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The problem presented involves finding the distribution of a transformed random variable \( Y \), where \( X \) follows a Pareto distribution with parameters \( \alpha \) and \( \theta \). The transformation given is \( Y = \ln\left(\frac{X + \theta}{\theta}\right) \).

Here’s a breakdown of the steps outlined in the image:

1. **Transformation Function:**
   \[
   Y = g(X) = \ln\left(\frac{X + \theta}{\theta}\right) \implies X = \theta \left(e^Y - 1\right) = k(Y)
   \]
   This represents the transformation between \( X \) and \( Y \).

2. **Jacobian (Derivative of Transformation):**
   \[
   k'(Y) = \frac{d}{dY} \left[ \theta (e^Y - 1) \right] = \theta e^Y
   \]

3. **Pareto Distribution of \( X \):**
   The probability density function (pdf) of \( X \), given that \( X \) is Pareto-distributed with parameters \( \alpha \) and \( \theta \), is:
   \[
   f_X(x) = \frac{\alpha \theta^\alpha}{(x + \theta)^{\alpha + 1}}
   \]

4. **Distribution of \( Y \):**
   To find the distribution of \( Y \), use the change of variables formula:
   \[
   f_Y(y) = f_X(k(y)) \cdot |k'(y)|
   \]
   Substituting the expressions for \( k(Y) \) and \( k'(Y) \):
   \[
   f_Y(y) = \frac{\alpha \theta^\alpha}{\left[\theta (e^y - 1) + \theta\right]^{\alpha + 1}} \cdot \theta e^y
   \]
   Simplifying:
   \[
   f_Y(y) = \frac{\alpha \theta^\alpha}{\left[\theta e^y\right]^{\alpha + 1}} \cdot \theta e^y = \alpha e^{-\alpha y}
   \]

5. **Result:**
   The pdf of \( Y \) is an exponential distribution with the form:
   \[
   f_Y(y) = \alpha e^{-\alpha y}
   \]
   which is the pdf of an exponential distribution with mean \( \frac{1}{\alpha} \).

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Would you like further details on any of these steps?

Here are five related questions to deepen your understanding:
1. How is the Pareto distribution defined and what are its key properties?
2. What is the general formula for changing variables in probability distributions?
3. Why does the exponential distribution arise from this transformation?
4. Can this transformation be applied to other distributions, and if so, how?
5. How can the mean and variance of an exponential distribution be derived?

**Tip:** When transforming variables in probability, the Jacobian (derivative) plays a crucial role in adjusting the probability density correctly.

Find the distribution of Y, where X follows a Pareto distribution with parameters α and θ, and the transformation is Y = ln((X + θ) / θ).

This page explains how to derive the distribution of Y from a Pareto-distributed X with a transformation Y = ln((X + θ) / θ). The process involves using Jacobians and results in an exponential distribution.

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