Chapter Notes: More On Expected Value
Expected value is one of the most powerful concepts in statistics and probability. It allows us to predict the long-term average outcome of a random process, even when individual results vary unpredictably. In this chapter, we extend our understanding of expected value beyond basic calculations to explore more complex scenarios, including multi-stage processes, conditional expectations, and applications in decision-making. By mastering these advanced techniques, you'll gain the tools to analyze real-world situations involving uncertainty, risk, and strategic choice.
Review: The Foundation of Expected Value
Before diving into more sophisticated applications, let's quickly review the fundamental concept. The expected value (also called the mean or expectation) of a discrete random variable is the weighted average of all possible outcomes, where each outcome is weighted by its probability.
For a random variable \( X \) with outcomes \( x_1, x_2, x_3, \ldots, x_n \) and corresponding probabilities \( P(x_1), P(x_2), P(x_3), \ldots, P(x_n) \), the expected value is:
\[ E(X) = x_1 \cdot P(x_1) + x_2 \cdot P(x_2) + x_3 \cdot P(x_3) + \cdots + x_n \cdot P(x_n) \]In summation notation, this is written as:
\[ E(X) = \sum_{i=1}^{n} x_i \cdot P(x_i) \]Here, \( E(X) \) represents the expected value, \( x_i \) represents each possible outcome, and \( P(x_i) \) represents the probability of that outcome occurring.
Example: A carnival game costs $2 to play.
You spin a wheel with three equally likely outcomes:
win $5, win $1, or win nothing.What is the expected value of your net winnings?
Solution:
First, identify the net winnings for each outcome. Since the game costs $2 to play, we subtract $2 from each prize:
Win $5: net = $5 - $2 = $3
Win $1: net = $1 - $2 = -$1
Win $0: net = $0 - $2 = -$2Each outcome has probability \( \frac{1}{3} \) since the wheel is equally likely to land on any of the three outcomes.
Now calculate the expected value:
\[ E(X) = 3 \cdot \frac{1}{3} + (-1) \cdot \frac{1}{3} + (-2) \cdot \frac{1}{3} \]\( E(X) = 1 - \frac{1}{3} - \frac{2}{3} = 1 - 1 = 0 \)
The expected value of your net winnings is $0, meaning over many plays, you would break even on average.
Linearity of Expectation
One of the most useful properties of expected value is called linearity of expectation. This property states that the expected value of a sum equals the sum of the expected values, regardless of whether the random variables are independent.
For any two random variables \( X \) and \( Y \), and any constants \( a \) and \( b \):
\[ E(aX + bY) = aE(X) + bE(Y) \]This property holds even when \( X \) and \( Y \) are not independent, which makes it extremely powerful for solving complex problems. You can break a complicated random process into simpler parts, find the expected value of each part, and then combine them.
Example: You roll two fair six-sided dice.
The first die pays you dollars equal to the number showing.
The second die costs you dollars equal to half the number showing.What is the expected value of your total winnings?
Solution:
Let \( X \) be the value shown on the first die and \( Y \) be the value shown on the second die.
Your total winnings are \( W = X - \frac{1}{2}Y \).
First, find \( E(X) \). For a fair six-sided die:
\[ E(X) = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5 \]Next, find \( E(Y) \). This is also a fair six-sided die, so \( E(Y) = 3.5 \).
Using linearity of expectation:
\[ E(W) = E\left(X - \frac{1}{2}Y\right) = E(X) - \frac{1}{2}E(Y) \]\( E(W) = 3.5 - \frac{1}{2}(3.5) = 3.5 - 1.75 = 1.75 \)
The expected value of your total winnings is $1.75.
Applications to Repeated Trials
Linearity of expectation is particularly useful when analyzing repeated trials of the same experiment. If you repeat an experiment \( n \) times, and each trial has expected value \( E(X) \), then the expected total from all trials is simply \( n \cdot E(X) \).
Example: A lottery ticket has a 1 in 20 chance of winning $50,
a 1 in 100 chance of winning $200,
and otherwise wins nothing.
Each ticket costs $5.If you buy 10 tickets, what is the expected value of your net winnings?
Solution:
First, find the expected value for one ticket.
The probability of winning $50 is \( \frac{1}{20} = 0.05 \).
The probability of winning $200 is \( \frac{1}{100} = 0.01 \).
The probability of winning nothing is \( 1 - 0.05 - 0.01 = 0.94 \).
Expected winnings from prizes:
\[ E(\text{prizes}) = 50(0.05) + 200(0.01) + 0(0.94) = 2.50 + 2.00 = 4.50 \]Net winnings per ticket = $4.50 - $5.00 = -$0.50.
For 10 tickets, using linearity of expectation:
\( E(\text{total net}) = 10 \times (-0.50) = -5.00 \)
The expected value of your net winnings after buying 10 tickets is -$5.00, meaning you expect to lose $5 on average.
Expected Value with Multi-Stage Processes
Many real-world situations involve multiple stages, where the outcome of one stage affects what happens in the next. To find the expected value in these situations, we use the law of total expectation, which breaks the calculation into conditional parts.
The law of total expectation states:
\[ E(X) = \sum_{i} E(X \mid A_i) \cdot P(A_i) \]This formula means: the overall expected value equals the sum of the expected values in each scenario, weighted by the probability of that scenario occurring. Here, \( A_i \) represents different possible conditions or scenarios, \( E(X \mid A_i) \) represents the expected value of \( X \) given that scenario \( A_i \) occurs, and \( P(A_i) \) represents the probability of scenario \( A_i \).
Example: A game has two stages.
First, you flip a fair coin.
If heads, you roll one six-sided die and win that many dollars.
If tails, you roll two six-sided dice and win the sum in dollars.What is the expected value of your winnings?
Solution:
Let \( H \) represent the event "flip heads" and \( T \) represent "flip tails."
The probability of each is \( P(H) = 0.5 \) and \( P(T) = 0.5 \).
If heads: You roll one die, so \( E(\text{winnings} \mid H) = 3.5 \).
If tails: You roll two dice. The expected value of one die is 3.5, so by linearity, the expected sum of two dice is \( E(\text{winnings} \mid T) = 2 \times 3.5 = 7 \).
Using the law of total expectation:
\[ E(\text{winnings}) = E(\text{winnings} \mid H) \cdot P(H) + E(\text{winnings} \mid T) \cdot P(T) \]\( E(\text{winnings}) = 3.5(0.5) + 7(0.5) = 1.75 + 3.5 = 5.25 \)
The expected value of your winnings is $5.25.
Expected Value in Decision-Making
Expected value is a critical tool for making rational decisions under uncertainty. When faced with multiple options, each with uncertain outcomes, calculating the expected value of each option helps identify which choice is most favorable in the long run.
Comparing Options
To make a decision based on expected value, calculate the expected value for each available option and choose the one with the highest expected value (if the outcomes represent gains) or lowest expected value (if the outcomes represent costs or losses).
Example: You must choose between two investment opportunities.
Investment A: 60% chance of earning $1000, 40% chance of losing $200.
Investment B: 30% chance of earning $2000, 70% chance of breaking even (earning $0).Which investment has the higher expected value?
Solution:
For Investment A:
\[ E(A) = 1000(0.60) + (-200)(0.40) = 600 - 80 = 520 \]For Investment B:
\[ E(B) = 2000(0.30) + 0(0.70) = 600 + 0 = 600 \]Compare: \( E(A) = 520 \) dollars and \( E(B) = 600 \) dollars.
Investment B has the higher expected value at $600, compared to Investment A's $520.
Expected Value vs. Risk Tolerance
While expected value provides a mathematical framework for decision-making, it's important to recognize its limitations. Expected value represents the average outcome over many repetitions, but in real life, you may only have one chance to make a decision. Additionally, people have different risk tolerances-some prefer safer options with lower variability, even if the expected value is slightly lower.
Think of it this way: an investment with a 1% chance of winning $1,000,000 and a 99% chance of losing $9,000 has a positive expected value of $1,000. However, most people would not risk $9,000 for such a small chance of winning, because the risk of loss is very real, even though the expected value is positive.
Expected Value with Dependent Events
When events are dependent-meaning the outcome of one affects the probabilities of another-calculating expected value requires careful attention to conditional probabilities. However, the linearity of expectation still holds, which can simplify calculations considerably.
Example: A bag contains 3 red marbles worth $5 each and 2 blue marbles worth $2 each.
You draw two marbles without replacement.What is the expected total value of the two marbles you draw?
Solution:
Let \( X_1 \) be the value of the first marble and \( X_2 \) be the value of the second marble.
The total value is \( T = X_1 + X_2 \).
By linearity of expectation: \( E(T) = E(X_1) + E(X_2) \).
For the first draw, there are 3 red out of 5 total marbles:
\[ E(X_1) = 5 \cdot \frac{3}{5} + 2 \cdot \frac{2}{5} = 3 + 0.8 = 3.8 \]For the second draw, by symmetry (each marble is equally likely to be drawn second as first), the expected value is the same:
\[ E(X_2) = 3.8 \]Therefore:
\[ E(T) = 3.8 + 3.8 = 7.6 \]The expected total value of the two marbles is $7.60.
Notice that even though the draws are dependent (removing one marble changes the composition of the bag), linearity of expectation still applies. We didn't need to calculate all possible combinations of draws and their probabilities.
Expected Value of Functions of Random Variables
Sometimes we need to find the expected value not of a random variable itself, but of some function of that random variable. For instance, we might want to know the expected value of the square of a die roll, or the expected profit given a random number of sales.
If \( Y = g(X) \) for some function \( g \), then:
\[ E(Y) = E(g(X)) = \sum_{i} g(x_i) \cdot P(x_i) \]In plain English: apply the function to each outcome, then multiply by the probability of that outcome, and sum everything up.
Example: You roll a fair six-sided die.
Your winnings equal the square of the number rolled.What is the expected value of your winnings?
Solution:
Let \( X \) be the number rolled. Your winnings are \( Y = X^2 \).
Calculate \( E(X^2) \) by squaring each outcome and weighting by probability:
\[ E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} + 3^2 \cdot \frac{1}{6} + 4^2 \cdot \frac{1}{6} + 5^2 \cdot \frac{1}{6} + 6^2 \cdot \frac{1}{6} \]\( E(X^2) = \frac{1}{6}(1 + 4 + 9 + 16 + 25 + 36) \)
\( E(X^2) = \frac{1}{6}(91) = \frac{91}{6} \approx 15.17 \)
The expected value of your winnings is approximately $15.17.
Important note: In general, \( E(g(X)) \neq g(E(X)) \). In the example above, \( E(X) = 3.5 \), but \( (E(X))^2 = 12.25 \), which is not equal to \( E(X^2) = 15.17 \). Always apply the function first to each outcome, then take the weighted average.
Expected Value and Fair Games
A game is called fair if the expected value of the net winnings is zero. In other words, over the long run, neither the player nor the house has an advantage. Most casino games and lotteries are designed to have negative expected value for the player, giving the house a long-term profit margin called the house edge.
Example: A game costs $10 to play.
You roll a fair die.
If you roll a 6, you win $50.
Otherwise, you win nothing.Is this a fair game?
Solution:
Calculate the expected net winnings.
Probability of rolling a 6: \( \frac{1}{6} \).
Net winnings if you roll a 6: $50 - $10 = $40.
Net winnings otherwise: $0 - $10 = -$10.
Expected value:
\[ E(\text{net}) = 40 \cdot \frac{1}{6} + (-10) \cdot \frac{5}{6} \]\( E(\text{net}) = \frac{40}{6} - \frac{50}{6} = \frac{-10}{6} \approx -1.67 \)
The expected net winnings are approximately -$1.67, so this is not a fair game; the player expects to lose money on average.
To make the game fair, the cost to play would need to equal the expected winnings from prizes. In the example above, the expected prize value is \( 50 \times \frac{1}{6} \approx 8.33 \), so a fair price would be about $8.33.
Applications in Insurance and Risk Management
Expected value plays a central role in the insurance industry. Insurance companies use expected value to set premiums that cover expected payouts plus administrative costs and profit. Customers pay premiums to transfer risk, even though the expected value of an insurance policy is typically negative for the policyholder.
For example, if a car insurance policy costs $1,200 per year, and the insurance company estimates a 2% chance the policyholder will file a claim averaging $15,000, the expected payout is $15,000 × 0.02 = $300. The $1,200 premium covers this expected payout plus the company's operating costs and profit margin. The policyholder accepts this negative expected value in exchange for protection against a large, unpredictable loss.
Example: A homeowner estimates a 0.5% annual chance of a fire causing $200,000 in damage.
An insurance company offers fire insurance for an annual premium of $1,500.What is the expected value of the insurance from the homeowner's perspective, considering both the premium paid and potential payout received?
Solution:
From the homeowner's perspective:
Cost of insurance = -$1,500 (paid with certainty).
Expected benefit from potential claim: With 0.5% = 0.005 probability, the insurance pays $200,000.
Expected payout = $200,000 × 0.005 = $1,000.
Net expected value = Expected payout - Premium paid:
\( E(\text{net}) = 1000 - 1500 = -500 \)
The expected value from the homeowner's perspective is -$500, meaning the homeowner expects to lose $500 on average by purchasing insurance, but gains protection against catastrophic loss.
Long-Term vs. Short-Term Thinking
Expected value is most meaningful when considering long-term, repeated scenarios. Over many trials, actual average outcomes converge toward the expected value (a principle formalized by the Law of Large Numbers). However, in a single trial or small number of trials, actual outcomes can vary widely from the expected value.
This distinction is crucial in decision-making. Businesses and casinos rely on expected value because they engage in thousands of similar transactions, allowing long-term averages to stabilize. Individual consumers making one-time decisions must balance expected value with other considerations like risk tolerance, worst-case scenarios, and personal circumstances.
Understanding expected value empowers you to analyze situations involving uncertainty with mathematical precision. Whether evaluating games of chance, investment opportunities, insurance policies, or strategic choices, expected value provides a rational foundation for decision-making. By mastering advanced applications-including linearity of expectation, multi-stage processes, and conditional expectations-you can tackle complex real-world problems where outcomes are uncertain but probabilities are known.
