Q 2.5. Let X and Y be independent Exponential(λ) random variables. Let M = min{X, Y }
and U = max{X, Y }.
1. Compute the distribution function and probability density of M.
2. Compute the distribution function and probability density of U.
3. Compute the joint distribution function F(M,U)
(m, u) of M and U, and thus deduce the
joint density fM,U (m, u).

To tackle this problem, let’s delve into each part step by step.

### 1. Distribution Function and Probability Density of \( M \)

Since \( X \) and \( Y \) are independent exponential random variables with parameter \( \lambda \):

- The cumulative distribution function (CDF) for an exponential random variable \( X \) is given by:
  \[
  F_X(x) = 1 - e^{-\lambda x}, \quad x \geq 0
  \]

The minimum \( M = \min(X, Y) \) has the CDF:
\[
F_M(m) = P(M \leq m) = P(X \leq m \text{ or } Y \leq m) = 1 - P(X > m)P(Y > m)
\]
\[
= 1 - (1 - F_X(m))(1 - F_Y(m)) = 1 - e^{-\lambda m} \cdot e^{-\lambda m} = 1 - e^{-2\lambda m}
\]

Thus, the probability density function (PDF) of \( M \) is the derivative of the CDF:
\[
f_M(m) = \frac{d}{dm} F_M(m) = 2\lambda e^{-2\lambda m}, \quad m \geq 0
\]

### 2. Distribution Function and Probability Density of \( U \)

For the maximum \( U = \max(X, Y) \):
\[
F_U(u) = P(U \leq u) = P(X \leq u \text{ and } Y \leq u) = P(X \leq u)P(Y \leq u) = (1 - e^{-\lambda u})^2
\]

Thus, the PDF of \( U \):
\[
f_U(u) = \frac{d}{du} F_U(u) = 2\lambda e^{-\lambda u}(1 - e^{-\lambda u}), \quad u \geq 0
\]

### 3. Joint Distribution Function \( F(M, U) \)

To find the joint distribution \( F(M, U) \):
\[
F(m, u) = P(M \leq m, U \leq u) = P(\min(X, Y) \leq m, \max(X, Y) \leq u)
\]

This implies:
\[
= P(X \leq u, Y \leq u) - P(X \leq m, Y \leq m) = F_U(u) - P(X \leq m, Y \leq m)
\]
\[
= (1 - e^{-\lambda u})^2 - (1 - e^{-\lambda m})^2
\]

Thus, the joint PDF \( f_{M,U}(m, u) \):
\[
f_{M,U}(m, u) = \frac{\partial^2}{\partial m \partial u} F(m, u)
\]

Calculating this leads to:
\[
f_{M,U}(m, u) = \frac{\partial}{\partial u}(2\lambda e^{-\lambda u}(1 - e^{-\lambda u})) = 2\lambda e^{-\lambda u} \cdot \lambda e^{-\lambda m} = 2\lambda^2 e^{-\lambda(m + u)}
\]

### Summary

1. \( F_M(m) = 1 - e^{-2\lambda m} \), \( f_M(m) = 2\lambda e^{-2\lambda m} \)
2. \( F_U(u) = (1 - e^{-\lambda u})^2 \), \( f_U(u) = 2\lambda e^{-\lambda u}(1 - e^{-\lambda u}) \)
3. \( F(M, U) \) leads to the joint density \( f_{M,U}(m, u) = 2\lambda^2 e^{-\lambda(m + u)} \)

Would you like to delve into any particular part or have more questions? Here are some suggestions:

1. How do independent random variables influence the behavior of \( M \) and \( U \)?
2. Can you provide examples of applications for these distributions?
3. What are the properties of exponential distributions?
4. How do you compute moments for \( M \) and \( U \)?
5. How does the joint density relate to other probability distributions? 

**Tip:** Familiarize yourself with the memoryless property of exponential distributions, as it simplifies many calculations.

Q 2.5. Let X and Y be independent Exponential(λ) random variables. Let M = min{X, Y} and U = max{X, Y}. 1. Compute the distribution function and probability density of M. 2. Compute the distribution function and probability density of U. 3. Compute the joint distribution function F(M,U)(m,u) of M and U, and thus deduce the joint density fM,U(m,u).

Explore the distribution functions and densities of M = min{X, Y} and U = max{X, Y}, where X and Y are independent Exponential(λ) random variables. This comprehensive guide includes key formulas and derivations suitable for advanced students.

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