Problem 20 - Calculus for Biology and Medicine

Question

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$ \begin{array}{cc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline 0 & 0.3 \\ 1 & 0.3 \\ 2 & 0.1 \\ 3 & 0.1 \\ 4 & 0.2 \\ \hline \end{array} $$ (a) Find \(E(X)\). (b) Find \(E\left(X^{2}\right)\). (c) Find \(E(2 X-1)\).

Step-by-Step Solution

Answer

(a) 1.6; (b) 4.8; (c) 2.2.

Step 1: Understanding the Expectation of X

The expectation of the random variable \(X\), denoted \(E(X)\), is calculated using the formula: \( E(X) = \sum xP(X=x) \). We will multiply each value of \(x\) by its probability and find the sum of these products.

Step 2: Calculating E(X)

Using the given table: \[ E(X) = 0(0.3) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.2) \]This simplifies to: \[ E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.8 = 1.6 \]

Step 3: Understanding the Expectation of X Squared

For finding \(E(X^2)\), use the formula: \( E(X^2) = \sum x^2P(X=x) \). This involves squaring each \(x\) value, then multiplying by its probability, and summing up these products.

Step 4: Calculating E(X²)

Using the given table: \[ E(X^2) = 0^2(0.3) + 1^2(0.3) + 2^2(0.1) + 3^2(0.1) + 4^2(0.2) \]Simplifying the expression: \[ E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 3.2 = 4.8 \]

Step 5: Understanding Expectation of a Linear Function of X

For \(E(2X-1)\), use the linearity of expectation: \(E(aX+b) = aE(X) + b\). Here \(a = 2\) and \(b = -1\), so we substitute and calculate.

Step 6: Calculating E(2X-1)

Substitute \(E(X) = 1.6\) from Step 2 into the formula: \[ E(2X-1) = 2 \times 1.6 - 1 = 3.2 - 1 = 2.2 \]

Key Concepts

Discrete Random Variable

A discrete random variable is a type of random variable that takes on a countable number of distinct values. Remember, discrete does not mean that the variable can't have infinitely many values, just that the values can be listed—like 0, 1, 2, and so on. You might think of rolling a die or flipping a coin when picturing examples of discrete random variables. These variables are often described by a probability mass function (PMF), which assigns a probability to each possible value. We need to keep in mind that the sum of all probabilities must equal 1.

In our example, we have a discrete random variable \(X\) with the following PMF: each value \(x\) of \(X\) is associated with its probability \(P(X = x)\). For example, \(P(X = 1) = 0.3\) and \(P(X = 3) = 0.1\). Each outcome is distinct and accumulates to a total probability of 1. These kinds of functions are fundamental in probability and statistics as they allow us to model and understand random behaviors in a structured manner.

Expectation in Probability

The expectation, or expected value, of a random variable is a foundational concept in probability that gives us a measure of the "center" or "average" outcome. For a discrete random variable, the expectation is calculated by multiplying each potential value by its probability and then summing up these products. The formula is: \( E(X) = \sum xP(X=x) \). This tells us the average value we would expect if we could repeat an experiment many times.

In our problem, the expectation for \(X\) is calculated as follows: Each \(x\) from the table is multiplied by its probability: \(0(0.3) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.2)\), which simplifies to an expected value of 1.6. This means, on average, the result of the random variable will be 1.6 after numerous trials. These calculations are handy in scenarios where averages inform decisions, such as in game theory, risk assessment, and any form of stochastic modeling.

Linearity of Expectation

The linearity of expectation is an incredibly useful property in probability, stating that the expected value of a sum of random variables is equal to the sum of their expected values, regardless of whether the variables are independent or not. Mathematically, this is expressed as \(E(aX + b) = aE(X) + b\), where \(a\) and \(b\) are constants and \(X\) is a random variable.

In this exercise, we explore this property by examining the expression \(E(2X-1)\). Using the given \(E(X) = 1.6\), we substitute into the formula: \(2E(X) - 1\). This results in \(2 \times 1.6 - 1 = 2.2\). This highlights that transformations of random variables such as scaling and shifting do not affect the core principles of their expectations. This principle simplifies many complex problems, offering a straightforward way to find expected values of transformed variables without complicated calculations.