Statistics 110- Lecture 11

1. Sympathetic Magic

• mistake of confusing r.v. with its distribution.
  • adding r.v.s are not the same as adding pmfs.
• $P(X=x) + P(Y=y)$
  • can be bigger than 1.
  • you can’t do things like this.
  • “Word is not the thing, the map is not the territory.”
  • r.v. = random house, dist. = blueprint
    • you can have many r.v.s with the same dist.
    • blueprint : specifying all the probabilities for randomly choosing the color of doors of the house.

2. Poisson Distribution

• Most important discrete distribution.
• PMF
  • $P(X=k) = {e^{-\lambda}\lambda^k \over k!}, k \in \{0,1,2,\dots\}$
  • $\lambda$ is the rate parameter. ($\lambda>0$) → param of poisson dist.
  • $X \sim Pois(\lambda)$
• Valid?
  • $\sum_{k=0}^\infty {e^{-\lambda}\lambda^k \over k!} = e^{-\lambda}e^\lambda = 1$ ( adding PMF = 1)
    • $\sum_{k=0}^\infty {\lambda^k \over k!} = e^\lambda$ by taylor series. check!!
  • expected value
    • $E(X) = e^{-\lambda} \sum_{k=1}^\infty k \ {\lambda^k \over k!}$ (when k=0, it’s 0 anyway)
      • $= e^{-\lambda} \sum_{k=1}^\infty \ {\lambda^k \over (k-1)!} = \lambda e^{-\lambda} \sum_{k=1}^\infty \ {\lambda^{k-1} \over (k-1)!} = \lambda\ e^{-\lambda}e^\lambda = \lambda$

Why do we care about Poisson?

• It’s the most used dist. as a model for discrete data in the real world.

1) General Applications

• Counting number of successes where there are a large number of trials each with small prob. of success.
• Examples ( Not exactly, but approximately Poisson)
  • number of emails in an hour
  • Number of chocolate chips in a chocolate chip cookie
  • Number of earthquakes in a year in some region
• There is no upper bound in $k \in \{0,1,2,\dots\}$ as it goes towards infinity, and the prob. being small is not that clear (how small?)
  • so it is useful to approximate a real world dist. with a poisson dist.

2) Poisson Paradigm (Poisson Approximation)

• Event $A_1,A_2, \dots, A_n$, $P(A_j) = p_j$, n large, $p_j$’s small.

  • large number of trials, each unlikely

• Events are independent or weakly dependent, then number of $A_j$’s that occur is approximately Poisson. → $Pois(\lambda)$

  • weakly dependent is hard to define.
    • we could get a little bit of information.
      • if we know $A_1,A_2$ happened, maybe it helps a little bit to know if $A_3$ happened.

• Expected Number of events occuring = $\lambda = \sum_{j=1}^n p_j$

  • If all events are independent & $p_j$’s are all the same, we have Bernoulli trials
    • Expected Value = $np$

3) Connection between $X \sim Bin(n,p)$ & $X \sim Pois(\lambda)$

• Let $n \rightarrow \infty$ (n is large number), $\lambda= np$ is held constant.
  • Each are expected values for Binomial & Poisson
• Find what happens to the PMF $P(X=k) = \binom {n}{k}p^k(1-p)^{n-k}$, k fixed.
  • $p = \lambda/n$, so plug it in.
  • $\binom {n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$ ( 조합의 정의)
• $P(X=k) = \frac{n(n-1)...(n-k+1)\lambda^k}{k!\ n^k} (1-{\lambda \over n})^n (1-{\lambda \over n})^{-k}$

→ 이제 limit을 취하자. ($n \rightarrow \infty, p \rightarrow 0$)

• $(1-{\lambda \over n})^n$은 $e^{-\lambda}$로 감.
• $= \frac{\lambda^k}{k!} \ e^{-\lambda}$ (limit 취하고 남은 값.)

→ Poisson PMF at k.

• So, Binomial converges to Poisson when $n \rightarrow \infty, p \rightarrow 0$.

Raindrop Example

• We have a piece of paper, and we want to know how many raindrops hit this paper in 1 minute.
  • let’s break the paper into millions of tiny squares. ( $n \rightarrow \infty, p \rightarrow 0$)
    • each square is unlikely to be hit by a raindrop.
• If we assume all squares are independent, probability is all the same, binomial is what we should use.
  • Probably not exactly independent.
• Also, binomial only takes Y/N.
  • what if a square gets hit 2 times? → no way to express this by binomial
• Poisson Dist. seems better.

Triple birthday problem

• Have n people, find the approx. prob. that there are 3 people that have the same birthday.
• Assume n is reasonably large number.
  • $\binom{n}{3}$ triplets of people, indicator r.v. for each, $I_{ijk}, i<j<k$
    • $I_{ijk}=1$ if they all have the same birthday.
  • $E(\text{number of triple matches}) = \binom n3 {1 \over 365^2}$
    • 1st person can have whatever birthday, and the 2nd person should match the 1st person (1/365), and the 3rd person should match the 2nd person.(1/365)
• Let $X = \text{number of triple matches}$ → this is approximately $Pois(\lambda), \lambda = \binom n3 {1 \over 365^2}$,
  • each triplets matching birthdays is unlikely, n is fairly large.
• We don’t exactly have independence.
  • $I_{123} , I_{124}$ are not completely independent. (weak dependence)
    • if $I_{123} =1,$ we already know that 1,2 have the same birthday when we compute $I_{124}$
• $P(X \ge 1) = 1-P(X=0) \approx 1-e^{-\lambda}\ \lambda^0/0! = 1-e^{-\lambda}$ (prob. of at least one triple match)

→ poisson is very useful for getting rough approximations.