Chapter 4: Problem 37
If \(Y\) is a continuous random variable with density function \(f(y)\) that is symmetric about 0 (that is, \(f(y)=f(-y)\) for all \(y\) ) and \(E(Y)\) exists, show that \(E(Y)=0 .\) [Hint: \(E(Y)=\int_{-\infty}^{0} y f(y) d y+\int_{0}^{\infty} y f(y) d y .\) Make the change of variable \(w=-y\) in the first integral.]
Short Answer
Step by step solution
Write the Expectation Formula
Change of Variable for the First Integral
Use Symmetry Property
Combine the Integrals
Conclude the Calculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous random variable
A continuous random variable is characterized by its probability density function (PDF), denoted by a function, such as \( f(y) \) in our context. The PDF describes the likelihood of the random variable taking on a specific value. It's important to note that the probability of the variable taking on exactly one specific value is actually zero; instead, probabilities are determined over intervals.
- Continuous range of values
- Uses a probability density function (PDF)
- Infinite possible outcomes within a range
Symmetric density function
For the random variable \(Y\) discussed here, the symmetry is defined mathematically as \( f(y) = f(-y) \) for all values of \(y\). This implies that if you were to plot the function, the two halves of the graph would be identical mirror images.
- Mirrored about a central point, here zero
- \( f(y) = f(-y) \) across all \(y\)
Change of variables technique
In the example given, we perform a substitution in the first integral by setting \( w = -y \), leading to \( dw = -dy \). This shifts the integration bounds and modifies the integral expression, simplifying the computation. It's a common technique for handling symmetric functions or dealing with integrals where recognizing symmetry or obtaining a function in its simpler form is beneficial.
- Facilitates integration
- Used here to reorganize symmetric components
- Simplifies symmetric function calculations
Expected value
For continuous random variables, the expected value \( E(Y) \) is given by the integral \( \int_{-\infty}^{\infty} y f(y) \, dy \). In the exercise, the symmetry of \( f(y) \) makes it possible to break this into two parts, which ultimately reveal that the overall expected value is zero.
- Represents a form of average
- Calculated through integration of \( y \times \text{PDF} \)
- Zero for symmetric functions about zero