Mathematical Statistics

Expected Values of Functions of Random Variables and Covariance

Samir Orujov, PhD

ADA University, School of Business

Information Communication Technologies Agency, Statistics Unit

2026-09-24

๐ŸŽฏ Learning Objectives

By the end of this lecture, you will be able to:

  • Compute \(E[g(Y_1, Y_2)]\) from a joint probability table or a joint density (Definition 5.9)

  • Use Theorems 5.6โ€“5.8 to take expectations of sums term by term, with or without independence

  • Apply Theorem 5.9 to factor \(E[g(Y_1)h(Y_2)]\) when \(Y_1\) and \(Y_2\) are independent

  • Compute \(\text{Cov}(Y_1, Y_2)\) and \(\rho\) by Theorem 5.10, and read their sign

  • Explain why independence implies zero covariance, but zero covariance does not imply independence

๐Ÿ—บ๏ธ Where We Are

Wackerly ยงยง5.5โ€“5.7

Wednesday (ยง5.4): independence means the joint distribution is the product of the marginals, everywhere. It ended on co-movement: at correlation 0.6, \(P(\text{both fall}) = 0.352\), not \(0.5 \times 0.5 = 0.25\).

We used that word, correlation, before defining it. Today we define it, as one expected value of a function of two variables: \(E[(Y_1 - \mu_1)(Y_2 - \mu_2)]\).

First, the general tool: \(E[g(Y_1, Y_2)]\) for total revenue, a product of shares, a spread.

โ“ Motivating Question

The appliance retailerโ€™s peak hour

A Baku home-appliance store sells air conditioners at 1,200 AZN and refrigerators at 900 AZN. In the Saturday peak hour it sells \(Y_1\) air conditioners and \(Y_2\) refrigerators.

What revenue should it expect in that hour, and do the two product lines sell together?

Revenue \(R = 1200Y_1 + 900Y_2\) is a function of two random variables. Chapter 4 told us how to find \(E[g(Y)]\) for one.

๐Ÿ“ Definition 5.9

Definition 5.9

Let \(g(Y_1, \ldots, Y_k)\) be a function of discrete random variables with probability function \(p(y_1, \ldots, y_k)\). Then \[E[g(Y_1, \ldots, Y_k)] = \sum_{\text{all } y_k} \cdots \sum_{\text{all } y_1} g(y_1, \ldots, y_k)\, p(y_1, \ldots, y_k).\] For continuous variables with joint density \(f\), the sums become integrals over \((-\infty, \infty)\).

In words: weight each value of \(g\) by the probability of the cell that produces it, and add. Taking \(g(Y_1, Y_2) = Y_1\) gives back Definition 4.5 through the marginal density.

๐Ÿฌ Worked Example: The Peak Hour

\(y_1 \backslash y_2\) 0 1 2 \(p_1\)
0 .30 .15 .05 .50
1 .10 .15 .10 .35
2 .02 .05 .08 .15
\(p_2\) .42 .35 .23 1

Each cell carries a revenue \(g(y_1, y_2) = 1200y_1 + 900y_2\): from 0 AZN (nothing sold) to 4200 AZN (two of each).

Definition 5.9, cell by cell: \[E(R) = 0(.30) + 900(.15) + 1800(.05) + \cdots + 4200(.08) = \mathbf{1509} \text{ AZN}\]

๐Ÿ“ Theorems 5.6โ€“5.8: Special Theorems

Theorems 5.6, 5.7, 5.8

For a constant \(c\) and functions \(g, g_1, \ldots, g_k\) of \(Y_1\) and \(Y_2\): \[E(c) = c, \qquad E[c\,g(Y_1, Y_2)] = c\,E[g(Y_1, Y_2)],\] \[E[g_1(Y_1, Y_2) + \cdots + g_k(Y_1, Y_2)] = E[g_1(Y_1, Y_2)] + \cdots + E[g_k(Y_1, Y_2)].\]

The proofs are the univariate ones: a sum or an integral is linear. Nothing here assumes independence. The expectation of a sum is the sum of the expectations, however the terms are related.

โšก Expected Revenue the Short Way

With \(g_1 = 1200Y_1\) and \(g_2 = 900Y_2\), Theorems 5.7 and 5.8 give \[E(R) = 1200E(Y_1) + 900E(Y_2) = 1200(0.65) + 900(0.81) = 780 + 729 = 1509 \text{ AZN}.\] Only the marginals were needed. The same 1509 as nine cells of arithmetic:

Code 
p <- matrix(c(.30, .15, .05,
              .10, .15, .10,
              .02, .05, .08), nrow = 3, byrow = TRUE)
y <- 0:2
R <- outer(1200 * y, 900 * y, "+")          # revenue in each cell
c(by_definition_5.9 = sum(R * p),
  by_linearity      = 1200 * sum(y * rowSums(p)) + 900 * sum(y * colSums(p)))
by_definition_5.9      by_linearity 
             1509              1509

โšก Worked Example: A Power Plant

A gas-fired plant sells a share \(Y_1\) of its monthly output on the spot market (the rest under contract), and a share \(Y_2\) of those spot sales clears at peak-hour prices. Model: \[f(y_1, y_2) = 2(1 - y_1), \qquad 0 \le y_1 \le 1,\; 0 \le y_2 \le 1.\]

\(Y_1 Y_2\) is the share of total output sold at the peak spot price. By Definition 5.9, \[E(Y_1Y_2) = \int_0^1\!\!\int_0^1 2y_1y_2(1 - y_1)\,dy_2\,dy_1 = \int_0^1 (y_1 - y_1^2)\,dy_1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.\]

On 180 GWh of monthly output, the plant should expect 30 GWh sold at the peak spot price.

๐Ÿ“ Theorem 5.9: Products Split

Theorem 5.9

If \(Y_1\) and \(Y_2\) are independent and \(g(Y_1)\), \(h(Y_2)\) are functions of \(Y_1\) only and \(Y_2\) only, then \[E[g(Y_1)h(Y_2)] = E[g(Y_1)]\,E[h(Y_2)],\] provided the expectations exist.

The plantโ€™s density is \(2(1 - y_1) \times 1\) on a rectangle, so by Wednesdayโ€™s Theorem 5.5 \(Y_1\) and \(Y_2\) are independent, with \(E(Y_1) = 1/3\) and \(E(Y_2) = 1/2\): \[E(Y_1Y_2) = \tfrac{1}{3} \cdot \tfrac{1}{2} = \tfrac{1}{6}, \text{ as before, with no double integral.}\]

๐Ÿ“ˆ The Sign of the Deviations

69% of 120 simulated months land where \((y_1 - \mu_1)(y_2 - \mu_2) > 0\), so the average product is positive.

๐Ÿ“ Definition 5.10 and Theorem 5.10

Definition 5.10

If \(Y_1\) and \(Y_2\) have means \(\mu_1\) and \(\mu_2\), the covariance of \(Y_1\) and \(Y_2\) is \[\text{Cov}(Y_1, Y_2) = E[(Y_1 - \mu_1)(Y_2 - \mu_2)].\] Its scale-free version is the correlation coefficient \(\rho = \dfrac{\text{Cov}(Y_1, Y_2)}{\sigma_1 \sigma_2}\), with \(-1 \le \rho \le 1\).

Theorem 5.10

\[\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - E(Y_1)E(Y_2).\]

Proof: expand the product, then apply Theorems 5.7 and 5.8 term by term.

๐Ÿฆ Worked Example: Two Bank Shares

Monthly returns (%) of two bank shares on the Baku Stock Exchange, \(Y_1\) for bank A and \(Y_2\) for bank B:

\(y_1 \backslash y_2\) \(-3\) \(1\) \(5\) \(p_1\)
\(-2\) .15 .08 .02 .25
\(1\) .08 .30 .12 .50
\(4\) .02 .07 .16 .25
\(p_2\) .25 .45 .30 1

\(E(Y_1) = 1.0\), \(E(Y_2) = 1.2\), and summing \(y_1y_2\,p(y_1, y_2)\) over the nine cells gives \(E(Y_1Y_2) = 4.44\).

\[\text{Cov}(Y_1, Y_2) = 4.44 - (1.0)(1.2) = 3.24, \qquad \rho = \frac{3.24}{\sqrt{4.5}\,\sqrt{8.76}} = 0.516.\]

๐Ÿ’ป Covariance and Correlation in R

y1 <- c(-2, 1, 4); y2 <- c(-3, 1, 5)
p  <- matrix(c(.15, .08, .02,
               .08, .30, .12,
               .02, .07, .16), nrow = 3, byrow = TRUE)
mu1 <- sum(y1 * rowSums(p)); mu2 <- sum(y2 * colSums(p))
cov_def  <- sum(outer(y1 - mu1, y2 - mu2) * p)       # Definition 5.10
cov_5.10 <- sum(outer(y1, y2) * p) - mu1 * mu2        # Theorem 5.10
s1 <- sqrt(sum(y1^2 * rowSums(p)) - mu1^2); s2 <- sqrt(sum(y2^2 * colSums(p)) - mu2^2)
round(c(cov_def = cov_def, cov_5.10 = cov_5.10, rho = cov_5.10 / (s1 * s2)), 3)
 cov_def cov_5.10      rho 
   3.240    3.240    0.516

The covariance of 3.24 is in โ€œpercent squaredโ€, hard to judge. \(\rho = 0.52\) says: moderate positive linear dependence.

๐Ÿ“ Theorem 5.11

Theorem 5.11

If \(Y_1\) and \(Y_2\) are independent, then \(\text{Cov}(Y_1, Y_2) = 0\).

Proof: Theorem 5.9 gives \(E(Y_1Y_2) = \mu_1\mu_2\), and Theorem 5.10 finishes it. For the power plant, \(\text{Cov} = \tfrac{1}{6} - \tfrac{1}{3}\cdot\tfrac{1}{2} = 0\).

Read it backwards as a test: the bank shares have \(\text{Cov} = 3.24 \ne 0\), so they cannot be independent.

The converse is false. Zero covariance does not make two variables independent.

๐Ÿ›ข๏ธ Zero Covariance, Still Dependent

An energy desk holds \(Y_1\) Brent and \(Y_2\) natural-gas contracts each day (\(-1\) short, \(0\) flat, \(+1\) long). The desk mandate forbids being flat in both (Wackerly Table 5.3):

\(y_1 \backslash y_2\) \(-1\) \(0\) \(+1\)
\(-1\) 1/16 3/16 1/16
\(0\) 3/16 0 3/16
\(+1\) 1/16 3/16 1/16

Dependent: \(p(0, 0) = 0 \ne p_1(0)p_2(0) = (6/16)^2\). Yet \(E(Y_1) = E(Y_2) = 0\), and the four corners give \(E(Y_1Y_2) = \tfrac{1}{16} - \tfrac{1}{16} - \tfrac{1}{16} + \tfrac{1}{16} = 0\), so Cov = 0.

๐Ÿ“‰ Dependent but Uncorrelated

The payoff is completely determined by the move, yet over 250 simulated days the sample correlation is \(r = -0.03\) and the fitted line (red) is almost flat. Covariance sees only linear dependence; a V shape cancels itself out.

๐Ÿง  Think-Pair-Share

\(y_1 \backslash y_2\) 0 1 2
0 .30 .15 .05
1 .10 .15 .10
2 .02 .05 .08

The peak hour again: \(E(Y_1) = 0.65\) air conditioners, \(E(Y_2) = 0.81\) refrigerators.

Four minutes, in pairs:

  1. Find \(E(Y_1Y_2)\).

  2. Find \(\text{Cov}(Y_1, Y_2)\). What sign did you expect, and why?

  3. Are \(Y_1\) and \(Y_2\) independent? Give two ways to tell.

โœ… Think-Pair-Share: Solution

  1. Only cells with \(y_1, y_2 \ge 1\) contribute: \[E(Y_1Y_2) = 1(.15) + 2(.10) + 2(.05) + 4(.08) = 0.77\]

  2. Theorem 5.10: \(\text{Cov}(Y_1, Y_2) = 0.77 - (0.65)(0.81) = 0.2435 > 0\). A customer refitting a kitchen or a flat buys both.

  1. Not independent. By Theorem 5.11, a nonzero covariance rules independence out. Directly: \(p(0, 0) = 0.30 \ne p_1(0)p_2(0) = 0.50 \times 0.42 = 0.21\).

Note that \(E(R) = 1509\) AZN did not need any of this: linearity ignores dependence.

๐Ÿ“ Quiz #1: Linearity

A telecom operatorโ€™s daily new mobile contracts \(Y_1\) and fibre contracts \(Y_2\) are dependent, with \(E(Y_1) = 2\) and \(E(Y_2) = 3\) (hundreds). What is \(E(4Y_1 - Y_2 + 10)\)?

  • \(15\)
  • \(21\)
  • \(9\)
  • Cannot be found without the joint distribution

๐Ÿ“ Quiz #2: From Moments to ฯ

\(E(Y_1Y_2) = 5\), \(E(Y_1) = 1\), \(E(Y_2) = 2\), \(V(Y_1) = 4\), \(V(Y_2) = 9\). What is \(\rho\)?

  • \(0.5\)
  • \(3\)
  • \(0.083\)
  • \(0.833\)

๐Ÿ“ Quiz #3: Reading a Zero

Two exchange-rate changes are found to have \(\text{Cov}(Y_1, Y_2) = 0\). Which conclusion is justified?

  • They are independent
  • There is no linear dependence; they may still be dependent
  • \(E(Y_1Y_2) = 0\)
  • Their correlation could still be positive

๐Ÿ“‹ Key Formulas

Statement
Definition 5.9 \(E[g(Y_1, Y_2)] = \sum\sum g(y_1, y_2)\,p(y_1, y_2)\) or \(\iint g\,f\,dy_1\,dy_2\)
Theorems 5.6โ€“5.8 \(E(c) = c\), \(E(cg) = cE(g)\), \(E(g_1 + g_2) = E(g_1) + E(g_2)\)
Theorem 5.9 independent: \(E[g(Y_1)h(Y_2)] = E[g(Y_1)]E[h(Y_2)]\)
Definition 5.10 \(\text{Cov}(Y_1, Y_2) = E[(Y_1 - \mu_1)(Y_2 - \mu_2)]\)
Theorem 5.10 \(\text{Cov}(Y_1, Y_2) = E(Y_1Y_2) - \mu_1\mu_2\)
correlation \(\rho = \text{Cov}(Y_1, Y_2)/(\sigma_1\sigma_2)\), \(\;-1 \le \rho \le 1\)
Theorem 5.11 independent \(\Rightarrow\) Cov \(= 0\) (not conversely)

๐Ÿ“‹ Summary

  • Definition 5.9: weight each value of \(g\) by its joint probability or density, and add

  • The expectation of a sum is the sum of the expectations, with no independence needed

  • Under independence, expectations of products split into products of expectations (Theorem 5.9)

  • Covariance is the average product of deviations; its sign says whether two variables move together

  • \(\rho\) removes the units, so it can be read as weak, moderate or strong

  • Independence forces zero covariance; zero covariance allows dependence that is not linear

๐Ÿ“š Practice Problems

Wackerly, 7th edition

  • ยงยง5.5โ€“5.6: Exercises 5.72 โ€“ 5.88; start with 5.72, 5.74, 5.77, 5.80 and 5.84

  • ยง5.7: Exercises 5.89 โ€“ 5.97; start with 5.89, 5.91, 5.93, 5.95 and 5.97

  • Change the bank-share table so that \(\rho < 0\) while keeping both marginals. Which cells must move?

Week 13, Problem Set 2 is open now and closes Sunday 13 December at 23:59 on WeBWorK, covering ยงยง5.5โ€“5.7.

Next class: 9 December, the expected value and variance of linear functions of random variables, and portfolio variance (Wackerly ยง5.8).

๐Ÿ™ Thank You

Dr. Samir Orujov

๐Ÿ“ง sorujov@ada.edu.az
๐Ÿข Building D, Room D325
๐Ÿ•“ Office hours: Wednesday, 16:00 โ€“ 18:00

Slides and readings: sorujov.net/teaching

โ“ Questions

  • The retailerโ€™s revenue needed only the marginals. Which quantity about \(R\) will need the covariance as well?

  • If you double every return of bank A, what happens to the covariance, and what happens to \(\rho\)?

  • Can you build a joint table where \(Y_2 = Y_1^2\) exactly and the covariance is zero? What must the distribution of \(Y_1\) look like?