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) \(E(X) = 1.6\), (b) \(E(X^2) = 4.8\), (c) \(E(2X - 1) = 2.2\).

Step 1: Understand the Expectation Formula

The expectation or expected value of a discrete random variable \(X\), denoted as \(E(X)\), is found by summing the products of each possible value of \(X\) and their corresponding probabilities.

Step 2: Calculate \(E(X)\)

Apply the formula for expected value: \(E(X) = \sum x_i P(X = x_i)\). Calculate each component:\(0 \cdot 0.3 = 0\)\(1 \cdot 0.3 = 0.3\)\(2 \cdot 0.1 = 0.2\)\(3 \cdot 0.1 = 0.3\)\(4 \cdot 0.2 = 0.8\)Then sum these products: \(0 + 0.3 + 0.2 + 0.3 + 0.8 = 1.6\). Thus, \(E(X) = 1.6\).

Step 3: Calculate \(E(X^2)\)

Use the formula for expected value, but for each value square \(x\): \(E(X^2) = \sum x_i^2 P(X = x_i)\). Compute each:\(0^2 \cdot 0.3 = 0\)\(1^2 \cdot 0.3 = 0.3\)\(2^2 \cdot 0.1 = 0.4\)\(3^2 \cdot 0.1 = 0.9\)\(4^2 \cdot 0.2 = 3.2\)Sum these: \(0 + 0.3 + 0.4 + 0.9 + 3.2 = 4.8\). Therefore, \(E(X^2) = 4.8\).

Step 4: Using the Linear Transformation in Expectation

To find \(E(2X - 1)\), use the property of expectation: \(E(aX + b) = aE(X) + b\). Here, \(a = 2\) and \(b = -1\).

Step 5: Find \(E(2X - 1)\)

Substitute \(a = 2\) and \(b = -1\) into the transformation formula: \(E(2X-1) = 2E(X) - 1\). We found \(E(X) = 1.6\) in Step 2. Thus, \(E(2X - 1) = 2(1.6) - 1 = 3.2 - 1 = 2.2\).

Key Concepts

Discrete Random Variables

A discrete random variable is a fundamental concept in probability and statistics. It's a variable that can take on a countable number of distinct outcomes. Unlike continuous random variables, which can take on any value in an interval, discrete random variables often represent counts or other distinct steps.
Consider a random variable \(X\) that represents the outcome of a rolled dice. The possible values are \(1, 2, 3, 4, 5, 6\), each with an equal probability of occurrence. This is a typical example of a discrete random variable since the outcomes are countable and distinct.
When working with discrete random variables, we often deal with a probability mass function (PMF). This function gives the probabilities that a discrete random variable is exactly equal to each possible value. For instance, the PMF of our dice example would assign a probability of \(\frac{1}{6}\) to each of the numbers \(1\) through \(6\).

• Discrete random variables are useful for modeling real-life phenomena that involve distinct steps or counts.
• The probability mass function helps in understanding how likely each outcome is, which is essential for further calculations like expected value.

Expected Value

The expected value, often denoted as \(E(X)\), is a measure of the center or average of a probability distribution. For discrete random variables, it is calculated by summing the products of each possible value of the variable and their respective probabilities.
Calculating the expected value can be thought of as finding a weighted average, where the weights are the probabilities of each outcome. In a given distribution with values \(x_1, x_2, ..., x_n\) and respective probabilities \(P(X = x_1), P(X = x_2), ..., P(X = x_n)\), the formula for the expected value is:
\[ E(X) = \sum x_i P(X = x_i) \]
For example, in our original exercise, the expected value \(E(X)\) was calculated as \(1.6\) by summing the products of each value of \(X\) and its probability.

• The expected value provides a single number summary of the distribution, representing a "center" of the distribution.
• It's important to remember that the expected value is not necessarily a value the random variable can actually take on in practice.

Linear Transformation

Linear transformations of random variables are common operations in probability and statistics. They involve changing the scale or location of the random variable through equations of the form \(aX + b\), where \(a\) and \(b\) are constants.
The expected value of a linear transformation uses this formula:
\[ E(aX + b) = aE(X) + b \]
This property of expectation makes it easy to calculate the expected value of transformed variables. For example, if you want to find the expected value of \(2X - 1\), you can compute it without needing to go back to the original probabilities. Simply multiply the expected value \(E(X)\) by \(a\) and then add \(b\).
As shown in the original step-by-step solution, the expected value of the linear transformation \(2X - 1\) was found to be \(2.2\).

• This process illustrates how linear transformations maintain a predictable relationship with expected values, aiding in simplifying complex probability problems.
• These transformations can effectively shift the random variable's distribution along the number line or change its scale.