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  • STA256 Lecture 19
  • Poisson Distribution

Poisson Distribution

• Not a fish

• Used for counts of data for $x=0,1,2,3,\dots$

• Involves a rate parameter $\lambda$ for the average # of objects per length, time, area, volume, etc…

• In the form rate: x per y

  • $\beta =\frac{1}{\lambda }$ - Scale
  • $\lambda =\frac{1}{\beta }$ - Rate

• Assume that occurances of objects in each interval are independent.

• Probability Mass Function:

  • $P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!}\text{ for }x=0,1,2,3,\dots$
  • This is obtained by taking the $\underset{n\to \mathrm{\infty }}{lim}$ of a Binomial Distribution

• Properties:

  • Mean: $\mu =E[X]=\lambda$
  • Variance: ${\sigma }^2=var(x)=\lambda$
  • MGF: $M_X(t)=e^{\lambda (e^t-1)}$
  • #tk practice deriving the MGF

• Derivation Proof:

  • Recall:
    • $e^x=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n$
  • The Binomial Distribution is $P(X=x)=(\genfrac{}{}{0ex}{}{n}{x})p^x(1-p{)}^{n-x}\text{ for }x=0,1,\dots ,n$
  • The mean of the Binomial Distribution is $\mu =np$
  • Now let $E(X)=\lambda$ so $\lambda =np$
  • Which means that $p=\frac{\lambda }{n}$
  • Sub in:
    • $P(X=x)=(\genfrac{}{}{0ex}{}{n}{x}){\left(\frac{\lambda }{n}\right)}^x{(1-\frac{\lambda }{n})}^{n-x}\text{ for }x=0,1,\dots ,n$
    • $P(X=x)=(\genfrac{}{}{0ex}{}{n}{x}){\left(\frac{\lambda }{n}\right)}^x{(\frac{n}{n}-\frac{\lambda }{n})}^{n-x}$
    • $P(X=x)=(\genfrac{}{}{0ex}{}{n}{x}){\left(\frac{\lambda }{n}\right)}^x{\left(\frac{n-\lambda }{n}\right)}^{n-x}$
  • Take the limit of both sides

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\underset{n\to \mathrm{\infty }}{lim}\left[(\genfrac{}{}{0ex}{}{n}{x}){\left(\frac{\lambda }{n}\right)}^x{\left(\frac{n-\lambda }{n}\right)}^{n-x}\right]$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\underset{n\to \mathrm{\infty }}{lim}\left[\left(\frac{n!}{x!(n-x)!}\right)\left(\frac{{\lambda }^x}{n^x}\right)\left(\frac{(n-\lambda {)}^{n-x}}{(n{)}^{n-x}}\right)\right]$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\underset{n\to \mathrm{\infty }}{lim}\left[\left(\frac{n!}{x!(n-x)!}\right)\left(\frac{{\lambda }^x}{n^x}\right){(1-\frac{\lambda }{n})}^{n-x}\right]$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\underset{n\to \mathrm{\infty }}{lim}\left[\left(\frac{n!}{n^x(n-x)!}\right)\left(\frac{{\lambda }^x}{x!}\right){(1-\frac{\lambda }{n})}^{n-x}\right]$

      • $\frac{n!}{n^x(n-x)!}=\frac{n(n-1)(n-2)\dots (n-x+1)(n-x)(n-x-1)\dots (2)(1)}{n^x(n-x)!}$
      • $\frac{n!}{n^x(n-x)!}=\frac{n(n-1)(n-2)\dots (n-x+1)\stackrel{(n-x)!}{\overbrace{(n-x)(n-x-1)\dots (2)(1)}}}{n^x(n-x)!}$
      • $\frac{n!}{n^x(n-x)!}=\frac{n(n-1)(n-2)\dots (n-x+1)\cancel{(n-x)(n-x-1)\dots (2)(1)}}{n^x\cancel{(n-x)!}}$
      • $\frac{n!}{n^x(n-x)!}=\frac{n(n-1)(n-2)\dots (n-x+1)}{n^x}$
      • Taking the limit of this
      • $\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{n^x(n-x)!}=\underset{n\to \mathrm{\infty }}{lim}\frac{\stackrel{x\text{ terms}}{\overbrace{n(n-1)(n-2)\dots (n-x+1)}}}{n^x}$
      • Since there's $x$ terms on the top and $n^x$, each of those on top is divided by $n$ on the bottom
      • $\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{n^x(n-x)!}=\underset{n\to \mathrm{\infty }}{lim}\frac{n(n-1)(n-2)\dots (n-x_{+}1)}{\underset{x\text{ terms}}{\underbrace{n\cdot n\cdot n\cdot \dots \cdot n}}}$
      • $\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{n^x(n-x)!}=\underset{n\to \mathrm{\infty }}{lim}\frac{n}{n}\cdot \frac{n\underset{}{\overset{\begin{array}{c}\text{this}\\ \text{doesn’t}\\ \text{matter}\end{array}}{\overbrace{-1}}}}{n}\cdot \frac{n-2}{n}\cdot \frac{n-3}{n}\cdot \dots \cdot \frac{n-x+1}{n}$
      • The constants doesn't matter when taking the limit
      • $\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{n^x(n-x)!}=\underset{n\to \mathrm{\infty }}{lim}\frac{n^x}{n^x}=1$
      • With this lemma

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\underset{n\to \mathrm{\infty }}{lim}\left[\left(\frac{{\lambda }^x}{x!}\right){(1-\frac{\lambda }{n})}^{n-x}\right]$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\left(\frac{{\lambda }^x}{x!}\right)\underset{n\to \mathrm{\infty }}{lim}{(1-\frac{\lambda }{n})}^{n-x}$

    • $x$ is a constant, and its effect is negligible

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\left(\frac{{\lambda }^x}{x!}\right)\underset{n\to \mathrm{\infty }}{lim}{(1-\frac{\lambda }{n})}^n$

    • Since $e^x=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\left(\frac{{\lambda }^x}{x!}\right)\underset{n\to \mathrm{\infty }}{lim}\underset{e^{-x}}{\underbrace{{(1-\frac{\lambda }{n})}^n}}$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\left(\frac{{\lambda }^x}{x!}\right)e^{-x}$

    • $\underset{n\to \mathrm{\infty }}{lim}P(X=x)=\left(\frac{{\lambda }^x{e}^{-x}}{x!}\right)\text{ for }x=0,1,2,\dots$

• Example:

  • Customers at a bank queue up at a rate of $5$ customers every 20 minutes.
    • Poisson Distribution, given away by $x$ per $y$ blah blah blah
  • Assume arrival times are independent.
  • 1
    • Find the probability that $7$ customers appear in a $20$ minute timeframe
    • $\lambda =\frac{5\text{ customers}}{20\text{ minutes}}$
    • $P(X=7)=\frac{e^{-\lambda }{\lambda }^x}{x!}=\frac{e^{-5}{5}^7}{7!}\approx 0.104$
  • 2
    • Find the probability that at least $1$ customer appears in a $20$ minute window
    • $P(X\ge 1)=P(X=1)+P(X=2)+\dots$
    • $P(X\ge 1)=1-P(X=0)$
    • $P(X=0)=\frac{e^{-5}{5}^0}{0!}\approx 0.006737947$
    • $P(X\ge 1)\approx 1-0.006737947\approx 0.993262053$
  • 3
    • What is the probability of observing $20$ customers in 1 hour
    • $\begin{array}{}\text{Customer}:\text{Time}\\ 5:20\\ 15:60\end{array}$
    • $P(X=20)=\frac{e^{-15}{15}^{20}}{20!}\approx 0.0418$
  • 4
    • What are the mean and standard deviation of the customers arriving in 20 minutes
    • $\mu =E[X]=\lambda =5$
    • ${\sigma }^2=var(x)=\lambda =5$
    • $\sigma =\sqrt{\lambda }=\sqrt{5}\approx 2.24$