Chapter 3: Problem 111
If $$ f_{Y}(y)=\frac{2 y}{k^{2}}, \quad 0 \leq y \leq k $$ for what value of \(k\) does \(\operatorname{Var}(Y)=2 ?\)
Short Answer
Expert verified
The value of k for which \(\operatorname{Var}(Y) = 2\) is \(k=\sqrt{2}\).
Step by step solution
01
Normalizing the density function
Firstly, we need to make sure that \(f_{Y}(y)\) is a valid density function. This requires that \(\int_{0}^{k} f_{Y}(y) dy = 1\). Solving the integral gives us \(\int_{0}^{k} \frac{2y}{k^2} dy = \frac{y^2}{k^2} \Biggr|_0^k = 1\), which ultimately implies \(k = \sqrt{2}\).
02
Calculating the expectation \(\mathbb{E}[Y]\)
After having found \(k\), we can calculate the expected value of the random variable \(Y\) using the formula \(\mathbb{E}[Y] = \int_{0}^{k} y f_{Y}(y) dy\). Substituting into the formula and evaluating the integral gives us \(\mathbb{E}[Y] = \int_{0}^{\sqrt{2}} \frac{2y^2}{(\sqrt{2})^2} dy = \frac{2(\sqrt{2})^3}{3}\).
03
Calculating the expectation \(\mathbb{E}[Y^2]\)
Next, the square of the expectation is required. This is given by \(\mathbb{E}[Y^2] = \int_{0}^{k} y^2 f_{Y}(y) dy\). Evaluating again gives us \(\mathbb{E}[Y^2] = \int_{0}^{\sqrt{2}} \frac{2y^3}{(\sqrt{2})^2} dy = \frac{({2})^2(\sqrt{2})^4}{5}\).
04
Variance calculation
Finally, we use the formula for the variance of a random variable, \(\operatorname{Var}(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2\). After conducting the calculations, we get \((\operatorname{Var})(Y)=2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Variance Calculation
Variance is a measure of how much a set of random variables spread out from their expected value. It quantifies the variability within a distribution. For a continuous random variable like \( Y \), its variance can be calculated using the expectation, or average, of its squared value minus the square of its expected value.
The formula for variance \( \operatorname{Var}(Y) \) is given by:
\[\operatorname{Var}(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2\]
Where:
The formula for variance \( \operatorname{Var}(Y) \) is given by:
\[\operatorname{Var}(Y) = \mathbb{E}[Y^2] - (\mathbb{E}[Y])^2\]
Where:
- \( \mathbb{E}[Y^2] \) is the expectation of the square of \( Y \).
- \( \mathbb{E}[Y] \) is the expectation of \( Y \).
Probability Density Function
A probability density function (PDF) describes the likelihood of a random variable to take on a particular value. It is a crucial part of working with continuous random variables, like those seen in this exercise. For a PDF to be valid, its integral over the entire space needs to equal 1.
Given a function \( f_{Y}(y) = \frac{2y}{k^2} \) for \(0 \leq y \leq k\), to ensure it represents a valid PDF, the integral must satisfy:
\[\int_{0}^{k} f_{Y}(y) \, dy = 1\]
This condition ensures the total area under the curve is 1, signifying 100% probability. When solving for \( k \) in this exercise, we set up the integral and solved for which \( k \) the condition is fulfilled. This vital step ensures that the function \( f_Y(y) \) can accurately describe a distribution of our random variable \( Y \).
Given a function \( f_{Y}(y) = \frac{2y}{k^2} \) for \(0 \leq y \leq k\), to ensure it represents a valid PDF, the integral must satisfy:
\[\int_{0}^{k} f_{Y}(y) \, dy = 1\]
This condition ensures the total area under the curve is 1, signifying 100% probability. When solving for \( k \) in this exercise, we set up the integral and solved for which \( k \) the condition is fulfilled. This vital step ensures that the function \( f_Y(y) \) can accurately describe a distribution of our random variable \( Y \).
Expectation Calculation
Expectation is a fundamental concept in probability that describes the average or mean value of a random variable. For continuous variables, the expectation is found using an integral of the random variable multiplied by its PDF.
The expectation of a random variable \( Y \), denoted \( \mathbb{E}[Y] \), is calculated using the formula:
The expectation of a random variable \( Y \), denoted \( \mathbb{E}[Y] \), is calculated using the formula:
- \( \mathbb{E}[Y] = \int_{0}^{k} y f_{Y}(y) \, dy \)
- \( \mathbb{E}[Y^2] = \int_{0}^{k} y^2 f_{Y}(y) \, dy \)