Math Problem Statement

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).

Solution

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

1. Distribution Function and Probability Density of MM

Since XX and YY are independent exponential random variables with parameter λ\lambda:

  • The cumulative distribution function (CDF) for an exponential random variable XX is given by: FX(x)=1eλx,x0F_X(x) = 1 - e^{-\lambda x}, \quad x \geq 0

The minimum M=min(X,Y)M = \min(X, Y) has the CDF: FM(m)=P(Mm)=P(Xm or Ym)=1P(X>m)P(Y>m)F_M(m) = P(M \leq m) = P(X \leq m \text{ or } Y \leq m) = 1 - P(X > m)P(Y > m) =1(1FX(m))(1FY(m))=1eλmeλm=1e2λm= 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 MM is the derivative of the CDF: fM(m)=ddmFM(m)=2λe2λm,m0f_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 UU

For the maximum U=max(X,Y)U = \max(X, Y): FU(u)=P(Uu)=P(Xu and Yu)=P(Xu)P(Yu)=(1eλu)2F_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 UU: fU(u)=dduFU(u)=2λeλu(1eλu),u0f_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)F(M, U)

To find the joint distribution F(M,U)F(M, U): F(m,u)=P(Mm,Uu)=P(min(X,Y)m,max(X,Y)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(Xu,Yu)P(Xm,Ym)=FU(u)P(Xm,Ym)= P(X \leq u, Y \leq u) - P(X \leq m, Y \leq m) = F_U(u) - P(X \leq m, Y \leq m) =(1eλu)2(1eλm)2= (1 - e^{-\lambda u})^2 - (1 - e^{-\lambda m})^2

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

Calculating this leads to: fM,U(m,u)=u(2λeλu(1eλu))=2λeλuλeλm=2λ2eλ(m+u)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. FM(m)=1e2λmF_M(m) = 1 - e^{-2\lambda m}, fM(m)=2λe2λmf_M(m) = 2\lambda e^{-2\lambda m}
  2. FU(u)=(1eλu)2F_U(u) = (1 - e^{-\lambda u})^2, fU(u)=2λeλu(1eλu)f_U(u) = 2\lambda e^{-\lambda u}(1 - e^{-\lambda u})
  3. F(M,U)F(M, U) leads to the joint density fM,U(m,u)=2λ2eλ(m+u)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 MM and UU?
  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 MM and UU?
  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.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Exponential Distributions
Joint Distribution

Formulas

F_X(x) = 1 - e^{-λx}, x ≥ 0
F_M(m) = 1 - e^{-2λm}
f_M(m) = 2λe^{-2λm}
F_U(u) = (1 - e^{-λu})^2
f_U(u) = 2λe^{-λu}(1 - e^{-λu})
F(m, u) = F_U(u) - (1 - e^{-λm})^2
f_{M,U}(m, u) = 2λ^2 e^{-λ(m + u)}

Theorems

Memoryless Property of Exponential Distributions
Properties of Independent Random Variables

Suitable Grade Level

Grades 11-12