```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 18)) ``` ## 🎯 Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **Compute** moments $\mu^{\text{ʹ}}_k$ and the moment-generating function of a continuous random variable - **Derive** the gamma and normal MGFs, and read moments off them - **Prove** Tchebysheff's theorem for a density, and use it when only $\mu$ and $\sigma$ are trusted - **Evaluate** $E[g(Y)]$ for a payout with a break in it, and for a loss with a point mass at zero - **Assemble** Chapter 4 into one toolkit ::: --- ## 🗺️ Where We Are ::: {style="font-size: 30px"} **Wackerly §§4.9–4.11** Wednesday finished the catalogue with the beta, and a rule for choosing among the densities: **theory, then support, then the histogram.** ::: {.fragment} Today: other expected values, Tchebysheff's theorem for continuous variables, and Chapter 4 in review. Three questions a Baku risk office asks every week: 1. A fund's log return is normal. What is the **expected value** of the money, $E(e^{R})$? 2. Only $\mu$ and $\sigma$ are known. How often can a day move **$3\sigma$**? 3. 90% of motor policies pay nothing. What does a policy cost **after a deductible**? ::: ::: --- ## 📐 Definitions 4.13 and 4.14 ::: {style="font-size: 29px"} ::: {.callout-note} ## Definitions 4.13 and 4.14 For a continuous $Y$, the $k$th moment **about the origin** is $\mu^{\text{ʹ}}_k = E(Y^k)$ and the $k$th **central** moment is $\mu_k = E[(Y - \mu)^k]$. The **moment-generating function** is $m(t) = E(e^{tY})$. It exists if $m(t)$ is finite for $|t| \le b$, for some $b > 0$. ::: ::: {.fragment} The integral replaces the sum, and nothing else changes. Expand $e^{ty}$ inside the integral: $$m(t) = 1 + t\mu^{\text{ʹ}}_1 + \frac{t^2}{2!}\mu^{\text{ʹ}}_2 + \cdots, \qquad m^{(k)}(0) = \mu^{\text{ʹ}}_k$$ so Theorem 3.12 holds for continuous variables as it stands. ::: ::: --- ## 📐 Example 4.13: The Gamma MGF ::: {style="font-size: 29px"} $$m(t) = \int_0^\infty e^{ty}\,\frac{y^{\alpha-1}e^{-y/\beta}}{\beta^\alpha\Gamma(\alpha)}\,dy = \frac{1}{\beta^\alpha\Gamma(\alpha)}\int_0^\infty y^{\alpha-1}\exp\!\left[\frac{-y}{\beta/(1-\beta t)}\right]dy$$ ::: {.fragment} For $t < 1/\beta$ the integrand is the **variable part of a gamma density** with scale $\beta/(1-\beta t)$. Its integral is the reciprocal of that density's constant, $[\beta/(1-\beta t)]^{\alpha}\,\Gamma(\alpha)$. ::: ::: {.fragment} $$m(t) = \frac{1}{(1-\beta t)^{\alpha}}, \qquad t < \frac{1}{\beta}$$ **Recognise a kernel; do not integrate.** The trick recurs all semester. ::: ::: --- ## 🔥 Example 4.14: Gamma Moments ::: {style="font-size: 29px"} Expand $(1-\beta t)^{-\alpha} = 1 + t(\alpha\beta) + \dfrac{t^2}{2!}\,\alpha(\alpha+1)\beta^2 + \cdots$, so $$\mu^{\text{ʹ}}_k = \alpha(\alpha+1)\cdots(\alpha+k-1)\,\beta^k$$ ::: {.fragment} Daily gas demand $Y$ at a distribution node (million m³) is gamma with $\alpha = 4$, $\beta = 2.5$: $$\mu = 10, \qquad \mu^{\text{ʹ}}_2 = 4 \times 5 \times 2.5^2 = 125, \qquad \sigma^2 = 125 - 10^2 = 25 = \alpha\beta^2$$ ::: ::: {.fragment} A balancing cost of $C = 0.8\,Y^2$ thousand AZN has $E(C) = 0.8 \times 125 = 100$, **not** $0.8 \times 10^2 = 80$. $E(Y^2) \ne [E(Y)]^2$. ::: ::: --- ## 📐 Example 4.16: The Normal MGF ::: {style="font-size: 29px"} **Theorem 4.12:** the MGF of $g(Y)$ is $E[e^{t g(Y)}] = \int e^{t g(y)} f(y)\,dy$. ::: {.fragment} Take $Y \sim N(\mu, \sigma^2)$ and $g(Y) = Y - \mu$. Complete the square in the exponent; what remains is a normal density with mean $\sigma^2 t$, which integrates to 1: $$m_{Y-\mu}(t) = e^{t^2\sigma^2/2}$$ ::: ::: {.fragment} Since $e^{tY} = e^{\mu t}e^{t(Y-\mu)}$, Exercise 4.138 gives $\; m_Y(t) = \exp\!\left(\mu t + \tfrac{1}{2}t^2\sigma^2\right)$. **Uniqueness:** identical MGFs mean identical distributions. An MGF of this form *is* a normal. ::: ::: --- ## 💰 The Value of a Fund ::: {style="font-size: 28px"} 100,000 AZN sits in an equity fund whose annual log return is $R \sim N(0.08,\, 0.20^2)$. The value in a year is $V = 100{,}000\,e^{R}$, and $E(e^{R})$ is the MGF at $t = 1$: $$E(V) = 100{,}000\; e^{0.08 + 0.02} = 110{,}517 \text{ AZN}$$ ::: ```{r} #| label: lognormal #| code-fold: false V <- 100000 * exp(rnorm(1e6, mean = 0.08, sd = 0.20)) round(c(mean = mean(V), median = median(V), exp_mu = 100000 * exp(0.08))) ``` ::: {style="font-size: 28px"} $100{,}000\,e^{0.08} = 108{,}329$ is the **median**. The extra $\sigma^2/2$ is what volatility adds to the mean. ::: --- ## 📐 Theorem 4.13: Tchebysheff ::: {style="font-size: 29px"} ::: {.callout-important} ## Theorem 4.13 Let $Y$ be a random variable with finite mean $\mu$ and variance $\sigma^2$. Then, for any $k > 0$, $$P(|Y - \mu| < k\sigma) \ge 1 - \frac{1}{k^2} \qquad \text{or} \qquad P(|Y - \mu| \ge k\sigma) \le \frac{1}{k^2}$$ ::: ::: {.fragment} Theorem 3.14 again, now with the continuous proof. A Baku bank's share has daily return mean 0.04% and sd 1.5%. A move of 4.5 points from the mean is $k = 3$: $$P(|R - \mu| \ge 3\sigma) \le 1/9 = 0.111 \quad \text{for any distribution; a normal gives } 0.0027.$$ ::: ::: --- ## ✏️ The Proof: Throw Away the Middle ::: {style="font-size: 29px"} Split the variance at $\mu \pm k\sigma$: $$\sigma^2 = \int_{-\infty}^{\mu-k\sigma}(y-\mu)^2 f(y)\,dy + \int_{\mu-k\sigma}^{\mu+k\sigma}(y-\mu)^2 f(y)\,dy + \int_{\mu+k\sigma}^{\infty}(y-\mu)^2 f(y)\,dy$$ ::: {.fragment} The middle integral is $\ge 0$: drop it. In both tails $(y-\mu)^2 \ge k^2\sigma^2$: replace it. $$\sigma^2 \ge k^2\sigma^2\left[P(Y \le \mu - k\sigma) + P(Y \ge \mu + k\sigma)\right] = k^2\sigma^2\,P(|Y-\mu| \ge k\sigma)$$ ::: ::: {.fragment} Divide by $k^2\sigma^2$. Two things were thrown away: the middle mass, and the tails' excess over $k^2\sigma^2$. **That waste is why the bound is loose.** ::: ::: --- ## 🏦 Example 4.17: A Slow Branch ::: {style="font-size: 28px"} Mortgage approval times $Y$ (days) at a Baku bank are gamma with $\alpha = 3.1$, $\beta = 2$: $\mu = 6.2$, $\sigma^2 = 12.4$, $\sigma = 3.52$. A new branch takes **22.5 days**. 22.5 exceeds $\mu$ by 16.3 days, or $k = 16.3/3.52 = 4.63$ standard deviations: $$P(|Y - 6.2| \ge 16.3) \le \frac{1}{4.63^2} = 0.0466$$ ::: ```{r} #| label: cheb-exact #| code-fold: false pgamma(22.5, shape = 3.1, scale = 2, lower.tail = FALSE) # exact P(Y >= 22.5) ``` ::: {style="font-size: 28px"} The bound says *under 5%* with no integral; the model says about **1 in 870**. Either way, look at the branch. ::: --- ## 📈 How Loose Is the Bound? ```{r} #| label: cheb-figure #| echo: false #| fig-width: 10 #| fig-height: 4.3 kk <- seq(1, 5, by = 0.01) g_mu <- 6.2; g_s <- sqrt(12.4) tails <- rbind( data.frame(k = kk, model = "Normal", p = 2 * pnorm(-kk)), data.frame(k = kk, model = "Gamma(3.1, 2)", p = pgamma(g_mu + kk * g_s, 3.1, scale = 2, lower.tail = FALSE) + pgamma(pmax(0, g_mu - kk * g_s), 3.1, scale = 2)), data.frame(k = kk, model = "Exponential", p = exp(-(1 + kk)))) tails$model <- factor(tails$model, levels = c("Exponential", "Gamma(3.1, 2)", "Normal")) ggplot(tails, aes(k, p, colour = model)) + geom_line(linewidth = 1.3) + geom_line(data = data.frame(k = kk, p = 1 / kk^2), aes(k, p), inherit.aes = FALSE, colour = "#8b2635", linewidth = 1.4, linetype = "dashed") + annotate("text", x = 4.2, y = 0.3, label = "Tchebysheff:~1/k^2", parse = TRUE, colour = "#8b2635", size = 6.5) + scale_y_log10(breaks = c(1e-4, 1e-3, 1e-2, 1e-1, 1), labels = c("0.0001", "0.001", "0.01", "0.1", "1")) + coord_cartesian(ylim = c(1e-4, 1)) + scale_colour_manual(values = c("#b3833a", "#3d6e8f", "#14130f")) + labs(x = "k (distance from the mean in standard deviations)", y = expression(P(abs(Y - mu) >= k * sigma)), colour = NULL) + theme(legend.position = "top", legend.text = element_text(size = 18)) ``` ::: {style="font-size: 28px"} Log scale. At $k = 3$ the bound is 0.111; the exponential's tail is 0.018, the normal's 0.0027. ::: --- ## 🧠 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 monthly data use per subscriber, $Y$ (GB), has MGF $$m(t) = (1 - 4t)^{-3}, \qquad t < 1/4$$ **Four minutes, in pairs:** 1. Name the distribution of $Y$, and find $\mu$ and $\sigma$. 2. Give an interval that holds at least 75% of subscribers. 3. The fair-use cap is 40 GB. What does Tchebysheff say about the share above it? ::: --- ## ✅ Think-Pair-Share: Solution ::: {style="font-size: 29px"} 1. By uniqueness, $Y$ is **gamma with $\alpha = 3$, $\beta = 4$**: $\mu = 12$ GB, $\sigma^2 = 48$, $\sigma = 6.93$. 2. $1 - 1/k^2 = 0.75$ gives $k = 2$: $12 \pm 13.86 = (-1.86,\ 25.86)$. Usage cannot be negative, so **$[0,\ 25.86)$ holds at least 75%**. The gamma itself puts 0.956 there. ::: {.fragment} 3. $40 - 12 = 28$ GB is $k = 28/6.93 = 4.04$ standard deviations: $$P(Y \ge 40) \le P(|Y - 12| \ge 28) \le \frac{48}{28^2} = 0.061$$ At most 6.1%, from two numbers. The gamma gives 0.0028: the cap binds for about **1 subscriber in 360**. ::: ::: --- ## 🧩 Optional §4.11: Point Masses ::: {style="font-size: 28px"} Two things Chapter 4 has not yet handled: - a **payout with a break** in it, $g(Y)$: Theorem 4.4 still gives $E[g(Y)] = \int g(y)f(y)\,dy$, split at the break - a variable with a **point mass**: most policies pay exactly 0, the rest pay a continuous amount ::: {.fragment} ::: {.callout-note} ## Definition 4.15 If $F(y) = c_1F_1(y) + c_2F_2(y)$, with $F_1$ the distribution function of a discrete $X_1$, $F_2$ that of a continuous $X_2$, and $c_1 + c_2 = 1$, then $$E[g(Y)] = c_1E[g(X_1)] + c_2E[g(X_2)]$$ ::: ::: ::: --- ## 🚗 Worked Example: A Motor Book ::: {style="font-size: 28px"} A policy pays nothing with probability $c_1 = 0.9$. Given a claim, the loss is exponential with mean 2,000 AZN. So $X_1 \equiv 0$ and $X_2 \sim$ Exp(2000): $$E(Y) = 0.1 \times 2000 = 200, \qquad E(Y^2) = 0.1 \times 2 \times 2000^2 = 800{,}000$$ $$V(Y) = 800{,}000 - 200^2 = 760{,}000, \qquad \sigma = 872 \text{ AZN}$$ ::: {.fragment} With a **500 AZN deductible** the insurer pays $g(y) = \max(y - 500, 0)$: $$E[g(X_2)] = \int_{500}^{\infty}(x - 500)\,\frac{e^{-x/2000}}{2000}\,dx = 2000\,e^{-0.25} = 1557.6$$ $E[g(Y)] = 0.9 \times 0 + 0.1 \times 1557.6 = 155.76$ **AZN**: the deductible cuts the pure premium by 22%. ::: ::: --- ## 💻 The Motor Book, Simulated ```{r} #| label: mixed-sim #| code-fold: false n <- 1e6 Y <- ifelse(runif(n) < 0.1, rexp(n, rate = 1 / 2000), 0) round(c(mean = mean(Y), sd = sd(Y), deductible = mean(pmax(Y - 500, 0)), franchise = mean(ifelse(Y > 500, Y, 0))), 1) ``` ```{r} #| label: mixed-cdf #| echo: false #| fig-width: 10 #| fig-height: 2.5 yy <- seq(0, 10000, by = 10) ggplot(data.frame(y = yy, F = 0.9 + 0.1 * (1 - exp(-yy / 2000))), aes(y, F)) + geom_segment(aes(x = -1500, xend = 0, y = 0, yend = 0), linewidth = 1.3, colour = "#14130f") + geom_line(linewidth = 1.3, colour = "#14130f") + annotate("point", x = 0, y = 0, shape = 21, size = 4, fill = "white") + annotate("point", x = 0, y = 0.9, size = 4) + annotate("text", x = 1900, y = 0.45, label = "jump of 0.9 at y = 0", colour = "#8b2635", size = 6) + scale_y_continuous(limits = c(0, 1), breaks = c(0, 0.5, 1)) + labs(x = "Annual payout per policy (AZN)", y = "F(y)") ``` ::: {style="font-size: 28px"} A **franchise** deductible pays the whole loss once it passes 500: $0.1 \times 2500\,e^{-0.25} = 194.70$ AZN. ::: --- ## 🧭 Chapter 4 in Review ::: {style="font-size: 28px"} | Model | Support | $E(Y)$ | $V(Y)$ | $m(t)$ | |---|---|---|---|---| | Uniform$(\theta_1, \theta_2)$ | $[\theta_1, \theta_2]$ | $(\theta_1+\theta_2)/2$ | $(\theta_2-\theta_1)^2/12$ | $\dfrac{e^{t\theta_2}-e^{t\theta_1}}{t(\theta_2-\theta_1)}$ | | Normal$(\mu, \sigma^2)$ | $(-\infty, \infty)$ | $\mu$ | $\sigma^2$ | $e^{\mu t + t^2\sigma^2/2}$ | | Gamma$(\alpha, \beta)$ | $(0, \infty)$ | $\alpha\beta$ | $\alpha\beta^2$ | $(1-\beta t)^{-\alpha}$ | | Exponential$(\beta)$ | $(0, \infty)$ | $\beta$ | $\beta^2$ | $(1-\beta t)^{-1}$ | | $\chi^2(\nu)$ | $(0, \infty)$ | $\nu$ | $2\nu$ | $(1-2t)^{-\nu/2}$ | | Beta$(\alpha, \beta)$ | $[0, 1]$ | $\alpha/(\alpha+\beta)$ | $\dfrac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}$ | no simple form | **The method:** pick the model by support and theory; any $E[g(Y)]$ by Theorem 4.4; moments from the MGF; Tchebysheff when only $\mu$, $\sigma$ are trusted. ::: --- ## 📝 Quiz #1: Reading an MGF {.quiz-question} A trader's quarterly FX gain (thousand AZN) has MGF $m(t) = e^{2t + 8t^2}$. What is its distribution? - [Normal with $\mu = 2$ and $\sigma^2 = 16$]{.correct data-explanation="✅ Match e^(μt + t²σ²/2): μ = 2 and σ²/2 = 8, so σ² = 16. By uniqueness the gain is normal."} - Normal with $\mu = 2$ and $\sigma^2 = 8$ - Normal with $\mu = 2$ and $\sigma^2 = 4$ - Gamma with $\alpha = 2$ and $\beta = 8$ --- ## 📝 Quiz #2: A Guarantee From Two Numbers {.quiz-question} A factory's monthly electricity bill has mean 42,000 AZN and sd 4,000 AZN, distribution unknown. At least what share of months fall within 10,000 AZN of the mean? - [$0.84$]{.correct data-explanation="✅ 10,000 = 2.5σ, so Tchebysheff gives 1 − 1/2.5² = 1 − 0.16 = 0.84."} - $0.16$ - $0.60$ - $0.99$ --- ## 📝 Quiz #3: A Call Option {.quiz-question} A Baku-listed bank's share price in one month is uniform on $(18, 22)$ AZN. A call struck at 20 pays $\max(S - 20, 0)$. What is its expected payoff? - [$0.50$ AZN]{.correct data-explanation="✅ The payout has a kink at the strike, so split there: ∫ from 20 to 22 of (s − 20)/4 ds = 2²/(2 × 4) = 0.50."} - $0$ AZN, since $E(S) = 20$ equals the strike - $1.00$ AZN - $2.00$ AZN --- ## 📋 Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Definition 4.13 | $\mu^{\text{ʹ}}_k = E(Y^k)$, $\quad \mu_k = E[(Y-\mu)^k]$ | | Definition 4.14, Theorem 3.12 | $m(t) = E(e^{tY})$, $\quad m^{(k)}(0) = \mu^{\text{ʹ}}_k$ | | gamma (Examples 4.13, 4.14) | $m(t) = (1-\beta t)^{-\alpha}$, $\quad \mu^{\text{ʹ}}_k = \alpha(\alpha+1)\cdots(\alpha+k-1)\beta^k$ | | normal (Example 4.16) | $m(t) = \exp(\mu t + t^2\sigma^2/2)$ | | Theorem 4.13 | $P(\lvert Y - \mu\rvert \ge k\sigma) \le 1/k^2$ | | Definition 4.15 | $E[g(Y)] = c_1E[g(X_1)] + c_2E[g(X_2)]$ | ::: --- ## 📋 Summary ::: {style="font-size: 30px"} - Moments and MGFs carry over from Chapter 3 with the sum replaced by an integral - The gamma MGF comes from **recognising a kernel**; the normal MGF from completing the square - $E(e^{R}) = e^{\mu + \sigma^2/2}$ for a normal log return: volatility raises the mean - Tchebysheff needs only $\mu$ and $\sigma$; it is loose because the proof throws mass away - Split $E[g(Y)]$ at a break; weight a point mass and a density by $c_1$, $c_2$ - Chapter 4 in one line: **support, model, Theorem 4.4, MGF, bound** ::: --- ## 📚 Practice Problems ::: {style="font-size: 28px"} **Wackerly, 7th edition** - §4.9: Exercises 4.136 – 4.145; start with 4.137, 4.138, 4.140 and 4.143 - §4.10: Exercises 4.146 – 4.154; start with 4.146, 4.147, 4.150 and 4.154(a, b) - §4.11 (optional): Exercises 4.155 – 4.157, the deductible logic in three settings **Week 11, Problem Set 2** is open now and closes **Sunday 29 November at 23:59** on WeBWorK, covering §§4.9–4.11. **Next class:** 25 November, Chapter 5 begins: bivariate and multivariate probability distributions (Wackerly §§5.1–5.2). ::: --- ## 🙏 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"} - The beta has no simple MGF. How would you find its third moment anyway? - Why does $E(e^{R})$ exceed $e^{E(R)}$ for any non-degenerate $R$, normal or not? - Tchebysheff gives the motor book and a normal with the same $\mu$ and $\sigma$ the same bound. For which is it closer to the truth, and why? ::: 