Problem 81
Question
Show that $$ f_{Y}(y)=\frac{1}{y^{2}}, \quad y \geq 1 $$ is a valid pdf but that \(Y\) does not have a finite expected value.
Step-by-Step Solution
Verified Answer
The function \( f_{Y}(y)=\frac{1}{y^{2}}, y \geq 1 \) is indeed a probability density function since it is always positive and that its integral over the entire space sums up to 1. However, the expected value of the random variable 'Y' under this pdf does not exist because it diverges to infinity.
1Step 1: Verification of the Non-negative Condition
Check if \( f_{Y}(y)=\frac{1}{y^{2}} \) is non-negative over all \( y \geq 1 \). Since the function \( f_{Y}(y)=\frac{1}{y^{2}} \) is always positive for \( y \geq 1 \), thus the non-negative condition is satisfied.
2Step 2: Verification of the Total Probability Condition
Check if the total integral of the function over all y is equal to 1. This is done by evaluating \( \int_{1}^{\infty} f_{Y}(y) dy \). The result of this integral is \( \left[-\frac{1}{y}\right]_{1}^{\infty}= 1 - 0 = 1 \) which shows the probabilistic integral condition is also satisfied.
3Step 3: Calculation of Expected Value
Try calculating the expected value by evaluating \( \int_{1}^{\infty} y \cdot f_{Y}(y) dy \), which comes out to be \( \int_{1}^{\infty} \frac{1}{y} dy \). The result of this integral is \[ \left[ \ln|y| \right]_{1}^{\infty} = \infty - 0 = \infty \], this shows that the expected value is infinite.
Key Concepts
Non-negative ConditionTotal Probability ConditionExpected Value of a Random Variable
Non-negative Condition
Understanding the non-negative condition is crucial when determining if a function can be a valid probability density function (pdf). In the context of probability, the pdf represents how the probability is distributed over various values of a random variable. For a pdf, one fundamental requirement is the non-negative condition. This states that the probability of the random variable assuming a particular value or within a range should always be non-negative.
This makes intuitive sense because you cannot have a negative probability; it's like saying there's less than no chance of something happening, which is a logical impossibility. In our example, the function given is
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
This function is always positive when \(y\) is greater than or equal to 1, meaning that we never encounter a scenario where the probability dips into the negatives. Therefore, for all intensively practical purposes, the function satisfies the non-negative condition, adhering to one of the basic properties of a probability density function.
This makes intuitive sense because you cannot have a negative probability; it's like saying there's less than no chance of something happening, which is a logical impossibility. In our example, the function given is
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
This function is always positive when \(y\) is greater than or equal to 1, meaning that we never encounter a scenario where the probability dips into the negatives. Therefore, for all intensively practical purposes, the function satisfies the non-negative condition, adhering to one of the basic properties of a probability density function.
Total Probability Condition
The total probability condition is another essential property that a function must satisfy to be considered a valid pdf. This condition implies that when you integrate (sum up) all the probabilities across all possible values of the random variable, the total must equal 1. This is based on the concept that a random variable has to fall into some value within the given range, and the sum of the probabilities of all these values must be certain or 100%.
In the case of the given function
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
we are asked to verify if the integral from 1 to infinity equals 1. By performing the calculation
\( \text{\( \int_{1}^{\infty} f_{Y}(y) dy \)} = \left[ -\frac{1}{y} \right]_{1}^{\infty} = 1 - 0 = 1 \),
this confirms that the integral across all the values of \(y\) sums to 1, satisfying the total probability condition. It's like confirming that our 'probability pie' is a whole pie, without any missing pieces.
In the case of the given function
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
we are asked to verify if the integral from 1 to infinity equals 1. By performing the calculation
\( \text{\( \int_{1}^{\infty} f_{Y}(y) dy \)} = \left[ -\frac{1}{y} \right]_{1}^{\infty} = 1 - 0 = 1 \),
this confirms that the integral across all the values of \(y\) sums to 1, satisfying the total probability condition. It's like confirming that our 'probability pie' is a whole pie, without any missing pieces.
Expected Value of a Random Variable
The expected value of a random variable is the average, or mean, value it would take over an infinite number of observations. In more precise terms, it's the long-run average outcome of a random phenomenon that's been modeled. The concept is similar to the idea of the center of mass in physics - a balancing point.
Calculating the expected value for a pdf involves integrating over all possible values of the variable with each value weighted by its probability. For our function
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
this requires evaluating the integral of \( y \cdot f_{Y}(y) \) with respect to \(y\) from 1 to infinity:
\( \left[ \ln|y| \right]_{1}^{\infty} = \infty - 0 = \infty \).
This results in infinity, which indicates that the expected value does not exist in the finite sense. This occurs with some pdfs; they may meet other requirements, like the non-negative and total probability conditions, but they can still have an infinite expected value. It's like trying to find the average of a set of numbers that include infinitely large values; the average becomes immeasurable.
Calculating the expected value for a pdf involves integrating over all possible values of the variable with each value weighted by its probability. For our function
\( f_{Y}(y)=\frac{1}{y^{2}} \text{ for } y \text{ \textgreater= } 1 \),
this requires evaluating the integral of \( y \cdot f_{Y}(y) \) with respect to \(y\) from 1 to infinity:
\( \left[ \ln|y| \right]_{1}^{\infty} = \infty - 0 = \infty \).
This results in infinity, which indicates that the expected value does not exist in the finite sense. This occurs with some pdfs; they may meet other requirements, like the non-negative and total probability conditions, but they can still have an infinite expected value. It's like trying to find the average of a set of numbers that include infinitely large values; the average becomes immeasurable.
Other exercises in this chapter
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