```{r} #| label: setup #| include: false set.seed(2026) ``` ## π¬ The Idea in 3 Minutes ::: {style="text-align:center"} [Watch this short intuition video before (or after) the slides. Captions: CC button.]{style="font-size:22px"} ::: --- ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Translate** a verbal description of an event into set notation, and read set notation back into words - **Distinguish** a simple event from a compound one, and write any event as a union of sample points - **State** the three axioms of probability and check whether a proposed assignment satisfies them - **Execute** the five steps of the sample-point method on a discrete sample space - **Recognise** that the axioms constrain an assignment of probabilities without ever choosing one ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly §§2.1 β 2.5** Last time we ended with a bound β Tchebysheff β that held for *any* data set. It bounded a proportion we could see. ::: {.fragment} Today we start on the quantity that proportion is supposed to estimate. Three axioms, and a method for computing with them. ::: ::: {.fragment} 1. Set notation as the language of events 2. Experiments, sample spaces, simple and compound events 3. The three axioms 4. The sample-point method ::: ::: --- ## π‘ Motivation: The Gambler's Argument ::: {style="font-size: 30px"} A trader claims his signal is worthless β a coin flip. You watch ten trades. **All ten win.** ::: {.fragment} You reject the claim. Not because ten wins is *impossible* under a fair coin, but because it is **improbable**. ::: ::: {.fragment} That single word carries the whole inference. Change the data to *six* wins out of ten and intuition stops answering: is that improbable or not? ::: ::: {.fragment} So we need a theory that returns a **number**. The number has to obey rules, or two analysts will compute different answers from the same experiment. ::: ::: --- ## π Set Notation: The Language of Events ::: {style="font-size: 30px"} ::: {.callout-note} ## Definitions $S$ β the **universal set**; $\varnothing$ β the **null set**; $A \subset B$ β $A$ is contained in $B$ $A \cup B$ β all points in $A$ **or** $B$ or both $A \cap B$ (or $AB$) β all points in $A$ **and** $B$ $\bar{A}$ β the **complement**: points of $S$ not in $A$, so $A \cup \bar{A} = S$ ::: ::: {.fragment} The two keywords are worth memorising: **union is "or"**, **intersection is "and"**. ::: ::: --- ## π« Mutually Exclusive Sets ::: {style="font-size: 32px"} ::: {.callout-note} ## Definition $A$ and $B$ are **disjoint**, or **mutually exclusive**, if $A \cap B = \varnothing$. ::: ::: {.fragment} For any set $A$, the pair $A$ and $\bar{A}$ is mutually exclusive. That is the pair we will use most. ::: ::: {.fragment} A bank's loan is *current* or *in arrears*, never both. A tender is *won* or *lost*. Mutually exclusive events are the ones we can add probabilities over β Axiom 3 says so, and nothing weaker does. ::: ::: --- ## βοΈ Four Identities You Will Use Constantly ::: {style="font-size: 30px"} **Distributive laws** $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$ $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$ **DeMorgan's laws** $$\overline{(A \cap B)} = \bar{A} \cup \bar{B} \qquad \overline{(A \cup B)} = \bar{A} \cap \bar{B}$$ ::: {.fragment} Read the second in words: "not (either covenant breached)" is "neither covenant breached". Translating between the two is most of the work in a probability exercise. ::: ::: --- ## π¬ Experiments and Events ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 2.1 An **experiment** is the process by which an observation is made. ::: ::: {.fragment} Note how wide that is. A controlled laboratory trial is an experiment; so is observing tomorrow's closing price of a share you do not control at all. ::: ::: {.fragment} Consider one trading day for a single share. Events of interest: $A$: the price rises. $\quad$ $B$: the price moves by more than 2%. $\quad$ $C$: the price is unchanged. ::: ::: --- ## 𧩠Simple and Compound Events ::: {style="font-size: 30px"} ::: {.callout-note} ## Definitions 2.2 β 2.3 A **simple event** cannot be decomposed, and corresponds to one and only one **sample point**, written $E_i$. The **sample space** $S$ is the set of all sample points. ::: ::: {.fragment} A **compound event** happens in more than one distinct way. Observing an odd number on a die is compound: it happens through $E_1$, $E_3$ or $E_5$. ::: ::: {.fragment} The test for membership is exact: $E_i$ belongs to $A$ **if and only if** $A$ occurs whenever $E_i$ occurs. ::: ::: --- ## π’ Discrete Sample Spaces ::: {style="font-size: 30px"} ::: {.callout-note} ## Definitions 2.4 β 2.5 A **discrete** sample space contains a finite or countable number of distinct sample points. An **event** in a discrete sample space is any subset of $S$. ::: ::: {.fragment} Counting defaults in a portfolio gives $S = \{E_0, E_1, E_2, \ldots\}$ β infinite, but countable, so discrete. ::: ::: {.fragment} Measuring a *yield* does not: the sample space is an interval of real numbers, which is uncountable. That case waits for Chapter 4. ::: ::: --- ## π The Three Axioms ::: {style="font-size: 30px"} ::: {.callout-important} ## Definition 2.6 To every event $A$ in $S$ assign a number $P(A)$ such that **Axiom 1.** $P(A) \geq 0$ **Axiom 2.** $P(S) = 1$ **Axiom 3.** If $A_1, A_2, \ldots$ are pairwise mutually exclusive, then $$P(A_1 \cup A_2 \cup \cdots) = \sum_{i=1}^{\infty} P(A_i)$$ ::: ::: --- ## π§ Where the Axioms Come From ::: {style="font-size: 30px"} Each one is a relative frequency written as a rule: - A relative frequency is never negative $\;\Rightarrow\;$ **Axiom 1** - Every repetition produces some outcome in $S$ $\;\Rightarrow\;$ **Axiom 2** - If a 1 occurs on $1/6$ of tosses and a 2 on $1/6$, then a 1 *or* a 2 occurs on $1/3$ $\;\Rightarrow\;$ **Axiom 3** ::: {.fragment} Axiom 3 stated for an infinite sequence implies the finite version, which is the one you will actually use. ::: ::: --- ## β οΈ What the Axioms Do *Not* Say ::: {style="font-size: 30px"} They state the conditions an assignment must satisfy. They do **not** tell you which assignment to make. ::: {.fragment} A coin has come up heads on 800 of 1000 tosses. Toss it once more: $S = \{H, T\}$. The axioms permit $P(H) = 1/2$ β and permit $P(H) = 0.8$. ::: ::: {.fragment} Both are legal models. Only one is a good one. **The axioms give consistency; the data must give the numbers.** ::: ::: --- ## πͺ The Sample-Point Method ::: {style="font-size: 28px"} 1. Define the experiment and say precisely what one simple event looks like 2. List the simple events; check that none can be decomposed. This is $S$ 3. Assign probabilities to the sample points, with $P(E_i) \geq 0$ and $\sum P(E_i) = 1$ 4. Define the event $A$ as a specific collection of sample points 5. Find $P(A)$ by **summing** the probabilities of the sample points in $A$ ::: {.fragment} Step 5 is legal only because of Axiom 3: distinct sample points are mutually exclusive. ::: ::: --- ## π Worked Example: A Credit File Audit ::: {style="font-size: 30px"} A branch has **five** loan files ready for review. Unknown to the reviewer, **two** were misclassified at origination. The auditor draws **two files at random**. Write the misclassified files $M_1, M_2$ and the sound ones $G_1, G_2, G_3$. ::: {.fragment} **Step 1.** One simple event = the unordered pair drawn. **Step 2.** There are ten pairs, so $S = \{E_1, \ldots, E_{10}\}$. ::: ::: --- ## π The Audit: Steps 2 and 3 ::: {style="font-size: 28px"} | | | | |---|---|---| | $E_1: \{M_1, M_2\}$ | $E_5: \{M_2, G_1\}$ | $E_8: \{G_1, G_2\}$ | | $E_2: \{M_1, G_1\}$ | $E_6: \{M_2, G_2\}$ | $E_9: \{G_1, G_3\}$ | | $E_3: \{M_1, G_2\}$ | $E_7: \{M_2, G_3\}$ | $E_{10}: \{G_2, G_3\}$ | | $E_4: \{M_1, G_3\}$ | | | ::: {.fragment} **Step 3.** The draw is at random, so no pair is favoured: $P(E_i) = 1/10$ for all $i$. ::: ::: --- ## π The Audit: Steps 4 and 5 ::: {style="font-size: 30px"} **Step 4.** Let $A$ be the event that the auditor sees **no** misclassified file. Checking every point, $$A = \{E_8, E_9, E_{10}\}$$ **Step 5.** By Axiom 3, $$P(A) = \tfrac{1}{10} + \tfrac{1}{10} + \tfrac{1}{10} = \tfrac{3}{10} = 0.3$$ ::: {.fragment} So the audit misses the problem entirely 30% of the time. That number, not a feeling, is what belongs in the report. ::: ::: --- ## π» The Audit, Enumerated in R ```{r} #| label: audit files <- c("M1", "M2", "G1", "G2", "G3") pairs <- combn(files, 2) # every unordered pair sound_only <- apply(pairs, 2, function(p) !any(startsWith(p, "M"))) data.frame( sample_points = ncol(pairs), points_in_A = sum(sound_only), P_A = sum(sound_only) / ncol(pairs) ) ``` ::: {style="font-size: 28px"} The machine lists $S$ and counts. Step 2 is where human error lives, and this is the check against it. ::: --- ## π¦ Unequal Sample Points: A Tender ::: {style="font-size: 30px"} A dealer bids in **two** consecutive treasury-bill auctions. From past auctions the odds she wins any one are **two to one**. ::: {.fragment} $S$ has four points, writing $WL$ for "won the first, lost the second": $$E_1: WW \quad E_2: WL \quad E_3: LW \quad E_4: LL$$ ::: ::: {.fragment} Equal probabilities would be wrong here. The assignment consistent with 2:1 odds is $$P(E_1) = \tfrac{4}{9}, \; P(E_2) = P(E_3) = \tfrac{2}{9}, \; P(E_4) = \tfrac{1}{9}$$ ::: ::: --- ## π¦ The Tender: At Least One Win ::: {style="font-size: 30px"} Check the assignment first: each $P(E_i) \geq 0$ and they sum to $9/9 = 1$. Axioms 1 and 2 hold. ::: {.fragment} Let $C$ be the event that she wins **at least one** auction. Then $C = E_1 \cup E_2 \cup E_3$ and $$P(C) = \tfrac{4}{9} + \tfrac{2}{9} + \tfrac{2}{9} = \tfrac{8}{9} \approx 0.889$$ ::: ::: {.fragment} Equally likely points would have given $3/4$. The sample-point method does not require equal probabilities β only *legal* ones. ::: ::: --- ## π» Three Trading Days ```{r} #| label: three-days S <- expand.grid(day1 = c("up", "down"), day2 = c("up", "down"), day3 = c("up", "down"), stringsAsFactors = FALSE) ups <- rowSums(S == "up") exact <- mean(ups == 2) # sample-point method sim <- mean(replicate(200000, sum(sample(c("up", "down"), 3, TRUE) == "up")) == 2) data.frame(sample_points = nrow(S), exactly_two_up = sum(ups == 2), by_enumeration = exact, by_simulation = round(sim, 4)) ``` ::: {style="font-size: 28px"} Eight equally likely points, three of them with exactly two up days: $P = 3/8$. The simulation agrees β as the relative-frequency reading of probability says it must. ::: --- ## π¬ Interactive: Why Relative Frequency Settles {.smaller} ```{ojs} //| echo: false viewof p_true = { const input = Inputs.range([0.05, 0.95], {value: 0.375, step: 0.005, label: "True P(event):"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false rng3 = { let s = 20260916; return () => (s = (s * 1103515245 + 12345) % 2147483648) / 2147483648; } // The running relative frequency after 1, 2, ..., 1500 repetitions. path = { const out = []; let hits = 0; for (let i = 1; i <= 1500; i++) { if (rng3() < p_true) hits++; out.push({n: i, freq: hits / i}); } return out; } md`After 1500 repetitions the relative frequency is **${path[1499].freq.toFixed(3)}**, against a true probability of **${p_true.toFixed(3)}**.` Plot.plot({ width: 1150, height: 320, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Number of repetitions", type: "log", domain: [1, 1500]}, y: {label: "Relative frequency", domain: [0, 1], tickFormat: ".1f"}, marks: [ Plot.ruleY([p_true], {stroke: "#8b2635", strokeWidth: 2.5}), Plot.line(path, {x: "n", y: "freq", stroke: "#14130f", strokeWidth: 1.6}), Plot.ruleY([0]) ] }) ``` --- ## π§ Think-Pair-Share ```{r} #| label: tps-timer #| echo: false countdown::countdown(minutes = 3, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) ``` ::: {style="font-size: 30px"} A risk officer assigns, for one loan reviewed next month: $$P(\text{current}) = 0.82, \quad P(\text{in arrears}) = 0.15, \quad P(\text{written off}) = 0.05$$ **Three minutes, in pairs:** 1. Is this assignment legal under Definition 2.6? 2. If not, which axiom fails, and by how much? 3. He then claims $P(\text{arrears or written off}) = 0.20$. On what does that step depend? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. **Not legal.** Axiom 1 holds β all three are non-negative. Axiom 2 fails: the three outcomes are exhaustive, so their probabilities must sum to $P(S) = 1$. They sum to $1.02$. 2. The excess is $0.02$. Small, and still fatal: every probability computed from this model can now exceed 1, and the model can no longer be called a probability at all. ::: {.fragment} An assignment that fails Axiom 2 is not "slightly wrong". It is outside the definition. ::: ::: --- ## β
Think-Pair-Share: The Addition ::: {style="font-size: 30px"} 3. Adding $0.15 + 0.05$ uses **Axiom 3**, and Axiom 3 requires the two events to be **mutually exclusive**. ::: {.fragment} Here they are: one loan, at one review date, cannot be both in arrears and written off. So the step is sound β once the assignment is repaired. ::: ::: {.fragment} The habit worth forming: **before adding two probabilities, say out loud why the events cannot both occur.** Next week's Β§2.8 is what to do when they can. ::: ::: --- ## π Quiz #1: Reading Set Notation {.quiz-question} Two covenants, $A$ and $B$, may be breached. Which expression says "neither is breached"? - [$\bar{A} \cap \bar{B}$]{.correct data-explanation="β
By DeMorgan this is the complement of (A βͺ B) β not (either one breached), which is exactly neither."} - $\bar{A} \cup \bar{B}$ - $\overline{A \cap B}$ - $A \cap B$ --- ## π Quiz #2: Simple or Compound? {.quiz-question} A share is watched for three days and each day recorded as *up* or *down*, so one simple event is a triple such as $(up, down, up)$. Which of these is a **compound** event? - [Exactly two of the three days are up]{.correct data-explanation="β
It occurs through three distinct sample points β UUD, UDU and DUU β so it decomposes. The other three each name a single triple."} - All three days are up - All three days are down - The first day is up, the second down, the third up --- ## π Quiz #3: Testing an Assignment {.quiz-question} A sample space has four points with $P(E_1) = 0.4$, $P(E_2) = 0.3$, $P(E_3) = 0.4$, $P(E_4) = -0.1$. Which axioms fail? - [Axiom 1 only β the four probabilities do sum to 1]{.correct data-explanation="β
0.4 + 0.3 + 0.4 β 0.1 = 1, so Axiom 2 holds; but P(Eβ) < 0 violates Axiom 1."} - Axiom 2 only - Axioms 1 and 2 - None of them fail --- ## π Quiz #4: The Sample-Point Method {.quiz-question} Three files are drawn at random from six, two of which are misclassified. What is the first step? - [State exactly what one simple event is β here, the unordered set of three files drawn]{.correct data-explanation="β
Step 1. Nearly every failed exercise fails here: list the wrong kind of object and every later step is wasted."} - Assign probability $1/6$ to each file - Compute the probability of the event of interest - Decide whether the sample points are equally likely --- ## π Summary ::: {style="font-size: 30px"} - Events are **sets**; union is "or", intersection is "and", and DeMorgan's laws move between them - A **simple event** is one sample point; a **compound event** is a union of several - The three axioms β non-negativity, $P(S) = 1$, additivity over mutually exclusive events β are the whole definition - The axioms **constrain** an assignment; they never **choose** one. Data does that - The **sample-point method**: describe, list, assign, collect, sum ::: --- ## π Practice Problems ::: {style="font-size: 30px"} **Wackerly, 7th edition** - Exercises at the end of Β§2.4 β especially 2.11, 2.15, 2.18 and 2.19 - Exercises 2.1 β 2.8 for set notation, and 2.3 for DeMorgan by Venn diagram **Problem Set 1** closes **26 September at 23:59** on WeBWorK, covering Chapter 1 and §§2.1 β 2.8. A tutorial runs that week, before the deadline. **Next class:** tools for counting sample points β the $mn$ rule, permutations, combinations, and partitions. ::: --- ## π Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** π§ sorujov@ada.edu.az π’ Building D, Room D325 π Office hours: Wednesday, 16:00 β 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## β Questions ::: {style="font-size: 32px"} - Can an event have probability 0 without being impossible? - The audit missed the problem 30% of the time with two files out of five. How many files must be drawn before that falls below 5%? - Why state Axiom 3 for an infinite sequence when every example in this lecture is finite? ::: 