Problem 4
Question
Let \(X\) be a random variable with \(\mathrm{E}[X]=2\), \(\operatorname{Var}(X)=4\). Compute the expectation and variance of \(3-2 X\).
Step-by-Step Solution
Verified Answer
Expectation is -1, and variance is 16.
1Step 1: Identify the Problem
We need to find the expectation and variance of a linear transformation of the random variable \(X\), specifically \(Y = 3 - 2X\).
2Step 2: Use Expectation Property
To find the expectation \(\mathrm{E}[Y]\), use the linearity property of expectation. For a linear transformation \(Y = aX + b\), the expectation is given by \(\mathrm{E}[Y] = a\mathrm{E}[X] + b\). Here, \(a = -2\) and \(b = 3\). Thus, compute \(\mathrm{E}[Y] = -2 \times 2 + 3 = -4 + 3 = -1\).
3Step 3: Use Variance Property
To find the variance \(\operatorname{Var}(Y)\), use the property of variance for a linear transformation, \(\operatorname{Var}(Y) = a^2 \operatorname{Var}(X)\) when \(Y = aX + b\). In this problem, \(a = -2\), so \(\operatorname{Var}(Y) = (-2)^2 \times 4 = 4 \times 4 = 16\).
4Step 4: Combine Results
Conclude the results using the calculations from previous steps. The expectation \(\mathrm{E}[3 - 2X]\) is \(-1\) and the variance \(\operatorname{Var}(3 - 2X)\) is \(16\).
Key Concepts
Linear TransformationExpectationVariance
Linear Transformation
Linear transformation is a fundamental concept when dealing with random variables. It involves reshaping or adjusting a random variable using a linear equation. In simple terms, a linear transformation has the form \( Y = aX + b \), where \( a \) and \( b \) are constants.
This equation alters the random variable \( X \) by scaling it with \( a \) and then shifting it by \( b \). Let's break this down with our example.
This equation alters the random variable \( X \) by scaling it with \( a \) and then shifting it by \( b \). Let's break this down with our example.
- Given \( Y = 3 - 2X \), the linear transformation here involves a scaling factor of \( a = -2 \) and a shift of \( b = 3 \).
- The transformation gives us a new variable \( Y \), derived from \( X \), which impacts both expectation and variance.
- It is crucial to apply the correct mathematical properties to find the expected results for these new transformations.
Expectation
Expectation, often represented as \( \mathrm{E}[X] \), describes the average or mean value of a random variable. It's like the practical center of a probability distribution, indicating what you'd expect to get, on average, if an experiment were performed many times.
In context of linear transformations, the expectation is calculated using a specific property of expectation:
This principle is immensely helpful as it allows us to understand how shifts and scales alter the expected outcomes of transformations.
In context of linear transformations, the expectation is calculated using a specific property of expectation:
- For a transformed variable \( Y = aX + b \), the new expectation is \( \mathrm{E}[Y] = a\mathrm{E}[X] + b \).
- In our problem, substituting \( a = -2 \) and \( b = 3 \), and knowing that \( \mathrm{E}[X] = 2 \), we find \( \mathrm{E}[Y] = -2 \times 2 + 3 = -1 \).
This principle is immensely helpful as it allows us to understand how shifts and scales alter the expected outcomes of transformations.
Variance
Variance is a measure of the spread, or how much the values of a random variable deviate from the mean. It's represented by \( \operatorname{Var}(X) \). Variance is crucial because it provides insight into the consistency of the data - a small variance means data points are near the mean, while a large variance shows more spread out values.
For linear transformations, variance involves a notable property:
Variance’s role in linear transformations is pivotal as it helps us understand how changes in scale affect the variability of the new distribution.
For linear transformations, variance involves a notable property:
- For a transformed variable \( Y = aX + b \), variance is given by \( \operatorname{Var}(Y) = a^2 \operatorname{Var}(X) \).
- In the exercise, with \( a = -2 \) and \( \operatorname{Var}(X) = 4 \), the new variance becomes \( \operatorname{Var}(Y) = (-2)^2 \times 4 = 16 \).
Variance’s role in linear transformations is pivotal as it helps us understand how changes in scale affect the variability of the new distribution.
Other exercises in this chapter
Problem 2
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
View solution Problem 3
For a certain random variable \(X\) it is known that \(\mathrm{E}[X]=2\), \(\operatorname{Var}(X)=3\). What is \(\mathrm{E}\left[X^{2}\right]\) ?
View solution Problem 5
Determine the expectation and variance of the \(\operatorname{Ber}(p)\) distribution.
View solution Problem 6
The random variable \(Z\) has probability density function \(f(z)=3 z^{2} / 19\) for \(2 \leq z \leq 3\) and \(f(z)=0\) elsewhere. Determine \(\mathrm{E}[Z]\).
View solution