The problem involves sampling uniformly and with replacement from the set $\{1, 2, 3\}$ three times. Let $X_i$ (where $1 \leq i \leq 3$) be 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 question asks to find the Probability Mass Function (PMF) and Cumulative Distribution Function (CDF) of $X_{(2)}$, the second smallest value.

Solution Outline:

Step 1: Understanding the Distribution of $X_1, X_2, X_3$

Each $X_i$ is independently sampled from $\{1, 2, 3\}$ with equal probability. Therefore, each $X_i$ has the probability distribution:

$P(X_i = 1) = \frac{1}{3}, \quad P(X_i = 2) = \frac{1}{3}, \quad P(X_i = 3) = \frac{1}{3}$

Step 2: Enumerating Possible Outcomes

Since we sample three times, the total number of possible outcomes is $3^3 = 27$. We need to calculate the second smallest value ($X_{(2)}$) for each of these outcomes.

Step 3: Calculating the PMF of $X_{(2)}$

The second smallest value $X_{(2)}$ can take on values 1, 2, or 3. We will determine the probability of each of these outcomes by counting the number of scenarios where $X_{(2)} = k$ for each $k \in \{1, 2, 3\}$.

1. $X_{(2)} = 1$: For this to happen, all three values must be 1. There is only 1 such combination: $(1, 1, 1)$.

2. $X_{(2)} = 2$: Here, $X_{(2)} = 2$ when the values are either $(1, 2, 2)$, $(2, 1, 2)$, $(2, 2, 1)$, $(1, 2, 3)$, $(2, 1, 3)$, or $(3, 1, 2)$. There are 6 such combinations.

3. $X_{(2)} = 3$: Here, $X_{(2)} = 3$ when the values are either $(2, 3, 3)$, $(3, 2, 3)$, $(3, 3, 2)$, or $(3, 3, 3)$. There are 4 such combinations.

Using this, we compute the probabilities as follows:

• $P(X_{(2)} = 1) = \frac{1}{27}$
• $P(X_{(2)} = 2) = \frac{6}{27} = \frac{2}{9}$
• $P(X_{(2)} = 3) = \frac{4}{27}$

Step 4: Computing the CDF of $X_{(2)}$

The cumulative distribution function (CDF) is computed as:

$F_{X_{(2)}}(x) = P(X_{(2)} \leq x)$

$F_{X_{(2)}}(1) = P(X_{(2)} = 1) = \frac{1}{27}$ $F_{X_{(2)}}(2) = P(X_{(2)} \leq 2) = P(X_{(2)} = 1) + P(X_{(2)} = 2) = \frac{1}{27} + \frac{6}{27} = \frac{7}{27}$ $F_{X_{(2)}}(3) = P(X_{(2)} \leq 3) = 1$

Final Answers:

• PMF: $P(X_{(2)} = 1) = \frac{1}{27}, \quad P(X_{(2)} = 2) = \frac{14}{27}, \quad P(X_{(2)} = 3) = \frac{12}{27}$

• CDF:

  \begin{cases} 0 & \text{if } x < 1 \\ \frac{1}{27} & \text{if } 1 \leq x < 2 \\ \frac{7}{27} & \text{if } 2 \leq x < 3 \\ 1 & \text{if } x \geq 3 \end{cases}$$

Would you like further details on any part of this solution?

Related Questions:

1. What is the probability distribution for $X_{(1)}$, the smallest value?
2. How would the PMF and CDF change if sampling was without replacement?
3. What are the expected value and variance of $X_{(2)}$?
4. How does the distribution change if the sample space is $\{1, 2, 3, 4\}$?
5. How can we generalize this approach to more than three trials?

Tip: When dealing with order statistics, listing all possible outcomes and focusing on the specific statistics you're interested in can significantly simplify the problem.