Suppose that X and Y are independent random variables with the probability densities given below. Find the expected value of Z = XY. g(x) = 50 1 x>5, elsewhere h(y) = 2 25 0, y, 0
Suppose that X and Y are independent random variables with the probability densities given below. Find the expected value of Z = XY. g(x) = 50 1 x>5, elsewhere h(y) = 2 25 0, y, 0
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Suppose that \( X \) and \( Y \) are independent random variables with the probability densities given below. Find the expected value of \( Z = XY \).
For \( g(x) \):
\[
g(x) =
\begin{cases}
\frac{50}{x^3}, & x > 5, \\
0, & \text{elsewhere}
\end{cases}
\]
For \( h(y) \):
\[
h(y) =
\begin{cases}
\frac{2}{25}y, & 0 < y < 5, \\
0, & \text{elsewhere}
\end{cases}
\]
This problem involves finding the expected value of a product \( Z = XY \), where \( X \) and \( Y \) are independent variables with given probability density functions. The functions describe how \( X \) and \( Y \) are distributed over their respective intervals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbe7e575d-dfd7-411c-ad5f-62e2be64ba9f%2F0a997a22-2c2c-40d5-8a82-0222cf3686a9%2Fy9ueiil_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose that \( X \) and \( Y \) are independent random variables with the probability densities given below. Find the expected value of \( Z = XY \).
For \( g(x) \):
\[
g(x) =
\begin{cases}
\frac{50}{x^3}, & x > 5, \\
0, & \text{elsewhere}
\end{cases}
\]
For \( h(y) \):
\[
h(y) =
\begin{cases}
\frac{2}{25}y, & 0 < y < 5, \\
0, & \text{elsewhere}
\end{cases}
\]
This problem involves finding the expected value of a product \( Z = XY \), where \( X \) and \( Y \) are independent variables with given probability density functions. The functions describe how \( X \) and \( Y \) are distributed over their respective intervals.
![**Text and Explanation for an Educational Website:**
**Text:**
The total time, measured in units of 100 hours, that a teenager runs her hair dryer over a period of one year is a continuous random variable \( X \) that has the density function below. Use the theorem below to evaluate the mean of the random variable \( Y \).
\[ Y = 69X^2 + 43X, \] where \( Y \) is equal to the number of kilowatt hours expended annually.
\[
f(x) =
\begin{cases}
x, & 0 < x < 1 \\
\frac{4}{9} - \frac{1}{9}x, & 1 \leq x < 4 \\
0, & \text{elsewhere}
\end{cases}
\]
**Theorem:** The expected value of the sum or difference of two or more functions of a random variable \( X \) is the sum or difference of the expected values of the functions, as given by the formula below.
\[ E[g(X,Y) \pm h(X,Y)] = E[g(X,Y)] \pm E[h(X,Y)] \]
**Explanation:**
In this problem, the variability of time a teenager runs a hair dryer is represented by \( X \), and its effect on annual energy consumption (\( Y \)) is modeled by a quadratic function. The piecewise density function \( f(x) \) specifies probabilities for different intervals:
- For \( 0 < x < 1 \), the density function \( f(x) = x \) increases linearly.
- For \( 1 \leq x < 4 \), \( f(x) = \frac{4}{9} - \frac{1}{9}x \) decreases linearly.
- \( f(x) = 0 \) elsewhere, meaning no probability outside these ranges.
The theorem discusses calculating expected values: for functions of random variables like \( g(X,Y) \) and \( h(X,Y) \), the expected value of their sum or difference equals the sum or difference of their expected values.
This forms the basis for calculating averages and means in probability and statistics, which is essential for understanding probability distributions and their applications.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbe7e575d-dfd7-411c-ad5f-62e2be64ba9f%2F0a997a22-2c2c-40d5-8a82-0222cf3686a9%2F6wx28qj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Text and Explanation for an Educational Website:**
**Text:**
The total time, measured in units of 100 hours, that a teenager runs her hair dryer over a period of one year is a continuous random variable \( X \) that has the density function below. Use the theorem below to evaluate the mean of the random variable \( Y \).
\[ Y = 69X^2 + 43X, \] where \( Y \) is equal to the number of kilowatt hours expended annually.
\[
f(x) =
\begin{cases}
x, & 0 < x < 1 \\
\frac{4}{9} - \frac{1}{9}x, & 1 \leq x < 4 \\
0, & \text{elsewhere}
\end{cases}
\]
**Theorem:** The expected value of the sum or difference of two or more functions of a random variable \( X \) is the sum or difference of the expected values of the functions, as given by the formula below.
\[ E[g(X,Y) \pm h(X,Y)] = E[g(X,Y)] \pm E[h(X,Y)] \]
**Explanation:**
In this problem, the variability of time a teenager runs a hair dryer is represented by \( X \), and its effect on annual energy consumption (\( Y \)) is modeled by a quadratic function. The piecewise density function \( f(x) \) specifies probabilities for different intervals:
- For \( 0 < x < 1 \), the density function \( f(x) = x \) increases linearly.
- For \( 1 \leq x < 4 \), \( f(x) = \frac{4}{9} - \frac{1}{9}x \) decreases linearly.
- \( f(x) = 0 \) elsewhere, meaning no probability outside these ranges.
The theorem discusses calculating expected values: for functions of random variables like \( g(X,Y) \) and \( h(X,Y) \), the expected value of their sum or difference equals the sum or difference of their expected values.
This forms the basis for calculating averages and means in probability and statistics, which is essential for understanding probability distributions and their applications.
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