Problem 19 - Calculus for Biology and Medicine

Question

Suppose that the probability mass function of a discrete random variable \(X\) is given by the following table: $$\begin{array}{cc} \hline \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) \\ \hline-2 & 0.1 \\ -1 & 0.4 \\ 0 & 0.3 \\ 1 & 0.2 \\ \hline \end{array}$$ (a) Find \(E(X)\). (b) Find \(E\left(X^{2}\right)\). (c) Find \(E[X(X-1)]\).

Step-by-Step Solution

Answer

(a) \(E(X) = -0.4\), (b) \(E(X^2) = 1.0\), (c) \(E[X(X-1)] = 1.4\).

Step 1: Understand the Probability Mass Function

The probability mass function (PMF) shows the probabilities of different outcomes for a discrete random variable \(X\). In this table, \(X\) can take the values -2, -1, 0, or 1, with corresponding probabilities of 0.1, 0.4, 0.3, and 0.2.

Step 2: Calculate the Expected Value, E(X)

To find the expected value \(E(X)\), use the formula \(E(X) = \sum{x_i \cdot P(X=x_i)}\). Apply this to each \(x_i\):\[-2 imes 0.1 + (-1) imes 0.4 + 0 imes 0.3 + 1 imes 0.2\]Calculating each term: \(-2 imes 0.1 = -0.2\),\(-1 imes 0.4 = -0.4\),\(0 imes 0.3 = 0\),\(1 imes 0.2 = 0.2\).Therefore, \(E(X) = -0.2 - 0.4 + 0 + 0.2 = -0.4\).

Step 3: Calculate the Expected Value of X squared, E(X²)

To find \(E(X^2)\), use the formula \(E(X^2) = \sum{x_i^2 \cdot P(X=x_i)}\). Calculate \(x_i^2\) for each outcome and multiply by the probability:\[(-2)^2 imes 0.1 + (-1)^2 imes 0.4 + 0^2 imes 0.3 + 1^2 imes 0.2\]This equals:\(4 imes 0.1 = 0.4\),\(1 imes 0.4 = 0.4\),\(0 imes 0.3 = 0\),\(1 imes 0.2 = 0.2\).Thus, \(E(X^2) = 0.4 + 0.4 + 0 + 0.2 = 1.0\).

Step 4: Calculate E[X(X-1)]

To find \(E[X(X-1)]\), apply the formula \(E[X(X-1)] = \sum{x_i(x_i-1) \cdot P(X=x_i)}\). Compute \(x_i(x_i-1)\) and multiply by the probability:\[-2(-2-1) imes 0.1 + (-1)(-1-1) imes 0.4 + 0(0-1) imes 0.3 + 1(1-1) imes 0.2\]This results in:\[-2(-3) imes 0.1 = 0.6\],\[-1(-2) imes 0.4 = 0.8\],\[0(-1) imes 0.3 = 0\],\[1(0) imes 0.2 = 0\].Therefore, \(E[X(X-1)] = 0.6 + 0.8 + 0 + 0 = 1.4\).

Key Concepts

Expected Value

The expected value, denoted as \(E(X)\), is a fundamental concept in probability and statistics. It represents the long-term average or mean of a random variable after many trials. For a discrete random variable, like in our problem, the expected value is calculated by summing the products of each possible value of the random variable and its corresponding probability. This can be represented by the formula: \(E(X) = \sum x_i \cdot P(X=x_i)\).

• In our example, the expected value \(E(X)\) involves calculating \(-2 \cdot 0.1 + (-1) \cdot 0.4 + 0 \cdot 0.3 + 1 \cdot 0.2\).
• Each term represents a potential outcome of \(X\) multiplied by its probability.
• Calculating these terms, we find \(E(X) = -0.4\).

This result tells us that, if we were to repeatedly sample from this distribution, on average, the value would tend toward \(-0.4\). Understanding expected value is crucial for predicting outcomes and strategic decision-making in various fields such as finance, insurance, and gambling.

Discrete Random Variable

A discrete random variable is a type of random variable that can take on a countable number of distinct values. This is contrasted with a continuous random variable, which can take on any value within an interval. Discrete random variables often arise from counting scenarios, such as the roll of a die or the number of students in a classroom.

• In the context of the exercise, \(X\) is a discrete random variable as it can only take on the specific values \(-2, -1, 0,\) and \(1\).
• The probability mass function (PMF) described in the problem provides the likelihood of each of these specific outcomes occurring.
• The PMF sums to 1, ensuring it covers all possible outcomes for the random variable \(X\).

Understanding discrete random variables is essential for analyzing situations where outcomes fall into distinct categories. They are a foundational element in studies involving discreet probabilities and their impacts across various practical and theoretical applications.

Expected Value of a Function of a Random Variable

The expected value of a function of a random variable extends the concept of expected value to functions applied to random variables. It is denoted as \(E[g(X)]\), where \(g(X)\) is any function applied to \(X\). This concept allows us to calculate the expected outcome of more complex expressions involving random variables.

• In our exercise, we compute \(E(X^2)\) and \(E[X(X-1)]\), both examples of expected values of functions.
• To find \(E(X^2)\), use \(E(X^2) = \sum x_i^2 \cdot P(X=x_i)\), leading to a result of \(1.0\).
• For \(E[X(X-1)]\), apply the formula: \(E[X(X-1)] = \sum x_i(x_i-1) \cdot P(X=x_i)\), resulting in \(1.4\).

These calculations help us understand not just the average outcome, but also the variation and potential interactions of outcomes. Functions of random variables are vital in scenarios where multiple features or non-linear transformations are part of analysis, such as in risk assessment or economic modeling.