```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## ๐ฌ The Idea in 2 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: - **Write** the probability-generating function $P(t) = E(t^Y)$ of a count, and read probabilities off its coefficients - **Differentiate** $P(t)$ at $t = 1$ to obtain factorial moments, and from them $E(Y)$ and $V(Y)$ - **Bound** $P(|Y - \mu| < k\sigma)$ with Tchebysheff's theorem when only $\mu$ and $\sigma$ are known - **Invert** the bound to size a buffer: find $C$ with $P(|Y-\mu| \ge C)$ below a target - **Choose** among the Chapter 3 models for a count on a bank's risk desk ::: --- ## ๐บ๏ธ Where We Are ::: {style="font-size: 30px"} **Wackerly ยงยง3.10โ3.11** Wednesday ended on **uniqueness**: recognise the MGF $m(t) = E(e^{tY})$, and you know the distribution. It stored the moments. ::: {.fragment} Today, two more tools, and then the chapter closes: 1. A **cousin** of $m(t)$ built for counts, $P(t) = E(t^Y)$, whose coefficients are the probabilities themselves. ยง3.10 is optional in the book, so we meet it briefly. 2. **Tchebysheff's theorem**, which needs no model at all: only $\mu$ and $\sigma$. This one is load-bearing. ::: ::: --- ## โ Motivating Question ::: {.callout-important} ## The risk desk's problem A Baku bank's SME book has averaged **12 defaults a month**, with a standard deviation of **3**, over five years. Nobody on the desk is willing to say the count is binomial, Poisson, or anything else. The regulator asks: *how confident are you that next month's defaults stay between 6 and 18?* ::: ::: {.fragment style="font-size: 30px"} Without a distribution there is no $p(y)$ to add up. By the end of today there will still be an answer, and it will be a guarantee. ::: --- ## ๐ Definition 3.15: The PGF ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.15 Let $Y$ be an integer-valued random variable with $P(Y = i) = p_i$, $i = 0, 1, 2, \ldots$ The **probability-generating function** of $Y$ is $$P(t) = E(t^Y) = p_0 + p_1 t + p_2 t^2 + \cdots = \sum_{i=0}^{\infty} p_i t^i$$ for all $t$ such that $P(t)$ is finite. ::: ::: {.fragment} The name is literal: **the coefficient of $t^i$ is $P(Y = i)$**. Expand $P(t)$ as a series and the probability function falls out. Compare $m(t)$, whose coefficients are moments. ::: ::: --- ## ๐ Theorem 3.13: Factorial Moments ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.16 The $k$th **factorial moment** of $Y$ is $\mu_{[k]} = E[Y(Y-1)(Y-2)\cdots(Y-k+1)]$. ::: ::: {.callout-important} ## Theorem 3.13 $$\left.\frac{d^k P(t)}{dt^k}\right|_{t=1} = P^{(k)}(1) = \mu_{[k]}$$ ::: ::: {.fragment} Setting $t = 1$ removes the powers. Then $\mu = \mu_{[1]}$ and $V(Y) = \mu_{[2]} + \mu - \mu^2$. ::: ::: --- ## ๐ Worked Example: Motor Claims ::: {style="font-size: 30px"} An insurer's claims per motor policy per year: $p_0 = 0.70$, $p_1 = 0.20$, $p_2 = 0.08$, $p_3 = 0.02$. $$P(t) = 0.70 + 0.20t + 0.08t^2 + 0.02t^3$$ ::: {.fragment} $P^{(1)}(t) = 0.20 + 0.16t + 0.06t^2$, so $\mu = P^{(1)}(1) = 0.42$ claims. ::: ::: {.fragment} $P^{(2)}(t) = 0.16 + 0.12t$, so $\mu_{[2]} = P^{(2)}(1) = 0.28$. ::: ::: {.fragment} $$V(Y) = 0.28 + 0.42 - 0.42^2 = 0.5236, \qquad \sigma = 0.724$$ ::: ::: --- ## ๐ป The Same Numbers, Two Ways ```{r} #| label: pgf-check #| code-fold: false y <- 0:3 p <- c(0.70, 0.20, 0.08, 0.02) mu <- sum(y * p) # P^(1)(1) mu_2f <- sum(y * (y - 1) * p) # P^(2)(1), the 2nd factorial moment c(mean = mu, fact2 = mu_2f, var = mu_2f + mu - mu^2) sims <- sample(y, 1e5, replace = TRUE, prob = p) # 100,000 simulated policies round(c(sim_mean = mean(sims), sim_var = var(sims)), 4) ``` ::: {style="font-size: 28px"} The PGF route and the simulation agree to about two decimal places. ::: --- ## ๐ The Geometric PGF ::: {style="font-size: 29px"} Examples 3.26โ3.27. A watch-listed borrower first misses a payment in month $Y$, geometric with $p = 0.25$. Since $p_0 = 0$, $$P(t) = \sum_{y=1}^{\infty} t^y q^{y-1} p = \frac{pt}{1 - qt}, \qquad t < 1/q$$ ::: {.fragment} Then $P^{(1)}(t) = \dfrac{p}{(1-qt)^2}$ gives $\mu = P^{(1)}(1) = 1/p = 4$ months, and $P^{(2)}(1) = 2q/p^2$ gives $$V(Y) = \frac{2q}{p^2} + \frac{1}{p} - \frac{1}{p^2} = \frac{q}{p^2} = 12$$ ::: ::: {.fragment} Why keep a second tool? Because $P(t)$ is sometimes the easier one to find. Note $m(t) = P(e^t)$. ::: ::: --- ## ๐ Theorem 3.14: Tchebysheff ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 3.14 Let $Y$ be a random variable with mean $\mu$ and finite variance $\sigma^2$. Then, for any constant $k > 0$, $$P(|Y - \mu| < k\sigma) \ge 1 - \frac{1}{k^2} \qquad \text{or} \qquad P(|Y - \mu| \ge k\sigma) \le \frac{1}{k^2}$$ ::: ::: {.fragment} - It holds for **any** distribution, bell-shaped or not - It is **conservative**: the true probability usually beats the bound comfortably - The proof is deferred to ยง4.10 ::: ::: --- ## ๐ฆ Answering the Risk Desk ::: {style="font-size: 30px"} $\mu = 12$ and $\sigma = 3$. The interval $(6, 18)$ is $\mu \pm k\sigma$ with $k = 2$: $$P(6 < Y < 18) = P(|Y - 12| < 2 \times 3) \ge 1 - \frac{1}{2^2} = \frac{3}{4}$$ ::: {.fragment} **At least 75%**, whatever the distribution. That is the sentence the desk can sign. ::: ::: {.fragment} If the book were steadier, $\sigma = 2$, the same interval is $k = 3$ and the bound rises to $1 - 1/9 = 8/9 \approx 0.889$. As in Example 3.28, **$\sigma$ drives the guarantee**. ::: ::: --- ## ๐งฎ Running It Backwards: a Provision ::: {style="font-size: 30px"} The desk wants a level $C$ with $P(|Y - 12| \ge C) \le 0.04$. ::: {.fragment} Set $1/k^2 = 0.04$, so $k = 5$ and $C = k\sigma = 5 \times 3 = 15$. ::: ::: {.fragment} Provision for $12 + 15 = 27$ defaults: the chance of **27 or more** is **at most 4%**, with no model assumed. ::: ::: {.fragment} The price of assuming nothing is a wide band. A model buys a tighter one, but only if the model is right. Exercise 3.167(b) asks exactly this question. ::: ::: --- ## ๐ป How Conservative Is the Bound? ```{r} #| label: cheb-table #| code-fold: false # Three models, each measured against its own mean and sd cover <- function(y, py, k) { mu <- sum(y * py); s <- sqrt(sum((y - mu)^2 * py)) sum(py[abs(y - mu) < k * s]) } k <- c(1.5, 2, 3) data.frame( k = k, binom_48_25 = sapply(k, \(k) cover(0:48, dbinom(0:48, 48, 0.25), k)), poisson_12 = sapply(k, \(k) cover(0:200, dpois(0:200, 12), k)), geom_25 = sapply(k, \(k) cover(1:500, dgeom(0:499, 0.25), k)), tchebysheff = 1 - 1 / k^2 ) |> round(4) ``` ::: {style="font-size: 28px"} All three beat the bound; at $k = 2$ by 18 to 19 percentage points. ::: --- ## ๐ The Bound Sits Under Every Model ```{r} #| label: cheb-figure #| echo: false #| fig-width: 10 #| fig-height: 4.4 kk <- seq(1, 4, by = 0.01) curve_df <- rbind( data.frame(k = kk, model = "Binomial(48, 0.25)", p = sapply(kk, \(k) cover(0:48, dbinom(0:48, 48, 0.25), k))), data.frame(k = kk, model = "Poisson(12)", p = sapply(kk, \(k) cover(0:200, dpois(0:200, 12), k))), data.frame(k = kk, model = "Geometric(0.25)", p = sapply(kk, \(k) cover(1:500, dgeom(0:499, 0.25), k)))) bound_df <- data.frame(k = kk, p = 1 - 1 / kk^2) ggplot(curve_df, aes(k, p, colour = model)) + geom_step(linewidth = 1.1) + geom_line(data = bound_df, aes(k, p), inherit.aes = FALSE, colour = "#8b2635", linewidth = 1.4, linetype = "dashed") + annotate("text", x = 3.1, y = 0.72, label = "Tchebysheff:~1 - 1/k^2", parse = TRUE, colour = "#8b2635", size = 6.5) + scale_colour_manual(values = c("#14130f", "#3d6e8f", "#b3833a")) + scale_y_continuous(limits = c(0, 1), breaks = seq(0, 1, 0.25)) + labs(x = "k (half-width in standard deviations)", y = expression(P(abs(Y - mu) < k * sigma)), colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` --- ## ๐ฏ But the Bound Cannot Be Improved ::: {style="font-size: 30px"} Exercise 3.169, as a policy-rate decision. The central bank cuts by 1 point, holds, or raises by 1 point: $$p(-1) = \tfrac{1}{18}, \qquad p(0) = \tfrac{16}{18}, \qquad p(1) = \tfrac{1}{18}$$ ::: {.fragment} $E(Y) = 0$ and $V(Y) = 2/18 = 1/9$, so $\sigma = 1/3$. Take $k = 3$, so $k\sigma = 1$: $$P(|Y| \ge 1) = \tfrac{2}{18} = \tfrac{1}{9} = \tfrac{1}{k^2}$$ ::: ::: {.fragment} Equality. For any $k > 1$ some distribution attains the bound, so **no better model-free statement exists**. ::: ::: --- ## 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 = 4, seconds = 0, top = 0, right = 0, font_size = "2em", warn_when = 30) } else { htmltools::HTML(paste0( '4:00
')) } ``` ::: {style="font-size: 30px"} A microfinance lender opens on average **40 new arrears cases a month**, with variance **25**. No model is assumed. **Four minutes, in pairs:** 1. Give a lower bound for $P(30 < Y < 50)$. 2. Find $C$ with $P(|Y - 40| \ge C) \le 1/16$. 3. A colleague says "it's 95%, by the empirical rule". When is she entitled to say that? ::: --- ## โ
Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. $\sigma = \sqrt{25} = 5$ and $(30, 50) = 40 \pm 2 \times 5$, so $k = 2$: $$P(30 < Y < 50) \ge 1 - \tfrac{1}{4} = 0.75$$ 2. $1/k^2 = 1/16$ gives $k = 4$, so $C = 4 \times 5 = 20$: outside $(20, 60)$ with probability at most $1/16$. ::: {.fragment} 3. Only if the distribution is **roughly mound-shaped**. The empirical rule is an approximation for bell shapes; Tchebysheff is a guarantee for every shape. They do not contradict: 95% is above 75%. ::: ::: --- ## ๐งญ Chapter 3 in Review: Which Model? ::: {style="font-size: 28px"} | Model | $Y$ counts | $E(Y)$ | $V(Y)$ | |---|---|---|---| | Binomial | successes in $n$ independent trials | $np$ | $npq$ | | Geometric | trial of the first success | $1/p$ | $q/p^2$ | | Neg. binomial | trial of the $r$th success | $r/p$ | $rq/p^2$ | | Hypergeometric | successes in $n$ drawn without replacement | $nr/N$ | $n\frac{r}{N}\frac{N-r}{N}\frac{N-n}{N-1}$ | | Poisson | rare events per unit of time or space | $\lambda$ | $\lambda$ | | No model | any count; only $\mu, \sigma$ known | $\mu$ | $\sigma^2$: Tchebysheff | ::: --- ## ๐ฆ The Risk Desk's Week ::: {style="font-size: 29px"} - 25 mortgage borrowers, each defaulting independently with probability 0.03: how many default? **Binomial(25, 0.03)** - A recovery agent phones debtors in turn: on which call does the first promise to pay come? **Geometric** - An examiner pulls 8 of 60 loan files, 5 of which are misfiled: how many misfiled in the sample? **Hypergeometric** - Card-fraud alerts in an hour, rare and independent: **Poisson** - Only five years of monthly means and standard deviations: **Tchebysheff** - The PGF and MGF sit behind every row: they deliver each $E(Y)$ and $V(Y)$ ::: --- ## ๐ Quiz #1: Reading a PGF {.quiz-question} A count has $P(t) = 0.5 + 0.3t + 0.2t^2$. What is $E(Y)$? - [$0.7$]{.correct data-explanation="โ
Theorem 3.13: E(Y) = Pโฝยนโพ(1) = 0.3 + 2(0.2)(1) = 0.7. The same as summing y p(y) = 0(0.5) + 1(0.3) + 2(0.2)."} - $1.0$ - $0.3$ - $0.5$ --- ## ๐ Quiz #2: The Guarantee {.quiz-question} Daily FX transfers at a branch have mean 50 and standard deviation 5, shape unknown. What does Tchebysheff guarantee for $P(40 < Y < 60)$? - [At least $0.75$]{.correct data-explanation="โ
(40, 60) is ฮผ ยฑ 2ฯ, so the bound is 1 โ 1/2ยฒ = 0.75. The 95% figure needs a mound shape, which we were not given."} - At least $0.95$ - Exactly $0.75$ - At most $0.25$ --- ## ๐ Quiz #3: Which Model? {.quiz-question} An auditor draws 10 of a bank's 200 guarantee contracts **without replacement** and counts how many are undocumented. Which model fits? - [Hypergeometric]{.correct data-explanation="โ
A fixed sample drawn without replacement from a finite pool: the draws are dependent, which rules out the binomial unless the pool is very large."} - Binomial - Poisson - Geometric --- ## ๐ Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Definition 3.15 | $P(t) = E(t^Y) = \sum_i p_i t^i$ | | Theorem 3.13 | $P^{(k)}(1) = \mu_{[k]} = E[Y(Y-1)\cdots(Y-k+1)]$ | | mean and variance | $\mu = P^{(1)}(1)$, $\;V(Y) = P^{(2)}(1) + \mu - \mu^2$ | | geometric, Poisson | $\dfrac{pt}{1-qt}$, $\;e^{\lambda(t-1)}$ | | Theorem 3.14 | $P(\lvert Y-\mu \rvert < k\sigma) \ge 1 - 1/k^2$ | | equivalently | $P(\lvert Y-\mu \rvert \ge k\sigma) \le 1/k^2$ | ::: --- ## ๐ Summary ::: {style="font-size: 30px"} - A PGF stores $P(Y = i)$ as the coefficient of $t^i$; its derivatives at $t = 1$ are factorial moments - $V(Y) = \mu_{[2]} + \mu - \mu^2$ turns them into a variance - Tchebysheff needs only $\mu$ and $\sigma$, and gives at least $1 - 1/k^2$ inside $\mu \pm k\sigma$ - The bound is conservative for familiar models but cannot be improved in general - Chapter 3 is a map: name the counting mechanism, pick the model, read off $E(Y)$ and $V(Y)$ ::: --- ## ๐ Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - ยง3.10 (optional): Exercises 3.164 โ 3.166, the binomial and Poisson PGFs - ยง3.11: Exercises 3.167 โ 3.174; start with 3.167, 3.169 and 3.173 - Redo the risk-desk provision with $\sigma = 4$: what does $C$ become? **Week 8, Problem Set 2** is open now and closes **Sunday 8 November at 23:59** on WeBWorK, covering ยงยง3.10โ3.11. **Next class:** 4 November, we leave counts behind for continuous random variables: distribution and density functions (Wackerly ยงยง4.1โ4.2). ::: --- ## 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"} - Tchebysheff says nothing for $k \le 1$. Why is $1 - 1/k^2$ useless there? - The Poisson PGF is $e^{\lambda(t-1)}$. What is its second factorial moment, and hence its variance? - A model gives a tighter band than Tchebysheff. When would a risk manager still report the Tchebysheff band? ::: 