [[Math/Calculus/Derivative/Chain Rule|chain rule]] and [[Product Rule|product rule]]. > > $ > \begin{align*} > \ev(X) &= \frac{d}{dt}M_X(t)|_{t = 0} \\ > &= \frac{d}{dt}\exp(\lambda(e^{t} - 1))|_{t = 0} \\ > &= \lambda e^{t} \exp(\lambda(e^{t} - 1))|_{t = 0} \\ > &= \lambda \\ > \end{align*} > $ > > $ > \begin{align*} > \var{X} &= \ev(X^2) - \ev(X)^2 \\ > &= \frac{d^2}{dt^2}M_X(t)|_{t = 0} - \ev(X)^2 \\ > &= \frac{d}{dt}\lambda e^{t} \exp(\lambda(e^t - 1)) - \ev(X)^2 \\ > &= \frac{d}{dt}(\lambda e^{t}) \exp(\lambda(e^t - 1)) + \lambda e^{t} \frac{d}{dt}(\exp(\lambda(e^t - 1))) - \ev(X)^2 \\ > &= \lambda e^{t} \exp(\lambda(e^t - 1)) + \lambda^2 e^{2t}\exp(\lambda(e^t - 1)) - \ev(X)^2 \\ > &= \lambda + \lambda^2 - \lambda^2 \quad\\ > &= \lambda\\ > \end{align*} > $ > [!theorem] > > If $X \sim Po(\lambda)$ and $Y \sim Po(\mu)$ are [[Probabilistic Independence|independent]] random variables, then $X + Y \sim Po(\lambda + \mu)$. > > *Proof*. Add the two variables using a [[Convolution of Functions|convolution]], and simplifying with the [[Binomial Theorem|binomial theorem]]: > $ > \begin{align*} > X + Y &\sim \paren{Po(\lambda) * Po(\mu)} \\ > &= \sum_{m = -\infty}^{\infty}Po(\lambda)(m) \cdot Po(\mu)(x - m) \\ > &= \sum_{m = 0}^{x}Po(\lambda)(m)\cdot Po(\mu)(x - m) \\ > &= \sum_{m = 0}^{x} e^{-\lambda}\frac{\lambda^{m}}{m!} \cdot e^{-\mu} \frac{\mu^{x - m}}{(x - m)!} \\ > &= e^{-(\lambda + \mu)} \sum_{m = 0}^{x} \frac{\lambda^{m}}{m!} \frac{\mu^{x - m}}{(x - m)!} \\ > &= e^{-(\lambda + \mu)} \sum_{m = 0}^{x} \frac{1}{m!(x - m)!}\lambda^{m}\mu^{x - m} \\ > &= e^{-(\lambda + \mu)} \frac{1}{x!} \sum_{m = 0}^{x} \frac{x!}{m!(x - m)!}\lambda^{m}\mu^{x - m} \\ > &= e^{-(\lambda + \mu)} \frac{(\lambda + \mu)^{x} }{x!} \\ > X + Y &\sim Po(\lambda + \mu) > \end{align*} > $ > [!theorem] > > Let $X \sim Po(\lambda)$ and $Y \sim Po(\mu)$ be independent Poisson random variables. Let $\tilde{X}_n$ be the random variable with the following distribution, representing the number of events coming from $X$, given that there are $n$ events in total. > $ > P(\tilde{X}_n = k) = P_{(X + Y = n)}(X = k) \quad \text{for}\ 1 \le k \le n > $ > $\tilde{X}_n$ follows the [[Binomial Random Variable|binomial distribution]] $\tilde{X}_n \sim B\paren{n, \frac{\lambda}{\lambda + \mu}}$. > > The expected value of $\tilde{X}_n$, or the [[Conditional Expectation|conditional expectation]] of $X$ given $X + Y = n$ is $\frac{n\lambda}{\lambda + \mu}$. > > *Proof*. Using the previous theorem, and a theorem from conditional expectations. > $ > \begin{align*} > \tilde{X}_n &\sim P_{(X + Y = n)}(X = k) \\ > &= \frac{P(X = k)P(Y = n - k)}{P(X + Y = n)} \\ > &= \frac{Po(\lambda)(k) \cdot Po(\mu)(n - k)}{Po(\lambda + \mu)(n)} \\ > &= \frac{\frac{\lambda^k}{k!}e^{-\lambda} \cdot \frac{\mu^{n - k}}{(n - k)!}e^{-\mu}}{\frac{(\lambda + \mu)^k}{n!}e^{-(\lambda + k)}} \\ > &= \frac{n!}{k!(n - k!)}\frac{\lambda^k \mu^{n - k}}{(\lambda + \mu)^k} > \frac{e^{-\lambda}e^{-\mu}}{e^{-\lambda - \mu}}\\ > &= {n \choose k}\paren{\frac{\lambda}{\lambda + \mu}}^k\paren{\frac{\mu}{\lambda + \mu}}^{n - k}\\ > &= {n \choose k}\paren{\frac{\lambda}{\lambda + \mu}}^k\paren{1 - \frac{\lambda}{\lambda + \mu}}^{n - k} \\ > \tilde{X}_n &\sim B\paren{n, \frac{\lambda}{\lambda + \mu}} > \end{align*} > $ > $ > \ev(\tilde{X}_n) = \ev\paren{B\paren{n, \frac{\lambda}{\lambda + \mu}}} = \frac{n\lambda}{\lambda + \mu} > $