The problem involves sampling uniformly and with replacement from the set $\{1, 2, 3\}$ three times. Let $X_1, X_2, X_3$ be the values of the three trials, and for $k \leq 3$, $X_{(k)}$ denotes the $k$-th smallest value among $X_1, X_2, X_3$. Specifically, you need to find the PMF (Probability Mass Function) and CDF (Cumulative Distribution Function) of $X_{(3)}$, which is the largest value among $X_1, X_2, X_3$.

Step 1: Understanding $X_{(3)}$

$X_{(3)}$ is the largest value among the three sampled values, $X_1, X_2,$ and $X_3$. Since the values are drawn from the set $\{1, 2, 3\}$, $X_{(3)}$ can only take on the values 1, 2, or 3.

Step 2: Finding the Probability Mass Function (PMF) of $X_{(3)}$

To find the PMF, we determine the probability that $X_{(3)}$ equals each possible value:

1. $P(X_{(3)} = 1)$: This occurs only if all samples are 1, which has a probability: $P(X_{(3)} = 1) = P(X_1 = 1, X_2 = 1, X_3 = 1) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

2. $P(X_{(3)} = 2)$: This occurs if the largest value is 2, which means at least one value is 2 and none of the values are 3. There are several configurations:

   • One 2 and two 1's: The probability of each configuration (e.g., $X_1 = 2, X_2 = 1, X_3 = 1$) is $\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} = \frac{1}{27}$, and there are 3 such configurations, leading to a total probability of $\frac{3}{27}$.
   • Two 2's and one 1: Probability of each such configuration is also $\frac{1}{27}$, and there are 3 configurations, contributing another $\frac{3}{27}$.

   Adding these, we get: $P(X_{(3)} = 2) = \frac{6}{27}.$

3. $P(X_{(3)} = 3)$: This occurs in all other cases, which are 19 out of 27 possibilities. This is the complement of the above cases: $P(X_{(3)} = 3) = 1 - \left(\frac{1}{27} + \frac{6}{27}\right) = \frac{20}{27}.$

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

The CDF $F(x)$ of $X_{(3)}$ is given by:

• For $x < 1$, $F(x) = 0$.
• For $1 \leq x < 2$, $F(x) = P(X_{(3)} = 1) = \frac{1}{27}$.
• For $2 \leq x < 3$, $F(x) = P(X_{(3)} \leq 2) = \frac{1}{27} + \frac{6}{27} = \frac{7}{27}$.
• For $x \geq 3$, $F(x) = 1$.

Summary:

• PMF: $P(X_{(3)} = 1) = \frac{1}{27}, \quad P(X_{(3)} = 2) = \frac{6}{27}, \quad P(X_{(3)} = 3) = \frac{20}{27}.$
• CDF: \begin{cases} 0, & x < 1, \\ \frac{1}{27}, & 1 \leq x < 2, \\ \frac{7}{27}, & 2 \leq x < 3, \\ 1, & x \geq 3. \end{cases}$$

Would you like more details or have any questions?

Follow-up Questions:

1. How would the PMF and CDF change if we sampled four times instead?
2. What are the probabilities of the other order statistics $X_{(1)}$ and $X_{(2)}$?
3. Can you explain how to find order statistics for a general set $\{1, 2, \ldots, n\}$?
4. How does sampling without replacement change the probabilities?
5. What is the expected value of $X_{(3)}$?

Tip:

When dealing with order statistics, listing possible scenarios systematically helps ensure you account for all configurations.