KindiakMath

The Poisson Distribution

Definition 1. The random variable X : \Omega \to \mathbb N_0 follows a Poisson distribution with rate parameter \lambda > 0, denoted X \sim \mathrm{Pois}(\lambda), if

\displaystyle \mathbb P(X=x) = c_\lambda \cdot \frac{\lambda^x}{x!}

for some c_\lambda > 0.

Problem 1. Evaluate c_\lambda.

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Solution. We require \sum_{x = 0}^\infty \mathbb P(X=x) = 1:

\begin{aligned}1 = \sum_{x = 0}^\infty \mathbb P(X=x) &= \sum_{x =0}^\infty c_\lambda \cdot \frac{\lambda^x}{x!} = c_\lambda \cdot \sum_{x = 0}^\infty \frac{\lambda^x}{x!} = c_\lambda \cdot e^{\lambda}. \end{aligned}

Therefore, c_\lambda = e^{-\lambda}.

Problem 2. Evaluate \mathbb E[X] and \mathrm{Var}(X).

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Solution. By definition of the expectation,

\begin{aligned} \mathbb E[X] &= \sum_{x = 0}^\infty x \cdot \mathbb P(X=x) \\ &= \sum_{x = 1}^\infty x \cdot c_\lambda \cdot \frac{\lambda^x}{x!} \\ &= \lambda \cdot \sum_{x = 1}^\infty c_\lambda \cdot \frac{\lambda^{x-1}}{(x-1)!} \\ &= \lambda \cdot \sum_{x = 0}^\infty c_\lambda \cdot \frac{\lambda^{x}}{x!} = \lambda \cdot 1 = \lambda. \end{aligned}

For the variance, we compute the term

\begin{aligned}\mathbb E[X(X-1)] &= \sum_{x = 0}^\infty x(x-1) \cdot \mathbb P(X=x) \\ &= \sum_{x = 2}^\infty x(x-1) \cdot c_\lambda \cdot \frac{\lambda^x}{x!} \\ &= \lambda^2 \cdot \sum_{x = 2}^\infty c_\lambda \cdot \frac{\lambda^{x-2}}{(x-2)!} \\ &= \lambda^2 \cdot 1 = \lambda^2. \end{aligned}

Therefore,

\begin{aligned} \mathrm{Var}(X) &= \mathbb E[X^2] - \mathbb E[X]^2 \\ &= \mathbb E[X(X-1)] + \mathbb E[X] - \mathbb E[X]^2 \\ &= \lambda^2 + \lambda - \lambda^2 = \lambda.\end{aligned}

Problem 3. Given that X \sim \mathrm{Pois}(\lambda) and Y \sim \mathrm{Pois}(\mu) are independent, determine the distribution of X + Y.

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Solution. Denoting W := X+Y, we take the discrete convolution of the p.d.f.s of X,Y to obtain

\begin{aligned}f_W (w) &= \sum_{x = 0}^w f_X(x) \cdot f_Y(w-x) \\ &= \sum_{x=0}^w c_\lambda \cdot \frac{\lambda^x}{x!} \cdot c_\mu \cdot \frac{\lambda^{w-x}}{(w-x)!} \\ &= \sum_{x=0}^w c_\lambda \cdot \frac{\lambda^x}{x!} \cdot c_\mu \cdot \frac{\mu^{w-x}}{(w-x)!} \\ &= c_\lambda \cdot c_\mu \cdot \sum_{x=0}^w \frac{\lambda^x}{x!} \cdot \frac{\mu^{w-x}}{(w-x)!} \\ &= c_\lambda \cdot c_\mu \cdot \frac 1{w!} \cdot \sum_{x=0}^w \frac{w!}{x! \cdot (w-x)!} \cdot \lambda^x \cdot \mu^{w-x} \\ &= c_\lambda \cdot c_\mu \cdot \frac 1{w!} \cdot \sum_{x=0}^w {w \choose x} \cdot \lambda^x \cdot \mu^{w-x} \\ &= c_\lambda \cdot c_\mu \cdot \frac {(\lambda + \mu)^w}{w!}.\end{aligned}

Furthermore, c_\lambda \cdot c_\mu = e^{-\lambda} \cdot e^{\mu} = e^{-(\lambda + \mu)} = c_{\lambda + \mu}. Hence,

\displaystyle \mathbb P(X+Y = w) = c_{\lambda + \mu} \cdot \frac{(\lambda + \mu)^w}{w!},

so that X +Y \sim \mathrm{Pois}(\lambda + \mu).

Problem 4. Fix \lambda > 0. Suppose X_n \sim \mathrm{Bin}(n, \lambda/n) and Y \sim \mathrm{Pois}(\lambda). Prove that f_{X_n} \to f_Y.

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Solution. Fix y \in \mathbb N_0. For each n,

\begin{aligned} f_{X_n}(y) &= {n \choose y} \left(1 - \frac{\lambda}{n}\right)^{n-y} \left( \frac{\lambda}n \right)^y \\ &= \frac{n(n-1) \cdot \dots \cdot (n-y+1)}{y!} \cdot \frac{\lambda^y }{n^y } \cdot \left(1 - \frac{\lambda}{n}\right)^{n-y} \\ &= \frac{n}{n} \cdot \frac{n-1}{n} \cdot \cdots \cdot \frac{n-y+1}n \cdot \left(1 - \frac{\lambda}{n}\right)^{-y} \cdot \left(1 - \frac{\lambda}{n}\right)^{n} \cdot \frac{\lambda^y }{y! } \end{aligned}

Taking n \to \infty,

\displaystyle \lim_{n \to \infty} f_{X_n}(y) = 1 \cdot 1 \cdot e^{-\lambda} \cdot \frac{\lambda^y}{y!} = e^{-\lambda} \cdot \frac{\lambda^y}{y!} = f_Y(y).

—Joel Kindiak, 1 Aug 25, 1751H

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