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To approach this problem, we need to use the **moment-generating function (MGF)** and **pivotal quantity** methods to derive the result and construct a confidence interval. Here's the step-by-step process:

### Step 1: Moment-Generating Function of Gamma Distribution

The moment-generating function (MGF) \( M_Y(t) \) of a gamma-distributed random variable \( Y \sim \text{Gamma}(\alpha, \beta) \) with shape parameter \( \alpha \) and rate parameter \( \beta \) is given by:

\[
M_Y(t) = \left( \frac{1}{1 - \beta t} \right)^\alpha \quad \text{for} \ t < \frac{1}{\beta}
\]

In our case, \( \alpha = 2 \), so the MGF becomes:

\[
M_Y(t) = \left( \frac{1}{1 - \beta t} \right)^2
\]

### Step 2: Apply the Moment-Generating Function to \( 2Y/\beta \)

We are tasked with proving that \( \frac{2Y}{\beta} \) follows a \( \chi^2 \) distribution with 4 degrees of freedom. Let's manipulate the MGF to demonstrate this result.

Consider the random variable \( Z = \frac{2Y}{\beta} \). The MGF of \( Z \) is the MGF of \( Y \) evaluated at \( t/\beta \):

\[
M_Z(t) = M_Y\left( \frac{t}{\beta} \right) = \left( \frac{1}{1 - \beta \cdot \frac{t}{\beta}} \right)^2 = \left( \frac{1}{1 - t} \right)^2
\]

This is the MGF of a chi-squared distribution with 4 degrees of freedom (since the MGF of \( \chi^2_4 \) is \( \left( 1 - 2t \right)^{-2} \)). Therefore:

\[
\frac{2Y}{\beta} \sim \chi^2_4
\]

### Step 3: Deriving the Confidence Interval for \( \beta \)

Now that we know \( \frac{2Y}{\beta} \sim \chi^2_4 \), we can use this fact to derive a 90% confidence interval for \( \beta \). The pivotal quantity \( Q = \frac{2Y}{\beta} \) follows a chi-squared distribution with 4 degrees of freedom. Let \( \chi^2_{4, \alpha/2} \) and \( \chi^2_{4, 1-\alpha/2} \) denote the critical values of the chi-squared distribution with 4 degrees of freedom at the 5th and 95th percentiles, respectively.

A 90% confidence interval for \( \frac{2Y}{\beta} \) is:

\[
\chi^2_{4, 0.05} \leq \frac{2Y}{\beta} \leq \chi^2_{4, 0.95}
\]

Solving for \( \beta \), we have:

\[
\frac{2Y}{\chi^2_{4, 0.95}} \leq \beta \leq \frac{2Y}{\chi^2_{4, 0.05}}
\]

Thus, the 90% confidence interval for \( \beta \) is:

\[
\left( \frac{2Y}{\chi^2_{4, 0.95}}, \frac{2Y}{\chi^2_{4, 0.05}} \right)
\]

### Conclusion

- The random variable \( \frac{2Y}{\beta} \) follows a chi-squared distribution with 4 degrees of freedom.
- A 90% confidence interval for \( \beta \) is \( \left( \frac{2Y}{\chi^2_{4, 0.95}}, \frac{2Y}{\chi^2_{4, 0.05}} \right) \).

Let me know if you would like further clarifications or additional details on any steps!

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### Questions to expand:
1. How is the moment-generating function derived for different distributions?
2. Can you explain how pivotal quantities help in constructing confidence intervals?
3. What are the properties of the gamma distribution?
4. How can we generalize this method for any degrees of freedom?
5. How does this approach change for other confidence levels (e.g., 95%)?

### Tip:
Remember that MGFs can uniquely determine the distribution of a random variable, making them a powerful tool for finding distributions of transformed variables!

Suppose that the random variable Y has a gamma distribution with parameters α=2 and an unknown β. Use the method of moment-generating functions to prove a general result implying that 2Y/β has a χ² distribution with 4 degrees of freedom (df). Using 2Y/β as a pivotal quantity, derive a 90% confidence interval for β.

This solution explores the use of moment-generating functions (MGF) to prove that 2Y/β follows a chi-squared distribution with 4 degrees of freedom for a gamma-distributed random variable Y with parameters α=2 and unknown β. It also derives a 90% confidence interval for β using the pivotal quantity.

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