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If \(X\) has the pdf \(f_{X}(x)=2 x, 0

Short Answer

Expert verified
The pdf of \(Y\) is \(f_Y(y) = \frac{2}{y^3}\), for \(y > 1\). It follows a special case of the Pareto distribution.

Step by step solution

01

Understand the Transformation

We need to find the probability density function (pdf) of the random variable \(Y\) where \(Y = \frac{1}{X}\). \(X\) has the pdf \(f_{X}(x)=2x\) for \(0 < x < 1\). First, we define \(Y\) in terms of \(X\), so \(y = \frac{1}{x}\) and hence \(x = \frac{1}{y}\).
02

Find the Support of Y

The support of \(Y\) is derived from the values \(X\) can take. Since \(0 < x < 1\), \(x = \frac{1}{y}\) implies that \(1 < y < \infty\). Thus, the support of \(Y\) is \(1 < y < \infty\).
03

Determine the Jacobian

To change variables, we need the derivative of \(x\) with respect to \(y\). Since \(x = \frac{1}{y}\), the derivative is \(\frac{d}{dy}(x) = \frac{d}{dy} \left( \frac{1}{y} \right) = -\frac{1}{y^2}\). The absolute value of this derivative is \(\left| -\frac{1}{y^2} \right| = \frac{1}{y^2}\).
04

Apply the Change of Variables Formula

Using the formula for transformation of random variables, the pdf of \(Y\) can be found as \(f_{Y}(y) = f_X(x) \cdot \left| \frac{dx}{dy} \right|\). Substitute \(x = \frac{1}{y}\), so \(f_X\left(\frac{1}{y}\right) = 2 \cdot \frac{1}{y}\).Thus, \(f_Y(y) = 2 \cdot \frac{1}{y} \cdot \frac{1}{y^2} = \frac{2}{y^3}\).
05

Verify the PDF is Valid

Finally, we verify that the obtained pdf integrates to 1 over the support \(1 < y < \infty\):\[\int_1^\infty \frac{2}{y^3} \, dy = \left[ -\frac{2}{2y^2} \right]_1^\infty = 1\]This confirms that \(f_Y(y) = \frac{2}{y^3}\) is indeed a valid pdf over \(1 < y < \infty\) and corresponds to a special case of the Pareto distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pareto Distribution
The Pareto distribution is a fascinating concept in probability and statistics, widely used in economics and social sciences to describe scenarios where a small number of entities hold the majority of a given resource. Named after Italian economist Vilfredo Pareto, this distribution helps model the idea that wealth, for example, is unevenly distributed.
The fundamental concept behind the Pareto distribution is the "80/20 rule." This idea suggests that 80% of the effects come from 20% of the causes. In the context of probability, the Pareto distribution is defined using a parameter, often denoted \(a\). It generally has the probability density function (pdf): \( f(x) = \frac{a}{x^{a+1}} \), for \( x > x_m \), where \( x_m \) is a scale parameter and \( a > 0 \).
Understanding the Pareto distribution helps explain phenomena such as wealth distribution, file size distribution in internet traffic, and natural phenomena like forest fire sizes. In our specific exercise, the transformed random variable \( Y = \frac{1}{X} \) represents a special case of the Pareto distribution.
Transformation of Variables
The transformation of variables is a crucial technique in probability and statistics, enabling the conversion of a probability distribution of one variable into another variable. Essentially, it helps us to understand how changing variables affects the overall distribution.
In our exercise, we start with a variable \(X\), described by the probability density function (pdf) \(f_X(x)=2x\) for \(0 < x < 1\). We desire to find the pdf for \(Y = \frac{1}{X}\). The transformation process requires expressing \(Y\) in terms of \(X\) and vice versa. Here, since \(y = \frac{1}{x}\), we rearrange to get \(x = \frac{1}{y}\).
Next, we identify the new support, which is the range of \(Y\). Given \(0 < x < 1\), this transformation leads to \(1 < y < \infty\). With these steps, we are equipped to apply further techniques, such as the Jacobian, to find the new pdf.
Jacobian
The Jacobian is a determinant used in transformation of variables, especially when dealing with functions of multiple variables. In the context of probability distributions, the Jacobian assists in calculating the new probability density function after transforming a random variable.
In our exercise, we use the Jacobian to adjust for the change from \(X\) to \(Y = \frac{1}{X}\). First, we need the derivative of \(x\) with respect to \(y\). Since \(x = \frac{1}{y}\), the derivative is \(\frac{d}{dy}(x) = -\frac{1}{y^2}\). The Jacobian is expressed as the absolute value of this derivative, leading to \(|-\frac{1}{y^2}| = \frac{1}{y^2}\).
This Jacobian corrects the scale of the probability distribution during the transformation, ensuring the new pdf \(f_Y(y)\) is accurately calculated. Incorporating the Jacobian is critical for maintaining the properties of probability distributions through variable transformation.
Probability Density Function
The probability density function (pdf) is a fundamental component of continuous probability distributions. It describes the likelihood of a random variable taking on certain values. Importantly, the area under the pdf across the possible values of the variable sums to 1, aligning with the properties of all probability distributions.
For a given random variable \(X\) with pdf \(f_X(x)=2x\) for \(0 By substituting the expressions, we find \(f_Y(y) = \frac{2}{y^3}\) for \(1 < y < \infty\). The final step in this calculation is to verify that this pdf integrates to 1 over its support. Indeed, the integral from 1 to infinity results in 1, confirming that \(f_Y(y)\) is a valid probability density function.

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