```{r} #| label: setup #| include: false set.seed(2026) ``` ## ๐ฌ 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"} ::: --- ## ๐ฏ Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **State** Definition 2.9 and compute $P(A \mid B)$ from a table or from stated probabilities - **Explain** the division by $P(B)$ as a rescaling of the sample space - **Distinguish** $P(A \mid B)$ from $P(B \mid A)$, and say why they differ - **Test** a pair of events for independence using Definition 2.10 - **Argue** why two mutually exclusive events of positive probability are necessarily *dependent* ::: --- ## ๐บ๏ธ Where We Are ::: {style="font-size: 32px"} **Wackerly ยง2.7** Saturday we counted. Every probability was $n_a/N$, and every sample point carried the same weight $1/N$. ::: {.fragment} That weighting was an assumption about **ignorance**: we knew nothing that separated one point from another. ::: ::: {.fragment} Today: information arrives. A bond is known to be speculative-grade; a borrower is known to have missed a payment. **What does the arrival of information do to a probability?** ::: ::: --- ## ๐ก The Motivating Question ::: {style="font-size: 30px"} Of 1,000 corporate bond issues tracked over five years, 80 defaulted. Unconditionally, $$P(D) = \frac{80}{1000} = 0.08$$ ::: {.fragment} Now you are told the issue was **speculative grade**. Is 0.08 still your answer? ::: ::: {.fragment} Obviously not โ and "obviously" is exactly what needs a definition. The intuition is that we are no longer working in a sample space of 1,000 issues. ::: ::: --- ## ๐ Definition 2.9 ::: {style="font-size: 32px"} ::: {.callout-note} ## Definition 2.9 The **conditional probability** of an event $A$, given that an event $B$ has occurred, is $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ provided $P(B) > 0$. ::: ::: {.fragment} The proviso is not a technicality. Conditioning on something that cannot happen is not a harder problem; it is not a problem. ::: ::: --- ## โ๏ธ Why Divide by $P(B)$? ::: {style="font-size: 30px"} Learning that $B$ occurred does two things. ::: {.fragment} **It discards** every sample point outside $B$. The ones left are exactly those in $A \cap B$ โ which is why the numerator is $P(A \cap B)$ and not $P(A)$. ::: ::: {.fragment} **It rescales** what remains. The surviving points carry total weight $P(B)$, not 1, so dividing by $P(B)$ restores Axiom 2: $P(B \mid B) = 1$. ::: ::: {.fragment} Conditioning is not a new kind of probability. It is the **same probability measure on a smaller sample space**. ::: ::: --- ## ๐ The Bond Book ::: {style="font-size: 28px"} 1,000 issues, classified by rating at issue and by whether they defaulted within five years. | | Defaulted ($D$) | Did not default | Total | |---|---|---|---| | Investment grade ($I$) | 12 | 688 | 700 | | Speculative grade ($S$) | 68 | 232 | 300 | | **Total** | **80** | **920** | **1000** | ::: {.fragment} One issue is drawn at random, every issue equally likely. Then $$P(D \mid S) = \frac{P(D \cap S)}{P(S)} = \frac{68/1000}{300/1000} = \frac{68}{300} = 0.2267$$ ::: ::: --- ## ๐ The Counts Were Enough ::: {style="font-size: 30px"} $$P(D \mid S) = \frac{68/1000}{300/1000} = \frac{68}{300}$$ The 1000s cancel. With equiprobable points, a conditional probability is just **the row read on its own**. ::: {.fragment} That is a computational convenience, not the definition. It fails the moment the sample points stop being equally likely โ and Definition 2.9 does not. ::: ::: --- ## โ๏ธ $P(A \mid B)$ Is Not $P(B \mid A)$ ::: {style="font-size: 30px"} $$P(D \mid S) = \frac{68}{300} = 0.2267 \qquad P(S \mid D) = \frac{68}{80} = 0.8500$$ ::: {.fragment} Same numerator, different denominator, and they answer different questions. - $P(D \mid S)$: *of the speculative issues, how many defaulted?* - $P(S \mid D)$: *of the defaults, how many were speculative?* ::: ::: {.fragment} Reading one as the other is the most expensive single error in applied probability. We will meet it again on 3 October under its own name. ::: ::: --- ## ๐ป Both Conditionals in R ```{r} #| label: bond-table bonds <- matrix(c(12, 688, 68, 232), nrow = 2, byrow = TRUE, dimnames = list(c("Investment", "Speculative"), c("Default", "No default"))) data.frame( quantity = c("P(D)", "P(D | S)", "P(D | I)", "P(S | D)"), value = round(c(sum(bonds[, 1]) / sum(bonds), bonds["Speculative", "Default"] / sum(bonds["Speculative", ]), bonds["Investment", "Default"] / sum(bonds["Investment", ]), bonds["Speculative", "Default"] / sum(bonds[, "Default"])), 4) ) ``` ::: {style="font-size: 28px"} $P(D \mid S) = 0.2267$ against $P(D \mid I) = 0.0171$. Rating carries information about default, and that comparison is the whole of the next definition. ::: --- ## ๐ Definition 2.10 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 2.10 Two events $A$ and $B$ are **independent** if any one of $$P(A \mid B) = P(A), \qquad P(B \mid A) = P(B), \qquad P(A \cap B) = P(A)P(B)$$ holds. If none holds, $A$ and $B$ are **dependent**. ::: ::: {.fragment} Any one โ the three are equivalent whenever both conditioning probabilities are defined. The third is the one to use, because it needs no proviso and is symmetric in $A$ and $B$. ::: ::: --- ## ๐งฎ Testing the Bond Book ::: {style="font-size: 30px"} $$P(S \cap D) = \frac{68}{1000} = 0.068 \qquad P(S)P(D) = 0.300 \times 0.080 = 0.024$$ ::: {.fragment} $0.068 \neq 0.024$, so rating class and default are **dependent** โ and the direction is informative: defaults among speculative issues run at nearly three times the rate the product rule would predict. ::: ::: {.fragment} Equivalently, $P(D \mid S) = 0.2267 \neq 0.08 = P(D)$. Conditioning **moved** the probability, and that movement *is* the dependence. ::: ::: --- ## ๐๏ธ Independence Is a Statement About a Pair ::: {style="font-size: 30px"} Independence is not a property of $A$, nor of $B$, nor of the experiment. It is a property of the **pair**. ::: {.fragment} A classification with three or more categories can be independent of an event in one category and dependent in another. Verifying one pair tells you nothing about the rest. ::: ::: {.fragment} Nor is it a physical claim about causation. Independence is an arithmetic relation among three numbers, and nothing more. ::: ::: --- ## โ ๏ธ Independence Is Not Mutual Exclusivity ::: {style="font-size: 30px"} These two are confused more often than any other pair of ideas in Chapter 2, and they are close to opposites. ::: {.fragment} - **Mutually exclusive**: $A \cap B = \varnothing$. If one happens, the other cannot. - **Independent**: $P(A \cap B) = P(A)P(B)$. If one happens, the other is *no less likely than before*. ::: ::: {.fragment} One says the events are maximally informative about each other. The other says they carry no information at all. ::: ::: --- ## ๐ The Argument in One Line ::: {style="font-size: 30px"} ::: {.callout-important} ## A small result worth stating If $P(A) > 0$, $P(B) > 0$ and $A \cap B = \varnothing$, then $A$ and $B$ are **dependent**. ::: ::: {.fragment} $$P(A \cap B) = P(\varnothing) = 0, \qquad P(A)P(B) > 0$$ so the product rule fails. Equivalently $P(A \mid B) = 0 \neq P(A)$: learning that $B$ occurred has driven $A$'s probability to zero, which is as much information as one event can carry about another. ::: ::: --- ## ๐๏ธ A Concrete Case ::: {style="font-size: 30px"} A ministry awards a single contract. Let $A$ be "firm A wins", $B$ be "firm B wins", with $P(A) = 0.45$ and $P(B) = 0.35$. ::: {.fragment} One contract, so $A \cap B = \varnothing$ and $P(A \cap B) = 0$. But $P(A)P(B) = 0.1575$. ::: ::: {.fragment} They are **dependent** โ strongly so. Told that firm B won, you know firm A did not. Two firms competing for the same contract are about as far from independent as events get. ::: ::: --- ## ๐ฌ Interactive: How Far From Independence? {.smaller} ```{ojs} //| echo: false viewof spec_def = { const input = Inputs.range([6, 140], {value: 68, step: 1, label: "Speculative issues that defaulted:"}); ['pointerdown','touchstart','mousedown','click','wheel','pointermove','touchmove'] .forEach(e => input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false tot = 1000 nS = 300 nI = 700 invDef = (s) => Math.max(0, 150 - s) pD = (s) => (s + invDef(s)) / tot bars = { const s = spec_def; return [ {what: "P(D | S)", p: s / nS}, {what: "P(D | I)", p: invDef(s) / nI}, {what: "P(D)", p: pD(s)} ]; } md`With **${spec_def}** speculative defaults: P(D | S) = **${(spec_def/nS).toFixed(4)}**, P(D) = **${pD(spec_def).toFixed(4)}**. The two agree exactly when the bars level out โ and that is independence.` Plot.plot({ width: 1150, height: 300, marginLeft: 120, marginBottom: 55, style: {fontSize: "18px"}, x: {label: "Probability", domain: [0, 0.5]}, y: {label: null, domain: ["P(D | S)", "P(D | I)", "P(D)"]}, marks: [ Plot.barX(bars, {y: "what", x: "p", fill: "#8b2635"}), Plot.ruleX([pD(spec_def)], {stroke: "#14130f", strokeDasharray: "4 4"}), Plot.ruleX([0]) ] }) ``` --- ## ๐ง Think-Pair-Share ```{r} #| label: tps-timer #| echo: false # The timer is the only thing in this deck that needs a package beyond base R. # Guarded so a machine without it renders the deck anyway, with a static # figure in the same corner, rather than halting the whole build. if (requireNamespace("countdown", quietly = TRUE)) { countdown::countdown(minutes = 3, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) } else { htmltools::HTML(paste0( '3:00
')) } ``` ::: {style="font-size: 30px"} A bank's currency desk records, for each trading day, whether the **euro book** made a profit ($E$) and whether the **dollar book** made a profit ($U$). Over 500 days: $E$ on 300, $U$ on 250, both on 150. **Three minutes, in pairs:** 1. Compute $P(U \mid E)$ and $P(U)$. 2. Are $E$ and $U$ independent? 3. Are they mutually exclusive? Could they be both? ::: --- ## โ
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. $P(U \mid E) = \dfrac{150/500}{300/500} = \dfrac{150}{300} = 0.50$, and $P(U) = \dfrac{250}{500} = 0.50$. 2. They are **equal**, so by Definition 2.10 the two books are **independent**. The product check agrees: $P(E \cap U) = 0.30 = 0.60 \times 0.50$. ::: {.fragment} 3. Not mutually exclusive โ they happened together on 150 days. And they could not be both: mutually exclusive events of positive probability are always dependent. ::: ::: --- ## ๐ Conditioning in Sequence ::: {style="font-size: 28px"} A placement of 12 mortgage loans contains 4 already in arrears. Two are drawn for review, without replacement. ::: {.fragment} $$P(A_1) = \frac{4}{12} = 0.3333$$ ::: ::: {.fragment} The second draw is made from what is left, so its probability is **conditional by construction**: $$P(A_2 \mid A_1) = \frac{3}{11} = 0.2727, \qquad P(A_2 \mid \bar{A_1}) = \frac{4}{11} = 0.3636$$ ::: ::: {.fragment} The two differ, so $A_1$ and $A_2$ are dependent. Sampling **without** replacement always produces dependence; sampling **with** replacement never does. ::: ::: --- ## ๐ Quiz #1: Reading the Definition {.quiz-question} $P(A \cap B) = 0.12$ and $P(B) = 0.40$. What is $P(A \mid B)$? - [$0.30$]{.correct data-explanation="โ
Definition 2.9: 0.12 / 0.40 = 0.30. The intersection goes on top, the conditioning event underneath."} - $0.048$ - $0.52$ - $3.33$ --- ## ๐ Quiz #2: Which Way Round? {.quiz-question} Of 200 flagged transactions, 30 were genuinely fraudulent; the bank processed 5,000 transactions in all. Which quantity is $30/200$? - [$P(\text{fraud} \mid \text{flagged})$]{.correct data-explanation="โ
The denominator is the flagged transactions, so that is the conditioning event: among those flagged, the proportion fraudulent."} - $P(\text{flagged} \mid \text{fraud})$ - $P(\text{fraud} \cap \text{flagged})$ - $P(\text{flagged})$ --- ## ๐ Quiz #3: Exclusive or Independent? {.quiz-question} $A$ and $B$ are mutually exclusive, with $P(A) = 0.3$ and $P(B) = 0.5$. Are they independent? - [No โ $P(A \cap B) = 0$ but $P(A)P(B) = 0.15$]{.correct data-explanation="โ
Mutually exclusive events of positive probability are always dependent: knowing B occurred rules A out entirely."} - Yes โ mutually exclusive events are always independent - Yes, because $P(A) + P(B) < 1$ - There is not enough information to say --- ## ๐ Quiz #4: What Independence Buys {.quiz-question} $A$ and $B$ are independent with $P(A) = 0.6$ and $P(B) = 0.25$. What is $P(A \mid B)$? - [$0.60$]{.correct data-explanation="โ
Under independence the conditioning does nothing: P(A | B) = P(A) = 0.60. That is precisely what Definition 2.10 says."} - $0.15$ - $0.25$ - $2.40$ --- ## ๐ Key Formulas ::: {style="font-size: 30px"} | | Statement | Requires | |---|---|---| | Definition 2.9 | $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$ | $P(B) > 0$ | | Definition 2.10 | $P(A \cap B) = P(A)P(B)$ | nothing | | equivalently | $P(A \mid B) = P(A)$ | $P(B) > 0$ | | mutually exclusive | $P(A \cap B) = 0$ | nothing | ::: {.fragment} The last two rows are compatible only when one of the events has probability zero. ::: ::: --- ## ๐ Summary ::: {style="font-size: 30px"} - Conditioning **discards** the points outside $B$ and **rescales** what is left by $P(B)$ - $P(A \mid B)$ and $P(B \mid A)$ share a numerator and answer different questions - Independence means the conditioning changes nothing: $P(A \mid B) = P(A)$ - Independence is a property of a **pair**, verified pair by pair - Mutually exclusive events of positive probability are **dependent**, never independent - Sampling without replacement is the everyday source of dependence ::: --- ## ๐ Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of ยง2.7 โ start with 2.69 โ 2.75, then 2.79 and 2.81 - Redo the bond table with the roles of the two classifications swapped, and check that the dependence verdict is unchanged **Week 3, Problem Set 1** opens today and closes **3 October at 23:59** on WeBWorK, covering ยง2.7. Twenty problems; unlimited attempts before the deadline. **Next class:** the multiplicative and additive laws โ Definition 2.9 rearranged into a tool, and the rule for the probability of a union. ::: --- ## ๐ Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** ๐ง sorujov@ada.edu.az ๐ข Building D, Room D325 ๐ Office hours: Wednesday, 16:00 โ 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## โ Questions ::: {style="font-size: 32px"} - If $A \subset B$, what are $P(A \mid B)$ and $P(B \mid A)$, without computing either? - Can an event be independent of itself? Which events, exactly? - The bond table's dependence was read as "rating carries information about default". What would have to be true of the table for the phrase to be reversed? ::: 