EE 364 Supplemental – Week 04b
Poisson Probability
Counting independent arrivals.
Suppose on average 120 packets/minute.
\(\therefore\) very small \(\delta t\) \(\implies\) \(P[\text{1 arrival in interval}] = p \ll 1\) (\(\approx 0\))
\(\therefore P[\text{2 arrivals}] = p^2\), \(P[\text{3 arrivals}] = p^3\), \(\ldots\) (independent arrivals)
\(\therefore\) \(\delta t \to 0\): \(P[\text{1 arrival}]\) small, \(P[\text{more than 1 arrival}] \approx 0\).
\(\therefore\) binomial structure: \(n\) flips, \(P[\text{1 arrival}] = p\), with \(n \cdot p = \lambda\).
\(n\) large, \(p\) small, \(\lambda = np\) constant rate.
Poisson Law
\[ b(n,k,p) \xrightarrow{d} P(\lambda) \quad \text{if } n \gg 1 \text{ and } p \ll 1 \text{ and } \lambda = np \]
Proof. \[ \begin{aligned} \lim_{n \to \infty} b(n,k,p) &= \lim_{n \to \infty} \binom{n}{k} p^k (1-p)^{n-k} \\[6pt] &= \lim_{n \to \infty} \frac{n!}{k!\,(n-k)!} \left(\frac{\lambda}{n}\right)^k \left(1 - \frac{\lambda}{n}\right)^{n-k} &&\text{[since } \lambda = np\text{]} \\[6pt] &= \frac{\lambda^k}{k!} \lim_{n \to \infty} \frac{n!}{(n-k)!\, n^k} \left(1 - \frac{\lambda}{n}\right)^n \left(1 - \frac{\lambda}{n}\right)^{-k} \\[6pt] &= \frac{\lambda^k}{k!} \lim_{n \to \infty} \frac{n \cdot (n-1) \cdots (n-k+1)}{n^k} \left(1 - \frac{\lambda}{n}\right)^n \left(1 - \frac{\lambda}{n}\right)^{-k} \\[6pt] &= \frac{\lambda^k}{k!} \cdot \underbrace{1 \cdot \left(1 - \tfrac{1}{n}\right) \cdots \left(1 - \tfrac{k-1}{n}\right)}_{\to 1} \cdot \underbrace{\left(1 - \tfrac{\lambda}{n}\right)^n}_{\to e^{-\lambda}} \cdot \underbrace{\left(1 - \tfrac{\lambda}{n}\right)^{-k}}_{\to 1} \\[6pt] &= \frac{\lambda^k}{k!}\, e^{-\lambda} = P(\lambda) \quad \square \end{aligned} \]
Poisson PMF
\(X \sim P(\lambda)\) for parameter \(\lambda > 0\) (“mean”).
\[ P(X = k) = \frac{e^{-\lambda}\, \lambda^k}{k!} \qquad k = 0, 1, 2, \ldots \]
Issue spot: counting structure. BEG CUP.
Example: Alpha-Decay of Radium to Radon
\[ {}^{226}_{88}\text{Ra} \longrightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He}^{2+} \quad (\alpha\text{-particle}) \]
Half-life of \({}^{226}\text{Ra}\) isotope: 1600 years.
Radium: 226 grams/mole. \(\therefore\) 1 gram \(\approx 10^{22}\) atoms (\(= n\)).
Binary event: atom “decays” spontaneously or not. “Vast inter-atomic distance” \(\implies\) independence.
\(\lambda = \alpha\)-emissions per gram per second \(\approx 10^{10}\) decays.
\(T = 1\) second:
\[ P(\text{1 } \alpha\text{-decay}) = b(10^{22}, 1) \approx \frac{10^{10}}{10^{22}} = 10^{-12} = \frac{1}{\text{trillion}} \]
\[ \therefore P(X(t) = k) \approx \frac{e^{-\lambda}\, \lambda^k}{k!} \quad \text{for } k = 0, 1, 2, \ldots \]
Example: Basketball Long Shots
\(X \sim b(70, 0.02)\). Assume independent shots.
\(n = 70 \gg 1\), \(p = 0.02 \ll 1\).
\(\therefore \lambda = np = 70 \cdot 0.02 = 1.4\)
Exact binomial probability: \[ P(\text{3 baskets in 70 trials}) = \binom{70}{3}(0.02)^3(0.98)^{67} = 0.1131 \]
Poisson approximation with \(\lambda = np = 1.4\): \[ P(X = 3) = \frac{(1.4)^3 e^{-1.4}}{3!} = 0.1128 \]
MGF Approach (Preview)
“moment generating function” – covered later.
\[ b(n,k,p) \leftrightarrow \mathcal{M}_B(s) = (1 - p + pe^s)^n \quad (= E[e^{sX}]) \]
\[ P(\lambda) \leftrightarrow \mathcal{M}_P(s) = e^{\lambda(e^s - 1)} \]
For \(np = \lambda\) constant:
\[ \begin{aligned} \lim_{n \to \infty} \mathcal{M}_B(s) &= \lim_{n \to \infty} \left(1 - \frac{\lambda}{n} + \frac{\lambda}{n} e^s\right)^n &&\text{[since } p = \tfrac{\lambda}{n}\text{]} \\ &= \lim_{n \to \infty} \left(1 - \frac{\lambda(e^s - 1)}{n}\right)^n \\ &= e^{\lambda(e^s - 1)} = \mathcal{M}_P(s) \end{aligned} \]
\(\therefore\) Again: \(b(n,k,p) \xrightarrow{d} P(\lambda)\) if \(\lambda = np\) (and Levy’s theorem, week 11).
Accuracy of Poisson Approximation
| Quality | Conditions |
|---|---|
| Okay | \(n \geq 20\) and \(p \leq 0.05\) |
| Excellent | \(n \geq 100\) and \(p \leq 0.1\) |
Measuring the Size of Sets
\(f: X \to Y\) is 1-to-1 iff \(\forall x\, \forall z: f(x) = f(z) \implies x = z\).
\(f: X \to Y\) is onto iff \(\forall y \in Y\; \exists x \in X: y = f(x)\). (i.e. \(f(X) = Y\).)
\(f: X \to Y\) is a bijection (“1-to-1 correspondence”) iff \(f\) is 1-to-1 and onto.
\(|A|\) = cardinality (“size”) of set \(A\).
| Set | Notation | Elements |
|---|---|---|
| Natural numbers | \(\mathbb{N} = \{1, 2, 3, \ldots\}\) | |
| Integers | \(\mathbb{Z} = \{0, \pm 1, \pm 2, \pm 3, \ldots\}\) | \(\mathbb{Z}^+ = \mathbb{N}\) |
| Reals | \(\mathbb{R} = (-\infty, \infty)\) |
\(A\) is finite iff \(A \xleftrightarrow[\text{onto}]{\text{1-to-1}} S\) for some \(S \subset \mathbb{N}\), such that \(A = \{a_1, \ldots, a_n\}\) for some \(n \in \mathbb{N}\) (or \(A = \emptyset\)). Else \(A\) is infinite.
Fact: \(A\) infinite \(\iff\) \(A \xleftrightarrow[\text{onto}]{\text{1-to-1}} B\) for some \(B \subset A\) and \(B \neq A\).
\(\therefore\) infinite \(\iff\) \(\sim\)finite.
Cantor’s Theorem and Cardinalities
\[ |X| < |2^X| \]
\(A\) is denumerable iff \(A \xleftrightarrow[\text{onto}]{\text{1-to-1}} \mathbb{N}\) (e.g. \(\mathbb{Z}\)).
\(A\) is countable iff (1) \(A\) is finite, or (2) \(A\) is denumerable.
Facts:
- \(|\mathbb{Z}| = |\mathbb{N}| = \aleph_0\) (aleph-nought)
- \(|\mathbb{R}| = |2^{\mathbb{Z}}| = \mathfrak{c} = \aleph_1\) (power of the continuum)
- \(|A| = \aleph_k \implies |2^A| = \aleph_{k+1}\) (\(k = 0, 1, \ldots\))
There is no \(\omega\) such that \(\aleph_k < \omega < \aleph_{k+1}\).
Ex: \((0, 1)\) same size as \(\mathbb{R}^+\).
\[ f(x) = \frac{1}{1 + e^{-x}} \]
Note: adding 2 points \(0, 1 \to [0,1]\). No 1-to-1 onto map with \(\mathbb{R}^+\).
Aside: \(\pm \infty\) not real numbers.