```{r} #| label: setup #| include: false set.seed(2026) # Primes in the maths are typed ^{\text{ʹ}} (U+02B9), not ' : MathJax 2 fetches its # AMS font for U+2032 and the headless QA pass logs that fetch as a page error. library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## 🎬 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"} ::: --- ## 🎯 Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Distinguish** moments about the origin $\mu^{\text{ʹ}}_k$ from central moments $\mu_k$ - **Derive** the moment-generating function of a Poisson or binomial count - **Recover** $E(Y)$ and $V(Y)$ from $m(t)$ by differentiating at $t = 0$ (Theorem 3.12) - **Transform** an MGF when a count becomes a P&L: $m_{aY+b}(t) = e^{bt}m_Y(at)$ - **Identify** a distribution from its MGF, using uniqueness ::: --- ## 🗺️ Where We Are ::: {style="font-size: 32px"} **Wackerly §3.9** · first lecture after Midterm I On 21 October we met the last two discrete models of the chapter, the **hypergeometric and the Poisson**, and then reviewed §§3.1–3.8 for Saturday's midterm. ::: {.fragment} For every model so far, $\mu$ and $\sigma^2$ cost a **new series trick**: the binomial theorem, a geometric sum, $E[Y(Y-1)]$ for the Poisson. ::: ::: {.fragment} Today: **one function per distribution** that hands over every moment by differentiation — and, more importantly, names the distribution. ::: ::: --- ## ❓ Motivating Question ::: {style="font-size: 30px"} A Baku bank's treasury counts large ($\geq$ 1M AZN) corporate payments cleared per hour at two branches. | Payments $y$ | 0 | 1 | 2 | 3 | 4 | |---|---|---|---|---|---| | Branch A, $p(y)$ | 0 | 1/2 | 0 | 1/2 | 0 | | Branch B, $p(y)$ | 1/8 | 0 | 3/4 | 0 | 1/8 | Both have $\mu = 2$ and $\sigma^2 = 1$. ::: {.callout-important} ## The question If two counts share a mean and a variance, is the risk the same? And what extra numbers **would** pin the distribution down? ::: ::: --- ## 📐 Definitions 3.12 and 3.13 ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.12 The $k$th moment of $Y$ taken **about the origin** is $E(Y^k)$, denoted $\mu^{\text{ʹ}}_k$. ::: ::: {.callout-note} ## Definition 3.13 The $k$th moment of $Y$ taken **about its mean**, the $k$th central moment, is $E[(Y-\mu)^k]$, denoted $\mu_k$. ::: So $\mu^{\text{ʹ}}_1 = \mu$, $\mu_2 = \sigma^2$, and Theorem 3.6 reads $\sigma^2 = \mu^{\text{ʹ}}_2 - (\mu^{\text{ʹ}}_1)^2$. ::: --- ## 💻 Two Branches, Four Moments ```{r} #| label: branch-moments yA <- c(1, 3); pA <- c(1/2, 1/2) yB <- c(0, 2, 4); pB <- c(1/8, 3/4, 1/8) mom <- function(y, p, k, about = 0) sum((y - about)^k * p) data.frame(k = 1:4, A_origin = sapply(1:4, \(k) mom(yA, pA, k)), B_origin = sapply(1:4, \(k) mom(yB, pB, k)), A_central = sapply(1:4, \(k) mom(yA, pA, k, 2)), B_central = sapply(1:4, \(k) mom(yB, pB, k, 2))) ``` ::: {style="font-size: 28px"} Identical up to $k = 3$; the **fourth** moment separates them. Branch B has more mass far from the mean — fatter tails. ::: --- ## 📐 Definition 3.14: The MGF ::: {style="font-size: 30px"} ::: {.callout-note} ## Definition 3.14 The moment-generating function $m(t)$ for a random variable $Y$ is $m(t) = E(e^{tY})$. We say that an MGF for $Y$ exists if there is a constant $b > 0$ such that $m(t)$ is finite for $|t| \leq b$. ::: ::: {.fragment} For a discrete $Y$ it is one sum: $$m(t) = \sum_y e^{ty}p(y)$$ It is a function of $t$, not of $y$; the $y$'s have been summed out. And always $m(0) = \sum_y p(y) = 1$. ::: ::: --- ## 🔑 Why "Moment-Generating"? ::: {style="font-size: 30px"} Expand $e^{ty} = 1 + ty + \dfrac{(ty)^2}{2!} + \dfrac{(ty)^3}{3!} + \cdots$ and take expectations term by term: $$m(t) = 1 + t\mu^{\text{ʹ}}_1 + \frac{t^2}{2!}\mu^{\text{ʹ}}_2 + \frac{t^3}{3!}\mu^{\text{ʹ}}_3 + \cdots$$ ::: {.fragment} **$\mu^{\text{ʹ}}_k$ is the coefficient of $t^k/k!$.** The MGF is every moment of $Y$ packed into one expression. ::: ::: {.fragment} Swapping the sum over $y$ with the infinite series is justified whenever $m(t)$ exists near 0. ::: ::: --- ## 📐 Theorem 3.12 ::: {style="font-size: 30px"} ::: {.callout-important} ## Theorem 3.12 If $m(t)$ exists, then for any positive integer $k$, $$\left.\frac{d^k m(t)}{dt^k}\right|_{t=0} = m^{(k)}(0) = \mu^{\text{ʹ}}_k$$ ::: ::: {.fragment} Differentiate the series $k$ times: every term below $t^k$ vanishes, the $t^k$ term leaves $\mu^{\text{ʹ}}_k$, and every higher term still carries a power of $t$, so it dies at $t = 0$. ::: ::: {.fragment} In practice: $\mu = m^{\text{ʹ}}(0)$ and $\sigma^2 = m^{\text{ʹʹ}}(0) - [m^{\text{ʹ}}(0)]^2$. ::: ::: --- ## 🚗 Example 3.23: Motor Claims ::: {style="font-size: 30px"} A Baku motor insurer's daily claim count is $Y \sim$ Poisson($\lambda$). $$m(t) = \sum_{y=0}^{\infty} e^{ty}\frac{\lambda^y e^{-\lambda}}{y!} = e^{-\lambda}\sum_{y=0}^{\infty}\frac{(\lambda e^t)^y}{y!}$$ ::: {.fragment} The sum is the Taylor series of $e^{\lambda e^t}$, so $$m(t) = e^{-\lambda}e^{\lambda e^t} = e^{\lambda(e^t - 1)}$$ ::: ::: {.fragment} The same trick as in Theorem 3.11: turn what is left into a Poisson p.f. with mean $\lambda e^t$, which sums to 1. ::: ::: --- ## 🧮 Example 3.24: Mean and Variance ::: {style="font-size: 30px"} $$m^{\text{ʹ}}(t) = e^{\lambda(e^t-1)}\,\lambda e^t, \qquad m^{\text{ʹʹ}}(t) = e^{\lambda(e^t-1)}(\lambda e^t)^2 + e^{\lambda(e^t-1)}\lambda e^t$$ ::: {.fragment} At $t = 0$: $\mu = m^{\text{ʹ}}(0) = \lambda$ and $\mu^{\text{ʹ}}_2 = m^{\text{ʹʹ}}(0) = \lambda^2 + \lambda$, so $$\sigma^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$$ ::: ::: {.fragment} With $\lambda = 4.5$ claims a day: $\mu^{\text{ʹ}}_2 = 24.75$, $\sigma^2 = 4.5$, $\sigma = 2.12$. **No $E[Y(Y-1)]$ detour needed.** ::: ::: --- ## 📈 The Binomial MGF: Winning Trades ::: {style="font-size: 30px"} An FX dealer books $n = 10$ independent trades a day, each profitable with $p = 0.6$. $Y$ = winning trades $\sim$ Bin$(10, 0.6)$. By the binomial theorem (Exercise 3.145): $$m(t) = \sum_{y=0}^{n}\binom{n}{y}(pe^t)^y q^{n-y} = (pe^t + q)^n$$ ::: {.fragment} $m^{\text{ʹ}}(0) = np = 6$, and $m^{\text{ʹʹ}}(0) = n(n-1)p^2 + np = 32.4 + 6 = 38.4$, so $$V(Y) = 38.4 - 6^2 = 2.4 = npq$$ ::: ::: --- ## 💰 From Trade Count to Daily P&L ::: {style="font-size: 30px"} Each winner earns 2 thousand AZN, each loser costs 1: $X = 2Y - (10 - Y) = 3Y - 10$. ::: {.callout-important} ## Exercise 3.156 If $X = aY + b$, then $m_X(t) = E(e^{t(aY+b)}) = e^{bt}\,m_Y(at)$. ::: ::: {.fragment} $$m_X(t) = e^{-10t}\,(0.6e^{3t} + 0.4)^{10}$$ Differentiating at 0: $E(X) = 3(6) - 10 = 8$ thousand AZN, $V(X) = 9(2.4) = 21.6$, $\sigma_X = 4.65$. ::: ::: --- ## 💻 Checking Theorem 3.12 Numerically ```{r} #| label: numeric-check mgf <- function(t, y, p) sapply(t, \(s) sum(exp(s * y) * p)) h <- 1e-4 d1 <- function(m) (m(h) - m(-h)) / (2 * h) # m'(0) d2 <- function(m) (m(h) - 2 * m(0) + m(-h)) / h^2 # m''(0) claims <- \(t) mgf(t, 0:200, dpois(0:200, 4.5)) x <- 3 * (0:10) - 10 # P&L support pnl <- \(t) mgf(t, x, dbinom(0:10, 10, 0.6)) round(rbind(claims = c(mean = d1(claims), var = d2(claims) - d1(claims)^2), pnl = c(mean = d1(pnl), var = d2(pnl) - d1(pnl)^2)), 3) ``` ::: {style="font-size: 28px"} Brute-force sums and finite differences agree with the calculus: 4.5 and 4.5, 8 and 21.6. ::: --- ## 🖼️ Same Slope, Different Curvature ```{r} #| label: mgf-figure #| echo: false #| fig-width: 10 #| fig-height: 4.1 tt <- seq(-0.8, 0.4, length.out = 300) df <- rbind( data.frame(t = tt, m = exp(4.5 * (exp(tt) - 1)), model = "Poisson(4.5): var 4.5"), data.frame(t = tt, m = (0.45 * exp(tt) + 0.55)^10, model = "Bin(10, 0.45): var 2.475")) ggplot(df, aes(t, m, colour = model)) + geom_abline(intercept = 1, slope = 4.5, linetype = "dashed", colour = "grey55") + geom_line(linewidth = 1.3) + annotate("point", x = 0, y = 1, size = 3.5) + scale_colour_manual(values = c("#8b2635", "#1f4e79")) + coord_cartesian(ylim = c(0, 5)) + scale_x_continuous(breaks = seq(-0.8, 0.4, by = 0.2)) + labs(x = "t", y = "m(t)", colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18), axis.text = element_text(size = 17), axis.title = element_text(size = 19)) ``` ::: {style="font-size: 28px"} Slope at 0 is the mean, curvature is $E(Y^2)$. Both counts average 4.5, so both curves leave $(0, 1)$ along the dashed tangent. The Poisson bends up faster: $\mu^{\text{ʹ}}_2 = 24.75$ against $22.725$. ::: --- ## 🔐 The Primary Use: Uniqueness ::: {style="font-size: 30px"} If $m(t)$ exists for $p(y)$, it is **unique**. If $m_Y(t) = m_Z(t)$ for all $|t| < b$, some $b > 0$, then $Y$ and $Z$ have the **same distribution**. ::: {.fragment} **Example 3.25, re-set.** A bank's monthly count of operational-loss events has $m_Y(t) = e^{3.2(e^t - 1)}$. ::: ::: {.fragment} That is the Poisson MGF with $\lambda = 3.2$, so $Y \sim$ Poisson(3.2): mean and variance 3.2, and a loss-free month has probability $e^{-3.2} = 0.0408$. ::: ::: --- ## 🧭 A Recognition Table ::: {style="font-size: 30px"} | Distribution | $m(t)$ | Source | |---|---|---| | Binomial$(n, p)$ | $(pe^t + q)^n$ | Exercise 3.145 | | Geometric$(p)$ | $\dfrac{pe^t}{1 - qe^t}$ | Exercise 3.147 | | Poisson$(\lambda)$ | $e^{\lambda(e^t - 1)}$ | Example 3.23 | ::: {.fragment} Of 12 SME loan applications, $Y$ approved has $m(t) = (0.3e^t + 0.7)^{12}$. So $Y \sim$ Bin(12, 0.3): $\mu = 3.6$, $\sigma^2 = 2.52$, $P(Y = 0) = 0.7^{12} = 0.0138$. ::: ::: --- ## 🧠 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 mobile operator's count of regional network outages per day has $$m(t) = 0.5 + 0.3e^{t} + 0.2e^{2t}$$ **Four minutes, in pairs:** 1. What is the distribution of $Y$? (Read it off: $m(t) = \sum_y e^{ty}p(y)$.) 2. Find $E(Y)$ and $V(Y)$ from $m(t)$. 3. Check both directly from $p(y)$. ::: --- ## ✅ Think-Pair-Share: Solution ::: {style="font-size: 30px"} 1. Match $\sum_y e^{ty}p(y)$ term by term: $p(0) = 0.5$, $p(1) = 0.3$, $p(2) = 0.2$. Uniqueness says that is **the** distribution. 2. $m^{\text{ʹ}}(t) = 0.3e^t + 0.4e^{2t}$, so $E(Y) = 0.7$. $m^{\text{ʹʹ}}(t) = 0.3e^t + 0.8e^{2t}$, so $E(Y^2) = 1.1$ and $V(Y) = 1.1 - 0.49 = 0.61$. ::: {.fragment} 3. Directly: $0(0.5) + 1(0.3) + 2(0.2) = 0.7$ and $0 + 0.3 + 4(0.2) = 1.1$. Same numbers — the MGF is just a disciplined way to do those sums. ::: ::: --- ## 📝 Quiz #1: Variance From an MGF {.quiz-question} Daily chargebacks at a card acquirer have $m(t) = e^{2.5(e^t - 1)}$. What is $V(Y)$? - [$2.5$]{.correct data-explanation="✅ This is the Poisson MGF with λ = 2.5, and a Poisson variance equals its mean. Via Theorem 3.12: m''(0) − m'(0)² = (2.5² + 2.5) − 2.5² = 2.5."} - $6.25$ - $8.75$ - $1.58$ --- ## 📝 Quiz #2: Name the Distribution {.quiz-question} Out of 15 export shipments, $Y$ are held at customs, with $m(t) = (0.2e^t + 0.8)^{15}$. What is the distribution of $Y$? - [Binomial with $n = 15$, $p = 0.2$]{.correct data-explanation="✅ (pe^t + q)^n with p = 0.2 multiplying e^t. By uniqueness Y ~ Bin(15, 0.2), with mean 3."} - Binomial with $n = 15$, $p = 0.8$ - Poisson with $\lambda = 3$ - Geometric with $p = 0.2$ --- ## 📝 Quiz #3: A Linear Change {.quiz-question} $Y$ has MGF $m_Y(t)$. A fee schedule charges $W = 2Y + 3$ AZN. What is $m_W(t)$? - [$e^{3t}\,m_Y(2t)$]{.correct data-explanation="✅ E(e^{t(2Y+3)}) = e^{3t} E(e^{(2t)Y}) = e^{3t} m_Y(2t). The shift comes out as a factor; the scale goes inside."} - $2\,m_Y(t) + 3$ - $e^{2t}\,m_Y(3t)$ - $m_Y(2t) + 3$ --- ## 📋 Key Formulas ::: {style="font-size: 30px"} | | Statement | |---|---| | Defs 3.12, 3.13 | $\mu^{\text{ʹ}}_k = E(Y^k)$, $\mu_k = E[(Y - \mu)^k]$ | | Def 3.14 | $m(t) = E(e^{tY}) = \sum_y e^{ty}p(y)$ | | Theorem 3.12 | $m^{(k)}(0) = \mu^{\text{ʹ}}_k$ | | mean, variance | $\mu = m^{\text{ʹ}}(0)$, $\sigma^2 = m^{\text{ʹʹ}}(0) - [m^{\text{ʹ}}(0)]^2$ | | linear change | $m_{aY+b}(t) = e^{bt}m_Y(at)$ | | binomial, Poisson | $(pe^t + q)^n$, $e^{\lambda(e^t - 1)}$ | ::: --- ## 📋 Summary ::: {style="font-size: 30px"} - Mean and variance do not determine a distribution; the full sequence of moments (usually) does - The MGF packs every moment into one function: $\mu^{\text{ʹ}}_k$ is the coefficient of $t^k/k!$ - Differentiate $k$ times and set $t = 0$ to read off $\mu^{\text{ʹ}}_k$ - A count turned into a P&L by $aY + b$ changes its MGF to $e^{bt}m_Y(at)$ - **Uniqueness** is the main use: recognise the MGF, and you know the distribution ::: --- ## 📚 Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - Exercises at the end of §3.9 — start with 3.145 and 3.146 (the binomial MGF), then 3.149, 3.153 – 3.155 - Exercises 3.156 – 3.157: the linear-change rule, and why a shift leaves the variance alone **Week 8, Problem Set 1** is open now and closes **Sunday 8 November at 23:59** on WeBWorK, covering §3.9. **Next class:** 31 October — probability-generating functions, Tchebysheff's theorem, and Chapter 3 in review (Wackerly §3.10–3.11). ::: --- ## 🙏 Thank You ::: {style="font-size: 34px"} **Dr. Samir Orujov** 📧 sorujov@ada.edu.az \ 🏢 Building D, Room D325 \ 🕓 Office hours: Wednesday, 16:00 – 18:00 Slides and readings: **sorujov.net/teaching** ::: --- ## ❓ Questions ::: {style="font-size: 32px"} - Branches A and B agreed on three moments. Can you build two counts that agree on four? - Why must every MGF satisfy $m(0) = 1$, and what does that say about the tangent line in today's figure? - If $Y_1$ and $Y_2$ are independent Poisson claim counts, what might $m_{Y_1 + Y_2}(t)$ be? ::: 