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The given problem involves a joint probability density function (PDF) for \(n\) random variables \(X_1, X_2, \dots, X_n\) and asks for the following:

### Part (a):
We are asked to find the **joint cumulative distribution function (CDF)**, \( F_{X_1, \dots, X_n}(x_1, \dots, x_n) \).

Given the joint PDF:
\[
f_{X_1, \dots, X_n}(x_1, \dots, x_n) = \begin{cases}
1 & \text{if } 0 \leq x_i \leq 1, \, i = 1, 2, \dots, n, \\
0 & \text{otherwise},
\end{cases}
\]
the joint CDF is the integral of the joint PDF over the range of interest, meaning:
\[
F_{X_1, \dots, X_n}(x_1, \dots, x_n) = \mathbb{P}(X_1 \leq x_1, \dots, X_n \leq x_n).
\]
Since the joint PDF is uniform over \([0,1]^n\), the joint CDF will simply be the product of the individual CDFs, which for each uniform variable \(X_i\) on \([0, 1]\) is just \(x_i\) (when \(0 \leq x_i \leq 1\)). Therefore:
\[
F_{X_1, \dots, X_n}(x_1, \dots, x_n) = x_1 x_2 \cdots x_n, \quad \text{for } 0 \leq x_i \leq 1 \text{ for all } i.
\]
Outside this range, the joint CDF is either 0 or 1 depending on whether the variables \(x_1, \dots, x_n\) are out of bounds.

Thus, 
\[
F_{X_1, \dots, X_n}(x_1, \dots, x_n) = \begin{cases}
x_1 x_2 \cdots x_n, & \text{if } 0 \leq x_i \leq 1 \text{ for all } i, \\
0, & \text{if any } x_i < 0, \\
1, & \text{if any } x_i > 1.
\end{cases}
\]

### Part (b):
We are asked to find the probability:
\[
\mathbb{P}\left(\min(X_1, X_2, X_3) \leq \frac{2}{3}\right),
\]
where \(n = 3\).

The event \(\min(X_1, X_2, X_3) \leq \frac{2}{3}\) is the complement of the event \(\min(X_1, X_2, X_3) > \frac{2}{3}\). Therefore:
\[
\mathbb{P}\left(\min(X_1, X_2, X_3) \leq \frac{2}{3}\right) = 1 - \mathbb{P}\left(\min(X_1, X_2, X_3) > \frac{2}{3}\right).
\]
Now, \(\mathbb{P}\left(\min(X_1, X_2, X_3) > \frac{2}{3}\right)\) is equivalent to:
\[
\mathbb{P}(X_1 > \frac{2}{3}, X_2 > \frac{2}{3}, X_3 > \frac{2}{3}),
\]
which, because \(X_1, X_2, X_3\) are independent and uniformly distributed on \([0, 1]\), is:
\[
\mathbb{P}(X_1 > \frac{2}{3}) \cdot \mathbb{P}(X_2 > \frac{2}{3}) \cdot \mathbb{P}(X_3 > \frac{2}{3}).
\]
For each \(X_i\), \(\mathbb{P}(X_i > \frac{2}{3}) = 1 - \frac{2}{3} = \frac{1}{3}\). Therefore:
\[
\mathbb{P}(X_1 > \frac{2}{3}, X_2 > \frac{2}{3}, X_3 > \frac{2}{3}) = \left(\frac{1}{3}\right)^3 = \frac{1}{27}.
\]
Thus, the desired probability is:
\[
\mathbb{P}\left(\min(X_1, X_2, X_3) \leq \frac{2}{3}\right) = 1 - \frac{1}{27} = \frac{26}{27}.
\]

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Would you like further details on these steps or additional explanations?

### Related Questions:
1. How do you calculate joint CDFs for independent random variables in other cases?
2. What is the interpretation of the minimum of several independent uniform random variables?
3. How would the result in part (b) change if the uniform distribution had a different range, say \([a, b]\)?
4. What if the random variables \(X_i\) were dependent? How would the approach change?
5. How do we generalize the result in part (b) to \(n > 3\) random variables?

### Tip:
For uniform distributions, the cumulative distribution function (CDF) is linear within the support, which simplifies calculations, especially for independent random variables.

The random variables X_1, ..., X_n have the joint PDF as shown. Find the joint CDF and, assuming n=3, compute P[min(X_1, X_2, X_3) <= 2/3].

Learn how to calculate the joint cumulative distribution function (CDF) for independent uniform random variables. This problem also explores how to compute the probability that the minimum of three independent uniform random variables is less than or equal to 2/3.

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