```{r} #| label: setup #| include: false set.seed(2026) library(ggplot2) theme_set(theme_minimal(base_size = 20)) ``` ## π― Learning Objectives ::: {style="font-size: 32px"} By the end of this lecture, you will be able to: - **State** Theorem 5.12 for the mean, variance and covariance of linear functions - **Compute** the expected return and variance of a two- and three-asset portfolio - **Explain** why the variance of an equally weighted portfolio falls towards a floor set by the average covariance - **Find** the hedge ratio that minimises the variance of a position - **Recognise** $\bar{Y}$ and $\hat{p}$ as linear functions, with $V(\bar{Y}) = \sigma^2/n$ ::: --- ## πΊοΈ Where We Are ::: {style="font-size: 32px"} **Wackerly Β§5.8** Last class ended on a question: the retailer's expected revenue needed only the marginals. **Which quantity about the revenue will need the covariance as well?** ::: {.fragment} The variance. Expectation is linear and ignores dependence; the variance of a sum is where $\text{Cov}(Y_1, Y_2)$ enters, and it enters with a factor of 2. ::: ::: {.fragment} In finance that one fact has a name: **portfolio variance**. It is why holding two risky assets can be less risky than holding either one. ::: ::: --- ## β Motivating Question ::: {style="font-size: 30px"} A Baku pension fund puts 100,000 AZN into two equity funds for a year: | | weight | mean return | s.d. of return | |---|---|---|---| | Energy fund $(Y_1)$ | 0.6 | 12% | 25% | | Telecom fund $(Y_2)$ | 0.4 | 8% | 15% | ::: {.fragment} The expected return is $0.6(12) + 0.4(8) = 10.4\%$, the weighted average. **Is the risk the weighted average too**, $0.6(25) + 0.4(15) = 21\%$? ::: ::: {.fragment} Only in one special case. Today's theorem says which, and what the risk is otherwise. ::: ::: --- ## π Theorem 5.12 ::: {style="font-size: 28px"} ::: {.callout-important} ## Theorem 5.12 Let $E(Y_i) = \mu_i$ and $E(X_j) = \xi_j$. Define $U_1 = \sum_{i=1}^n a_iY_i$ and $U_2 = \sum_{j=1}^m b_jX_j$ for constants $a_i$, $b_j$. Then a. $E(U_1) = \sum_{i=1}^n a_i\mu_i$ b. $V(U_1) = \sum_{i=1}^n a_i^2V(Y_i) + 2\sum\sum_{1 \le i < j \le n} a_ia_j\text{Cov}(Y_i, Y_j)$ c. $\text{Cov}(U_1, U_2) = \sum_{i=1}^n\sum_{j=1}^m a_ib_j\text{Cov}(Y_i, X_j)$ ::: ::: {.fragment} Part (a) is Theorems 5.7 and 5.8 again. Parts (b) and (c) are new, and (b) is a special case of (c), since $\text{Cov}(Y_i, Y_i) = V(Y_i)$. ::: ::: --- ## π Reading Part (b) for Two Variables ::: {style="font-size: 30px"} $$V(a_1Y_1 + a_2Y_2) = a_1^2V(Y_1) + a_2^2V(Y_2) + 2a_1a_2\text{Cov}(Y_1, Y_2)$$ ::: {.fragment} - Constants come out **squared**, so $V(-Y) = V(Y)$ - The sign of the covariance term depends on the signs of $a_1a_2$ **and** of the covariance - With $a_1 = 1$, $a_2 = -1$: $\; V(Y_1 - Y_2) = V(Y_1) + V(Y_2) - 2\text{Cov}(Y_1, Y_2)$ ::: ::: {.fragment} The proof expands $\left[\sum a_i(Y_i - \mu_i)\right]^2$: the $n$ squared terms give the variances, and the $n(n-1)$ cross terms pair up into $2\sum\sum_{i input.addEventListener(e, ev => ev.stopPropagation())); return input; } ``` ```{ojs} //| echo: false sdP = (w, r) => Math.sqrt(w*w*625 + (1-w)*(1-w)*225 + 2*w*(1-w)*r*375) riskCurve = Array.from({length: 101}, (_, i) => ({w: i/100, s: sdP(i/100, rho_ET)})) wStar = Math.min(1, Math.max(0, (225 - rho_ET*375) / (625 + 225 - 2*rho_ET*375))) best = [{w: wStar, s: sdP(wStar, rho_ET)}] md`Lowest risk: **${best[0].s.toFixed(1)}%** with **${(100*wStar).toFixed(0)}%** in energy.` Plot.plot({ width: 1150, height: 300, marginLeft: 78, marginBottom: 58, style: {fontSize: "18px"}, x: {label: "Weight w in the energy fund", domain: [0, 1]}, y: {label: "Portfolio s.d. (%)", domain: [0, 26]}, marks: [ Plot.line([{w: 0, s: 15}, {w: 1, s: 25}], {x: "w", y: "s", stroke: "#cbb8a9", strokeDasharray: "4 4"}), Plot.line(riskCurve, {x: "w", y: "s", stroke: "#14130f", strokeWidth: 2.5}), Plot.dot(best, {x: "w", y: "s", r: 8, fill: "#8b2635"}), Plot.ruleY([0]) ] }) ``` --- ## ποΈ Three Assets: Adding a Bond Fund ::: {style="font-size: 28px"} The fund rebalances to 50% energy, 30% telecom and 20% an AZN government bond fund $(Y_3)$: mean 6%, s.d. 5%. | | $\text{Cov}$ with telecom | $\text{Cov}$ with bonds | |---|---|---| | **Energy** | $0.3(25)(15) = 112.5$ | $-0.2(25)(5) = -25$ | | **Telecom** | | $0.1(15)(5) = 7.5$ | ::: {.fragment} Theorem 5.12(b): three variance terms and **three** pairs, $$V(R_p) = 156.25 + 20.25 + 1 + 2(.5)(.3)(112.5) + 2(.5)(.2)(-25) + 2(.3)(.2)(7.5)$$ $$= 177.5 + 33.75 - 5 + 0.9 = 207.15, \qquad \sigma_{R_p} = 14.39\%$$ ::: ::: --- ## π» The Same Sum as a Matrix ```{r} #| label: three-asset w <- c(energy = 0.5, telecom = 0.3, bonds = 0.2) mu <- c(12, 8, 6) sd <- c(25, 15, 5) R <- matrix(c( 1.0, 0.3, -0.2, 0.3, 1.0, 0.1, -0.2, 0.1, 1.0), 3, 3) Sigma <- diag(sd) %*% R %*% diag(sd) # covariance matrix c(mean = sum(w * mu), var = drop(t(w) %*% Sigma %*% w), sd = sqrt(drop(t(w) %*% Sigma %*% w))) ``` ::: {style="font-size: 28px"} $w^\top\Sigma w$ is Theorem 5.12(b) with all $n^2$ terms written out: each off-diagonal pair appears twice, which is the factor 2. ::: --- ## π $n$ Equally Weighted Assets ::: {style="font-size: 28px"} Take $n$ assets, each with variance $\sigma^2$ and every pair with covariance $\rho\sigma^2$. Put $a_i = 1/n$ in Theorem 5.12(b): $$V(R_p) = n\cdot\frac{1}{n^2}\sigma^2 + 2\cdot\frac{n(n-1)}{2}\cdot\frac{1}{n^2}\rho\sigma^2 = \frac{\sigma^2}{n} + \left(1 - \frac{1}{n}\right)\rho\sigma^2$$ ::: {.fragment} - With $\rho = 0$ this is **Example 5.27**: $V(\bar{Y}) = \sigma^2/n$, which falls to zero - With $\rho > 0$ it falls only to $\rho\sigma^2$, the risk **no amount of diversification removes** ::: ::: {.fragment} The first term is risk specific to each asset; the second is risk shared by the whole market. ::: ::: --- ## π Diversification Has a Floor ```{r} #| label: diversification #| fig-width: 10 #| fig-height: 4 n <- 1:40; s <- 30 df <- rbind(data.frame(n = n, sd = sqrt(s^2 / n + (1 - 1/n) * 0.25 * s^2), case = "r25"), data.frame(n = n, sd = sqrt(s^2 / n), case = "r0")) ggplot(df, aes(n, sd, colour = case)) + geom_hline(yintercept = 15, linetype = "dashed", colour = "grey55") + geom_line(linewidth = 1.4) + scale_colour_manual(values = c(r25 = "#8b2635", r0 = "#3d6e8f"), labels = c(r25 = expression(rho == 0.25), r0 = expression(rho == 0)), name = NULL) + scale_x_continuous(breaks = c(1, 5, 10, 20, 30, 40)) + scale_y_continuous(breaks = seq(0, 30, 5), limits = c(0, 30)) + labs(x = "Number of shares held", y = "Portfolio s.d. (%)") + theme(legend.position = c(0.8, 0.8), legend.text = element_text(size = 22, hjust = 0), legend.key.width = unit(1.6, "cm")) ``` ::: {style="font-size: 28px"} Shares with $\sigma = 30\%$: at $\rho = 0.25$, ten shares give 17.1% and thirty give 15.7%, never below $\sqrt{0.25 \times 900} = 15\%$. ::: --- ## π‘οΈ Hedging With a Negative Covariance ::: {style="font-size: 30px"} A Sumqayit fuel distributor's monthly margin $Y_1$ (thousand AZN) shrinks when crude prices rise. It buys $h$ oil-linked contracts, each with value change $Y_2$. The position is $H = Y_1 + hY_2$. ::: {.fragment} $$V(H) = V(Y_1) + h^2V(Y_2) + 2h\,\text{Cov}(Y_1, Y_2)$$ A parabola in $h$. Setting the derivative to zero: $$h^* = -\frac{\text{Cov}(Y_1, Y_2)}{V(Y_2)}, \qquad V(H^*) = V(Y_1)(1 - \rho^2)$$ ::: ::: {.fragment} Negative covariance makes $h^* > 0$: **buy** the asset that moves against you. ::: ::: --- ## β½ Worked Example: The Hedge Ratio ::: {style="font-size: 30px"} $\sigma_1 = 40$, $\sigma_2 = 8$ (thousand AZN), $\rho = -0.8$, so $\text{Cov}(Y_1, Y_2) = -0.8(40)(8) = -256$. ::: {.fragment} $$h^* = \frac{256}{64} = 4 \text{ contracts}$$ $$V(H^*) = 1600 + 16(64) + 2(4)(-256) = 1600 + 1024 - 2048 = 576$$ ::: ::: {.fragment} The s.d. falls from 40 to $\sqrt{576} = 24$ thousand AZN, as $40\sqrt{1 - 0.64} = 24$ predicts. ::: ::: {.fragment} A hedge removes the **correlated** part of the risk. The remaining 24 comes from what the contract does not track. ::: ::: --- ## π§ 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"} Back to the energy fund $Y_1$ and telecom fund $Y_2$: means 12 and 8, variances 625 and 225, $\text{Cov}(Y_1, Y_2) = 112.5$. Client A holds $U = 0.5Y_1 + 0.5Y_2$. Client B runs a long-short book, $W = Y_1 - Y_2$. **Four minutes, in pairs:** 1. Find $E(W)$ and $V(W)$. 2. Use Theorem 5.12(c) to find $\text{Cov}(U, W)$. 3. Can $U$ and $W$ be independent? ::: --- ## β
Think-Pair-Share: Solution ::: {style="font-size: 28px"} 1. $E(W) = 12 - 8 = 4\%$ and $V(W) = 625 + 225 - 2(112.5) = 625$, so $\sigma_W = 25\%$. 2. With $a = (0.5, 0.5)$ and $b = (1, -1)$, four terms: $$\text{Cov}(U, W) = 0.5(625) - 0.5(112.5) + 0.5(112.5) - 0.5(225) = 200$$ ::: {.fragment} 3. No. Theorem 5.11 says independence forces zero covariance, and $200 \ne 0$. With $V(U) = 268.75$, $\rho_{UW} = 200/(16.39 \times 25) = 0.49$. The long-short book looks "market neutral", yet its return still moves with client A's portfolio. ::: ::: --- ## π Quiz #1: A Difference {.quiz-question} $V(Y_1) = 4$, $V(Y_2) = 9$ and $\text{Cov}(Y_1, Y_2) = 2$. What is $V(Y_1 - Y_2)$? - [$9$]{.correct data-explanation="β
Theorem 5.12(b) with a = (1, β1): 4 + 9 β 2(2) = 9. The constants are squared, and the cross term takes the sign of the product of the constants, (1)(β1) = β1."} - $17$ - $13$ - $5$ --- ## π Quiz #2: Sixteen Independent Shares {.quiz-question} Sixteen independent shares each have s.d. 20%. What is the s.d. of the equally weighted portfolio? - [$5\%$]{.correct data-explanation="β
With zero covariances V = ΟΒ²/n = 400/16 = 25, so the s.d. is Ο/βn = 20/4 = 5%."} - $20\%$ - $1.25\%$ - $10\%$ --- ## π Quiz #3: A Default-Rate Estimate {.quiz-question} A bank reviews $n = 400$ loans, each defaulting independently with $p = 0.05$, and estimates the rate by $\hat{p} = Y/n$. What is the s.d. of $\hat{p}$? - [$0.0109$]{.correct data-explanation="β
Example 5.28: V(pΜ) = pq/n = (0.05)(0.95)/400 = 0.00011875, and its square root is 0.0109."} - $0.00012$ - $0.2179$ - $0.0475$ --- ## π Key Formulas ::: {style="font-size: 28px"} | | Statement | |---|---| | Theorem 5.12(a) | $E\left(\sum a_iY_i\right) = \sum a_i\mu_i$ | | Theorem 5.12(b) | $V\left(\sum a_iY_i\right) = \sum a_i^2V(Y_i) + 2\sum\sum_{i 