Suppose the random variable Y has density h(·; β) where
h(y; β) = βy−(1+β)1(y ≥ 1)
for parameter β ∈ (0, ∞). Answer the following two questions about the random variable Y .
Both solutions may be left in terms of β.
(a) (2.5pts) What is E(Y ^−1)?
(b) (2.5pts) What is P (2 ≤ Y ≤ 4)?


### Problem Setup
The given probability density function (PDF) of the random variable \( Y \) is:  
\[
h(y; \beta) = \beta y^{-(1+\beta)} \cdot 1(y \geq 1),
\]  
where \( \beta > 0 \) is a parameter and \( 1(y \geq 1) \) is the indicator function that ensures the PDF is nonzero only for \( y \geq 1 \).

---

### Part (a) \( \mathbb{E}(Y^{-1}) \)

The expected value of \( Y^{-1} \) is calculated as:
\[
\mathbb{E}(Y^{-1}) = \int_{1}^{\infty} y^{-1} h(y; \beta) \, dy.
\]
Substituting \( h(y; \beta) = \beta y^{-(1+\beta)} \), we get:
\[
\mathbb{E}(Y^{-1}) = \int_{1}^{\infty} y^{-1} \cdot \beta y^{-(1+\beta)} \, dy = \beta \int_{1}^{\infty} y^{-(2+\beta)} \, dy.
\]

#### Simplify the Integral
The integral of \( y^{-k} \) for \( k > 1 \) is given by:
\[
\int_{1}^{\infty} y^{-k} \, dy = \frac{1}{k-1}.
\]
Here, \( k = 2 + \beta > 1 \), so:
\[
\int_{1}^{\infty} y^{-(2+\beta)} \, dy = \frac{1}{(2+\beta) - 1} = \frac{1}{1+\beta}.
\]

Thus:
\[
\mathbb{E}(Y^{-1}) = \beta \cdot \frac{1}{1+\beta} = \frac{\beta}{1+\beta}.
\]

---

### Part (b) \( P(2 \leq Y \leq 4) \)

The probability \( P(2 \leq Y \leq 4) \) is:
\[
P(2 \leq Y \leq 4) = \int_{2}^{4} h(y; \beta) \, dy.
\]
Substituting \( h(y; \beta) = \beta y^{-(1+\beta)} \), we have:
\[
P(2 \leq Y \leq 4) = \int_{2}^{4} \beta y^{-(1+\beta)} \, dy = \beta \int_{2}^{4} y^{-(1+\beta)} \, dy.
\]

#### Solve the Integral
The integral of \( y^{-k} \) is:
\[
\int y^{-k} \, dy = \frac{y^{1-k}}{1-k}, \quad \text{for } k \neq 1.
\]
Here, \( k = 1+\beta > 1 \), so:
\[
\int_{2}^{4} y^{-(1+\beta)} \, dy = \left[ \frac{y^{1-(1+\beta)}}{1-(1+\beta)} \right]_{2}^{4}.
\]

Simplify:
\[
1-(1+\beta) = -\beta, \quad \text{and } 1-(1+\beta) = -\beta.
\]
\[
\int_{2}^{4} y^{-(1+\beta)} \, dy = \left[ \frac{y^{-\beta}}{-\beta} \right]_{2}^{4} = -\frac{1}{\beta} \left[ 4^{-\beta} - 2^{-\beta} \right].
\]

Substitute back into the probability:
\[
P(2 \leq Y \leq 4) = \beta \cdot \left( -\frac{1}{\beta} \right) \cdot \left( 4^{-\beta} - 2^{-\beta} \right) = 2^{-\beta} - 4^{-\beta}.
\]

---

### Final Answers
(a) \( \mathbb{E}(Y^{-1}) = \frac{\beta}{1+\beta}. \)  
(b) \( P(2 \leq Y \leq 4) = 2^{-\beta} - 4^{-\beta}. \)

---

Do you need further clarifications or details?  
Here are five follow-up questions:  
1. How do we ensure \( h(y; \beta) \) is a valid probability density function?  
2. Can you compute \( \mathbb{E}(Y) \) for the same random variable?  
3. What is the variance of \( Y \)?  
4. How does \( P(2 \leq Y \leq 4) \) behave as \( \beta \to 0 \)?  
5. Can you generalize \( P(a \leq Y \leq b) \) for arbitrary \( a, b \geq 1 \)?  

**Tip:** When solving integrals of power functions, remember the formula \( \int x^{-k} dx = \frac{x^{1-k}}{1-k} \) for \( k \neq 1 \), and check the convergence criteria carefully!

Suppose the random variable Y has density h(·; β) where h(y; β) = βy−(1+β)1(y ≥ 1) for parameter β ∈ (0, ∞). Answer the following two questions about the random variable Y: (a) What is E(Y^−1)? (b) What is P(2 ≤ Y ≤ 4)?

Solve problems involving the probability density function h(y; β) = βy^−(1+β)1(y ≥ 1) for parameter β ∈ (0, ∞). Learn how to calculate the expected value E(Y^−1) and the probability P(2 ≤ Y ≤ 4) using integration techniques. Detailed solutions included for students and professionals in statistics and calculus.

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