To solve the problem, let's break it down step by step.

Problem Summary

You are sampling uniformly and with replacement from the set $\{1, 2, 3\}$ three times. Let $X_i$ represent the value of the $i$th trial. For $k \leq 3$, denote by $X_{(k)}$ the $k$th smallest of $X_1, X_2, X_3$. The task is to find the PMF (Probability Mass Function) and CDF (Cumulative Distribution Function) of $X_{(1)}$, the minimum of $X_1, X_2, X_3$.

Step 1: Understanding $X_{(1)}$

$X_{(1)}$ represents the minimum value of three independent and identically distributed (i.i.d.) random variables $X_1, X_2, X_3$, each of which takes values in the set $\{1, 2, 3\}$.

Step 2: Finding the PMF of $X_{(1)}$

To find the PMF of $X_{(1)}$, we need to calculate the probabilities of $X_{(1)} = 1$, $X_{(1)} = 2$, and $X_{(1)} = 3$.

1. $P(X_{(1)} = 1)$:

   • $X_{(1)} = 1$ if and only if all three trials result in the value 1. The probability of each trial being 1 is $\frac{1}{3}$, and since trials are independent, $P(X_{(1)} = 1) = 1 - P(\text{All trials are greater than 1}) = 1 - \left(\frac{2}{3}\right)^3 = \frac{19}{27}$.

2. $P(X_{(1)} = 2)$:

   • $X_{(1)} = 2$ if the minimum value among the three trials is 2. This happens when at least one trial is 2 and the others are 2 or 3. The probability is: $P(X_{(1)} = 2) = P(\text{At least one trial is 2 and no trial is 1})$ The complement is that all trials are 3, so: $P(X_{(1)} = 2) = \left(\frac{2}{3}\right)^3 - \left(\frac{1}{3}\right)^3 = \frac{8}{27}$

3. $P(X_{(1)} = 3)$:

   • $X_{(1)} = 3$ occurs if all the trials result in 3, which has a probability of: $P(X_{(1)} = 3) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

Step 3: Finding the CDF of $X_{(1)}$

The CDF $F_{X_{(1)}}(x)$ is the probability that $X_{(1)} \leq x$.

1. For $x < 1$:

   • $F_{X_{(1)}}(x) = 0$.

2. For $1 \leq x < 2$:

   • $F_{X_{(1)}}(x) = P(X_{(1)} \leq 1) = P(X_{(1)} = 1) = \frac{19}{27}$.

3. For $2 \leq x < 3$:

   • $F_{X_{(1)}}(x) = P(X_{(1)} \leq 2) = P(X_{(1)} = 1) + P(X_{(1)} = 2) = \frac{19}{27} + \frac{8}{27} = 1 - \frac{1}{27} = \frac{26}{27}$.

4. For $x \geq 3$:

   • $F_{X_{(1)}}(x) = P(X_{(1)} \leq 3) = 1$.

Summary of Results

• PMF of $X_{(1)}$: $P(X_{(1)} = 1) = \frac{19}{27}, \quad P(X_{(1)} = 2) = \frac{8}{27}, \quad P(X_{(1)} = 3) = \frac{1}{27}$
• CDF of $X_{(1)}$: 0, & \text{if } x < 1 \\ \frac{19}{27}, & \text{if } 1 \leq x < 2 \\ \frac{26}{27}, & \text{if } 2 \leq x < 3 \\ 1, & \text{if } x \geq 3 \end{cases}$$

Would you like more details on any part of this solution, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the PMF change if the set were $\{1, 2, 3, 4\}$ and you sampled 4 times instead of 3?
2. What is the expected value of $X_{(1)}$?
3. How do you compute the PMF of $X_{(2)}$, the second smallest value?
4. How does the concept of order statistics generalize for larger sample sizes?
5. What is the probability that $X_{(1)} = X_{(2)}$?

Tip: When calculating CDFs, remember that they are non-decreasing functions and reach 1 as the argument goes to the maximum value in the distribution.