Problem 2 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

The probability distribution of a discrete random variable \(X\) is given by $$ \mathrm{P}(X=-1)=\frac{1}{5}, \quad \mathrm{P}(X=0)=\frac{2}{5}, \quad \mathrm{P}(X=1)=\frac{2}{5} . $$ a. Compute \(\mathrm{E}[X]\). b. Give the probability distribution of \(Y=X^{2}\) and compute \(\mathrm{E}[Y]\) using the distribution of \(Y\). c. Determine \(\mathrm{E}\left[X^{2}\right]\) using the change-of-variable formula. Check your answer against the answer in \(\mathbf{b}\). d. Determine \(\operatorname{Var}(X)\).

Step-by-Step Solution

Answer

E[X] = \(\frac{1}{5}\), E[Y] = E[X^2] = \(\frac{3}{5}\), Var(X) = \(\frac{14}{25}\).

Step 1: Compute E[X]

The expected value (or mean) of a discrete random variable \( X \) is given by the formula \( E[X] = \sum x_i \cdot P(x_i) \), where \( x_i \) are the possible values of \( X \) and \( P(x_i) \) is the probability of each value. For our problem: \[ E[X] = (-1) \cdot \frac{1}{5} + 0 \cdot \frac{2}{5} + 1 \cdot \frac{2}{5} = -\frac{1}{5} + 0 + \frac{2}{5} = \frac{1}{5} \]

Step 2: Determine the Distribution of Y

Since \( Y = X^2 \), we need to find the probability distribution of \( Y \). For each value of \( X \), calculate \( Y \): 1. If \( X = -1 \), then \( Y = (-1)^2 = 1 \). 2. If \( X = 0 \), then \( Y = 0^2 = 0 \). 3. If \( X = 1 \), then \( Y = 1^2 = 1 \). Now, compute the probabilities: - \( P(Y = 0) = P(X = 0) = \frac{2}{5} \) - \( P(Y = 1) = P(X = -1) + P(X = 1) = \frac{1}{5} + \frac{2}{5} = \frac{3}{5} \)

Step 3: Compute E[Y]

Using the probability distribution of \( Y \) from Step 2, the expected value of \( Y \) is \[ E[Y] = 0 \cdot \frac{2}{5} + 1 \cdot \frac{3}{5} = 0 + \frac{3}{5} = \frac{3}{5} \]

Step 4: Compute E[X^2] Using Change of Variable

To find \( E[X^2] \) directly using \( X \), apply the formula for expected value to \( X^2 \): \[ E[X^2] = (-1)^2 \cdot \frac{1}{5} + 0^2 \cdot \frac{2}{5} + 1^2 \cdot \frac{2}{5} = 1 \cdot \frac{1}{5} + 0 + 1 \cdot \frac{2}{5} = \frac{1}{5} + \frac{2}{5} = \frac{3}{5} \] This confirms our calculation of \( E[Y] \) in Step 3.

Step 5: Compute Var(X)

The variance of \( X \), \( \operatorname{Var}(X) \), is given by the formula \( \operatorname{Var}(X) = E[X^2] - (E[X])^2 \). From Steps 1 and 4, we have \( E[X^2] = \frac{3}{5} \) and \( E[X] = \frac{1}{5} \). Thus, \[ \operatorname{Var}(X) = \frac{3}{5} - \left( \frac{1}{5} \right)^2 = \frac{3}{5} - \frac{1}{25} = \frac{15}{25} - \frac{1}{25} = \frac{14}{25} \]

Key Concepts

Understanding Discrete Random Variables

A discrete random variable is a type of random variable that takes on a countable number of distinct values. This means that you can list all possible outcomes, like the roll of a die or the number of goals scored in a sports match. In our exercise, the discrete random variable is denoted by \( X \) and can take values -1, 0, or 1.
Each value of \( X \) has a corresponding probability. These probabilities must add up to 1, ensuring that one of the possible outcomes will occur. For our exercise, the probability distribution is given by:

• \( P(X = -1) = \frac{1}{5} \)
• \( P(X = 0) = \frac{2}{5} \)
• \( P(X = 1) = \frac{2}{5} \)

This distribution allows us to calculate expectations and variances by applying formulas specific to discrete random variables.

Expected Value Calculation

The expected value is a central concept in probability, helping us to find the average or mean expected outcome of a random variable after many trials. To calculate the expected value \( \mathrm{E}[X] \) of a discrete random variable, use the formula:
\[ \mathrm{E}[X] = \sum x_i \cdot P(x_i) \]
Where \( x_i \) represents each possible value of \( X \) and \( P(x_i) \) is the probability that \( X \) equals \( x_i \).
Applying it to our exercise:

• \( \mathrm{E}[X] = (-1) \cdot \frac{1}{5} + 0 \cdot \frac{2}{5} + 1 \cdot \frac{2}{5} \)
• \( \mathrm{E}[X] = -\frac{1}{5} + 0 + \frac{2}{5} = \frac{1}{5} \)

This shows that our expected average value after numerous trials of our random variable \( X \) is \( \frac{1}{5} \).
The expected value is an important measure that provides insight into the balance point of the probability distribution.

Variance Computation

Variance measures the spread or variability of a random variable from its expected value. For a discrete random variable \( X \), variance helps us understand how much the values of \( X \) deviate from its expected value on average. It is calculated with the formula:
\[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 \]
To use this, we first need the expected value of \( X^2 \), which is computed similarly to the expected value of \( X \):

• \( E[X^2] = (-1)^2 \cdot \frac{1}{5} + 0^2 \cdot \frac{2}{5} + 1^2 \cdot \frac{2}{5} \)
• \( E[X^2] = 1 \cdot \frac{1}{5} + 0 + 1 \cdot \frac{2}{5} = \frac{3}{5} \)

Now, using \( \mathrm{E}[X] = \frac{1}{5} \):

• \( \operatorname{Var}(X) = \frac{3}{5} - \left( \frac{1}{5} \right)^2 \)
• \( \operatorname{Var}(X) = \frac{3}{5} - \frac{1}{25} = \frac{15}{25} - \frac{1}{25} = \frac{14}{25} \)

The variance of \( \frac{14}{25} \) illustrates that the values of \( X \) vary from its mean, which in this case is \( \frac{1}{5} \). A larger variance indicates more spread out values.