Understanding Discrete Random Variables and Expected Value

A discrete random variable is a variable whose value can only take on a finite or countable number of distinct values. In this problem, the random variable X represents possible outcomes that are whole numbers from 0 to 8. The probability function, denoted by \(f(x)\) or \(P(X=x)\), gives the probability that the random variable X takes on a specific value \(x\).

The question asks us to find the expected value of X, denoted as \(E(X)\). The expected value is essentially the average value of the random variable over a large number of trials. For a discrete random variable, it is calculated using the formula:

\(E(X) = \sum_{x} x \cdot f(x)\)

Before we can calculate \(E(X)\), we first need to determine the value of the constant \(k\) in the probability function \(f(x)\). A fundamental property of probability distributions is that the sum of probabilities for all possible values of the random variable must equal 1.

Finding the Value of Constant k

We are given the probability function values for X = 0, 1, 2, ..., 8:

Assuming K is the same as k, the sum of all probabilities is:

\(\sum_{x=0}^{8} f(x) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8) = 1\)

\(k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1\)

Let's sum the coefficients of \(k\):

\((1 + 2 + 3 + 5 + 5 + 4 + 3 + 2 + 1)k = 1\)

\(26k = 1\)

Solving for \(k\):

\(k = \frac{1}{26}\)

Calculating the Expected Value E(X)

Now that we have the value of \(k\), we can calculate the expected value \(E(X)\) using the formula \(E(X) = \sum x \cdot f(x)\).

\(E(X) = (0 \cdot f(0)) + (1 \cdot f(1)) + (2 \cdot f(2)) + (3 \cdot f(3)) + (4 \cdot f(4)) + (5 \cdot f(5)) + (6 \cdot f(6)) + (7 \cdot f(7)) + (8 \cdot f(8))\)

Substitute the values of \(x\) and \(f(x)\):

\(E(X) = (0 \cdot k) + (1 \cdot 2k) + (2 \cdot 3k) + (3 \cdot 5k) + (4 \cdot 5k) + (5 \cdot 4k) + (6 \cdot 3k) + (7 \cdot 2k) + (8 \cdot k)\)

Simplify each term:

\(E(X) = 0k + 2k + 6k + 15k + 20k + 20k + 18k + 14k + 8k\)

Sum the coefficients of \(k\):

\(E(X) = (0 + 2 + 6 + 15 + 20 + 20 + 18 + 14 + 8)k\)

\(E(X) = 103k\)

Now substitute the value of \(k = \frac{1}{26}\) into the expression for \(E(X)\):

\(E(X) = 103 \cdot \frac{1}{26}\)

\(E(X) = \frac{103}{26}\)

Thus, the expected value of the discrete random variable X is \(\frac{103}{26}\).

Summary of Expected Value Calculation

Here's a quick summary of the steps:

• Sum all probabilities to find the value of the constant \(k\).
• Use the formula \(E(X) = \sum x \cdot f(x)\).
• Multiply each possible value of X by its corresponding probability \(f(x)\).
• Sum these products to get the expected value \(E(X)\).

Revision Table: Discrete Random Variable & Expected Value

Additional Information: Properties of Expected Value

The expected value has several important properties:

• For a constant \(c\), \(E(c) = c\).
• For a constant \(c\) and a random variable X, \(E(cX) = cE(X)\).
• For two random variables X and Y, \(E(X + Y) = E(X) + E(Y)\). This holds true even if X and Y are not independent.
• For two independent random variables X and Y, \(E(XY) = E(X)E(Y)\). This property does not hold for dependent variables in general.

Understanding the expected value is crucial in probability and statistics for analyzing the average outcome of random processes, making decisions under uncertainty, and in fields like finance (e.g., expected return on an investment) and insurance (e.g., calculating premiums based on expected claims).