The task asks to construct a 90% confidence interval for the parameter $\theta$ using the statistic $X_1$, where the sample size is 1 and the distribution is a uniform distribution $R(1; \theta + 1)$. You are also given the condition that $X_1 = 1$.

Let's break down the steps:

Step 1: Understanding the distribution

The distribution is uniform $R(1; \theta + 1)$, which means that the random variable $X_1$ is uniformly distributed between 1 and $\theta + 1$. The probability density function (PDF) of $X_1$ is: $f(x; \theta) = \frac{1}{\theta}, \quad \text{for} \quad 1 \leq x \leq \theta + 1.$

Step 2: Confidence interval construction

For a uniform distribution, a pivotal quantity for constructing confidence intervals is: $P(1 \leq X_1 \leq \theta + 1) = 1.$ Given that $X_1 = 1$, we can use the following inequality to construct the confidence interval: $P(X_1 \leq \theta + 1) = 0.9.$ That is, we are looking for the value of $\theta$ such that the probability of observing $X_1 \leq \theta + 1$ is 0.9.

Since $X_1 = 1$, solving for $\theta$ gives: $\theta = 1.$

Therefore, the exact 90% confidence interval for $\theta$ when $X_1 = 1$ is $[1, \theta + 1]$.

Do you want further details on the calculation or any additional clarification?

Here are some related questions:

1. What is the general approach to constructing confidence intervals for uniform distributions?
2. How does the size of the sample impact the width of a confidence interval?
3. Why is a pivotal quantity used in confidence interval calculations?
4. Can you explain the relation between probability and confidence levels in statistics?
5. What would change in the confidence interval calculation if the sample size increased?

Tip: For uniform distributions, finding the confidence interval often involves finding boundaries based on the order statistics of the sample.