The problem provided involves sampling uniformly and with replacement from the set $\{1, 2, 3\}$ three times. Let $X_i$ for $i = 1, 2, 3$ be the value of the $i$th trial. The problem then defines $X_{(k)}$ as the $k$th smallest of $X_1, X_2, X_3$. Specifically, part (a) asks to find the Probability Mass Function (PMF) and Cumulative Distribution Function (CDF) of $X_{(2)}$, the second smallest value among $X_1, X_2, X_3$.

1. Understanding the Distribution of $X_{(2)}$

Since the sampling is done with replacement from $\{1, 2, 3\}$, each $X_i$ can independently take the value of 1, 2, or 3 with equal probability, $P(X_i = x) = \frac{1}{3}$ for $x \in \{1, 2, 3\}$.

2. Identifying the Possible Values of $X_{(2)}$

The second smallest value $X_{(2)}$ can be:

• $X_{(2)} = 1$, when at least two of the values among $X_1, X_2, X_3$ are 1.
• $X_{(2)} = 2$, when at least one value is 1, and at least one value is 2 (with none being 3).
• $X_{(2)} = 3$, when all values are 3.

3. Calculating the PMF

Let's compute the probability of each possible outcome:

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

This is the probability that at least two of $X_1, X_2, X_3$ are 1. The only way this can happen is if all three $X_i$'s are 1: $P(X_{(2)} = 1) = P(X_1 = 1, X_2 = 1, X_3 = 1) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

b. $P(X_{(2)} = 2)$

This happens in two scenarios:

• One value is 1 and the other two are 2.
• Two values are 1 and the third value is 2.

The probabilities for each case are: $\text{Case 1: } P(\text{one 1, two 2's}) = \binom{3}{1} \left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^2 = 3 \cdot \frac{1}{3} \cdot \frac{1}{9} = \frac{3}{27} = \frac{1}{9}$ $\text{Case 2: } P(\text{two 1's, one 2}) = \binom{3}{2} \left(\frac{1}{3}\right)^2 \left(\frac{1}{3}\right) = 3 \cdot \frac{1}{9} \cdot \frac{1}{3} = \frac{3}{27} = \frac{1}{9}$ So, $P(X_{(2)} = 2) = \frac{1}{9} + \frac{1}{9} = \frac{2}{9}$.

c. $P(X_{(2)} = 3)$

This happens when all three values are 3: $P(X_{(2)} = 3) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

4. Constructing the PMF of $X_{(2)}$

Summarizing the PMF: $P(X_{(2)} = 1) = \frac{1}{27}, \quad P(X_{(2)} = 2) = \frac{2}{9}, \quad P(X_{(2)} = 3) = \frac{1}{27}$

5. Constructing the CDF of $X_{(2)}$

The CDF $F_{X_{(2)}}(x)$ is given by: $F_{X_{(2)}}(x) = P(X_{(2)} \leq x)$ $F_{X_{(2)}}(1) = P(X_{(2)} \leq 1) = P(X_{(2)} = 1) = \frac{1}{27}$ $F_{X_{(2)}}(2) = P(X_{(2)} \leq 2) = P(X_{(2)} = 1) + P(X_{(2)} = 2) = \frac{1}{27} + \frac{2}{9} = \frac{1}{27} + \frac{6}{27} = \frac{7}{27}$ $F_{X_{(2)}}(3) = P(X_{(2)} \leq 3) = P(X_{(2)} = 1) + P(X_{(2)} = 2) + P(X_{(2)} = 3) = \frac{7}{27} + \frac{1}{27} = \frac{8}{27}$

Would you like further details on any of these steps, or do you have any questions?

Relative Questions:

1. How would the PMF and CDF change if we sampled without replacement?
2. What if we increased the number of trials from three to four? How would this affect the distribution of $X_{(2)}$?
3. Can you compute the expectation and variance of $X_{(2)}$?
4. How does the distribution of the maximum value $X_{(3)}$ compare to that of $X_{(2)}$?
5. What is the joint distribution of $X_{(1)}$, $X_{(2)}$, and $X_{(3)}$?

Tip:

When dealing with order statistics, it's important to carefully consider the arrangement of the sampled values and how they contribute to the overall distribution.